Lecture Notes on Engineering Statics

Engineering Mechanics Statics Supported with MATLAB Codes

Dr. Ahmed Momtaz Hosny PhD in Aircraft Dynamics and Control, BUAA

Lecturer at KMA

Lecture Notes & Solved Examples with MATLAB Applications

List of Contents

Chapter 1 Introduction to Mechanics

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Statics of Particles

Statics of Rigid Bodies, Equivalent Systems of Forces

Equilibrium of Rigid Bodies

Distributed Forces, Centroids and Center of Gravity

Analysis of Structures, Simple Truss (Pin Joints & Two-Forces Members)

Chapter 7 Shear and Bending Moment of Beams

Chapter 8

Friction, Pulleys, Tension Belts and Journal Bearings

Chapter 1 Introduction to Mechanics

1.1 WHAT IS MECHANICS? Mechanics can be defined as that science which describes and predicts the conditions of rest or motion of bodies under the action of forces. It is divided into three parts: mechanics of rigid bodies, mechanics of deformable bodies, and mechanics of fluids. The mechanics of rigid bodies is subdivided into statics and dynamics, the former dealing with bodies at rest, the latter with bodies in motion. In this part of the study of mechanics, bodies are assumed to be perfectly rigid. Actual structures and machines, however, are never absolutely rigid and deform under the loads to which they are subjected. But these deformations are usually small and do not appreciably affect the conditions of equilibrium or motion of the structure under consideration. They are important, though, as far as the resistance of the structure to failure is concerned and are studied in mechanics of materials, which is a part of the mechanics of deformable bodies. The third division of mechanics, the mechanics of fluids, is subdivided into the study of incompressible fluids and of compressible fluids. An important subdivision of the study of incompressible fluids is hydraulics, which deals with problems involving water. Mechanics is a physical science, since it deals with the study of physical phenomena. However, some associate mechanics with mathematics, while many consider it as an engineering subject. Both these views are justified in part. Mechanics is the foundation of most engineering sciences and is an indispensable prerequisite to their study. However, it does not have the empiricism found in some engineering sciences, i.e., it does not rely on experience or observation alone; by its rigor and the emphasis it places on deductive reasoning it resembles mathematics. But, again, it is not an abstract or even a pure science; mechanics is an applied science. The purpose of mechanics is to explain and predict physical phenomena and thus to lay the foundations for engineering applications.

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1.2 SYSTEMS OF UNITS With the four fundamental concepts introduced in the preceding section are associated the so-called kinetic units, i.e., the units of length, time, mass, and force. International System of Units (SI Units). In this system, which will be in universal use after the United States has completed its conversion to SI units, the base units are the units of length, mass, and time, and they are called, respectively, the meter (m), the kilogram (kg), and the second (s). All three are arbitrarily defined. The unit of force is a derived unit. It is called the newton (N) and is defined as the force which gives an acceleration of 1 m/s 2 to a mass of 1 kg.

1 N = (1 kg)(1 m/s2

)

Multiples and submultiples of the fundamental SI units may be obtained through the use of the prefixes defined in Table 1.1 . The multiples and submultiples of the units of length, mass, and force most frequently used in engineering are, respectively, the kilometer (km) and the millimeter (mm); the megagram (Mg) and the gram (g); and the kilonewton (kN). According to Table 1.1 ,

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Other derived SI units used to measure the moment of a force, the work of a force, etc., are shown in Table 1.2 . While these units will be introduced in later chapters as they are needed.

U.S. Customary Units. Most practicing American engineers still commonly use a system in which the base units are the units of length, force, and time. These units are, respectively, the foot (ft), the pound (lb), and the second (s). The U.S. customary units most frequently used in mechanics are listed in Table 1.3 with their SI equivalents.

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Chapter 2 Statics of Particles

2.1 INTRODUCTION In this chapter you will study the effect of forces acting on particles. First you will learn how to replace two or more forces acting on a given particle by a single force having the same effect as the original forces. This single equivalent force is the resultant of the original forces acting on the particle. Later the relations which exist among the various forces acting on a particle in a state of equilibrium will be derived and used to determine some of the forces acting on the particle. The use of the word “particle” does not imply that our study will be limited to that of small corpuscles. What it means is that the size and shape of the bodies under consideration will not significantly affect the solution of the problems treated in this chapter and that all the forces acting on a given body will be assumed to be applied at the same point. Since such an assumption is verified in many practical applications, you will be able to solve a number of engineering problems in this chapter. The first part of the chapter is devoted to the study of forces contained in a single plane, and the second part to the analysis of forces in three-dimensional space. 2.2 FORCE ON A PARTICLE. RESULTANT OF TWO FORCES A force represents the action of one body on another and is generally characterized by its point of application, its magnitude, and its direction. Forces acting on a given particle, however, have the same point of application. Each force considered in this chapter will thus be completely defined by its magnitude and direction. The magnitude of a force is characterized by a certain number of units. As indicated in Chap. 1, the SI units used by engineers to measure the magnitude of a force are the newton (N) and its multiple the kilonewton (kN), equal to 1000 N, while the U.S. customary units used for the same purpose are the pound (lb) and its multiple the kilopound (kip), equal to 1000 lb. The direction of a force is defined by the line of action and the sense of the force. The line of action is the infinite straight line along which the force acts; it is characterized by the angle it forms with some fixed

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axis (Fig. 2.1). The force itself is represented by a segment of that line; through the use of an appropriate scale, the length of this segment may be chosen to represent the magnitude of the force.

Fig. 2.1 Finally, the sense of the force should be indicated by an arrowhead. It is important in defining a force to indicate its sense. Two forces having the same magnitude and the same line of action but different sense, such as the forces shown in Fig. 2.1a and b, will have directly opposite effects on a particle. Experimental evidence shows that two forces P and Q acting on a particle A (Fig. 2.2a) can be replaced by a single force R which has the same effect on the particle (Fig. 2.2c). This force is called the resultant of the forces P and Q and can be obtained, as shown in Fig. 2.2b, by constructing a parallelogram, using P and Q as two adjacent sides of the parallelogram. The diagonal that passes through A represents the resultant. This method for finding the resultant is known as the parallelogram law for the addition of two forces. This law is based on experimental evidence; it cannot be proved or derived mathematically.

Fig. 2.2

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2.3 VECTORS It appears from the above that forces do not obey the rules of addition defined in ordinary arithmetic or algebra. For example, two forces acting at a right angle to each other, one of 4 lb and the other of 3 lb, add up to a force of 5 lb, not to a force of 7 lb. Forces are not the only quantities which follow the parallelogram law of addition. As you will see later, displacements, velocities, accelerations, and momenta are other examples of physical quantities possessing magnitude and direction that are added according to the parallelogram law. All these quantities can be represented mathematically by vectors, while those physical quantities which have magnitude but not direction, such as volume, mass, or energy, are represented by plain numbers or scalars. Vectors are defined as mathematical expressions possessing magnitude and direction, which add according to the parallelogram law. Vectors are represented by arrows in the illustrations and will be distinguished from scalar quantities in this text through the use of boldface type (P). In longhand writing, a vector may be denoted by drawing a short arrow above the letter used to represent it (𝑷𝑷��⃗ ) or by underlining the letter (P)

Two vectors which have the same magnitude and the same direction are said to be equal, whether or not they also have the same point of application (Fig. 2.4); equal vectors may be denoted by the

. The last method may be preferred since underlining can also be used on a typewriter or computer. The magnitude of a vector defines the length of the arrow used to represent the vector. In this text, italic type will be used to denote the magnitude of a vector. Thus, the magnitude of the vector P will be denoted by P. A vector used to represent a force acting on a given particle has a well-defined point of application, namely, the particle itself. Such a vector is said to be a fixed, or bound, vector and cannot be moved without modifying the conditions of the problem. Other physical quantities, however, such as couples (see Chap. 3), are represented by vectors which may be freely moved in space; these vectors are called free vectors. Still other physical quantities, such as forces acting on a rigid body (see Chap. 3), are represented by vectors which can be moved, or slid, along their lines of action; they are known as sliding vectors.

same letter.

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The negative vector of a given vector P is defined as a vector having the same magnitude as P and a direction opposite to that of P (Fig. 2.5); the negative of the vector P is denoted by -P. The vectors P and -P are commonly referred to as equal and opposite vectors. Clearly, we have

P + (-P) = 0

2.4 ADDITION OF VECTORS We saw in the preceding section that, by definition, vectors add according to the parallelogram law. Thus, the sum of two vectors P and Q is obtained by attaching the two vectors to the same point A and constructing a parallelogram, using P and Q as two sides of the parallelogram (Fig. 2.6). The diagonal that passes through A represents the sum of the vectors P and Q, consequently we can write

R = P + Q = Q + P

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PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

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PROBLEM 2.1

Two forces are applied at point B of beam AB. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

(a) Parallelogram law:

(b) Triangle rule:

We measure: 3.30 kN, 66.6R α= = ° 3.30 kN=R 66.6°

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2.1

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PROBLEM 2.2

The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

We measure: 51.3

59.0

αβ

= °= °


(b) Triangle rule:

We measure: 139.1 lb,R = 67.0γ = ° 139.1 lbR = 67.0°

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PROBLEM 2.3

Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 10 kN and Q = 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

SOLUTION


(b) Triangle rule:

We measure: 20.1 kN,R = 21.2α = ° 20.1 kN=R 21.2°

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PROBLEM 2.4

Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

SOLUTION


(b) Triangle rule:

We measure: 8.03 kips, 3.8R α= = ° 8.03 kips=R 3.8°

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PROBLEM 2.5

A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant.

SOLUTION

Using the triangle rule and the law of sines:

(a) 120 N

sin 30 sin 25

P=° °

101.4 NP =

(b) 30 25 180

180 25 30

125

ββ

° + + ° = °= ° − ° − °= °

120 N

sin 30 sin125=

° °R

196.6 N=R

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PROBLEM 2.6

A trolley that moves along a horizontal beam is acted upon by two forces as shown. (a) Knowing that α = 25°, determine by trigonometry the magnitude of the force P so that the resultant force exerted on the trolley is vertical. (b) What is the corresponding magnitude of the resultant?

SOLUTION


(a) 1600 N

sin 25° sin 75

P=°

3660 NP =

(b) 25 75 180

180 25 75

80

ββ

° + + ° = °= ° − ° − °= °

1600 N

sin 25° sin80

R=°

3730 NR =

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PROBLEM 2.11

A steel tank is to be positioned in an excavation. Knowing that α = 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION


(a) 50 60 180

180 50 60

70

ββ

+ ° + ° = °= ° − ° − °= °

425 lb

sin 70 sin 60

P=° °

392 lbP =

(b) 425 lb

sin 70 sin 50

R=° °

346 lbR =

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PROBLEM 2.12

A steel tank is to be positioned in an excavation. Knowing that the magnitude of P is 500 lb, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION


(a) ( 30 ) 60 180

180 ( 30 ) 60

90

sin (90 ) sin 60

425 lb 500 lb

α ββ αβ α

α

+ ° + ° + = °= ° − + ° − °= ° −

° − °=

90 47.402α° − = ° 42.6α = °

(b) 500 lb

sin (42.598 30 ) sin 60

R =° + ° °

551 lbR =

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%%%%%%%%%Problem 2.12 How to calculate angle Alfa for different values of P %%%%%%%%%Considering Vertical Resultant R and Constant Force of 425 with %%%%%%%%%angle 30 degree %% Calculating the regular problem %% Beta=180 - (Alfa+30)- 60 ie Beta= 90- Alfa ; %% Substituting in Sines Law sin(Beta)/425 = sin(pi/3)/500; B =(sin(pi/3)/500)*425; Beta = asin(B) Beta = Beta * 57.3 %%%% in Radians Alfa=90 - Beta %%%%%%%%% Calculating the resultant Force R=(500/sin(pi/3))* sin((Alfa+30)/57.3) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% Performing a loop calculation for P=[500 550 600 650 700 750 800 850 900] P=[500 550 600 650 700 750 800 850 900] for I=0:1:8 I=I+1 B =(sin(pi/3)/P(I))*425; Beta = asin(B) Beta = Beta * 57.3 %%%% in Degrees Beta1(I)=Beta Alfa(I)=90 - Beta R(I)=(P(I)/sin(pi/3))* sin((Alfa(I)+30)/57.3) end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% Plotting Alfa and Beta for Different P

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figure(1); plot(P,Alfa); xlabel('P in LB'),grid on;;ylabel('Alfa in Degrees'),grid on;; figure(2); plot(P,Beta1); xlabel('P in LB'),grid on;;ylabel('Beta in Degrees'),grid on;; figure(3); plot(P,R); xlabel('P in LB'),grid on;;ylabel('R Resultant'),grid on;; figure(4); plot(P,Alfa,'+',P,Beta1,'o'); xlabel('P in LB'),grid on;;ylabel('Alfa Beta in Degrees'),grid on;;

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PROBLEM 2.15

Solve Problem 2.2 by trigonometry.

PROBLEM 2.2 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

8tan

1038.66

6tan

1030.96

α

α

β

β

=

= °

=

= °

Using the triangle rule: 180

38.66 30.96 180

110.38

α β ψψψ

+ + = °° + ° + = °

= °

Using the law of cosines: 2 22 (120 lb) (40 lb) 2(120 lb)(40 lb)cos110.38

139.08 lb

R

R

= + − °=

Using the law of sines: sin sin110.38

40 lb 139.08 lb

γ °=

15.64

(90 )

(90 38.66 ) 15.64

66.98

γφ α γφφ

= °= ° − += ° − ° + °= ° 139.1 lb=R 67.0°

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%%%%%%%%%Problem 2.15 How to calculate Resultant R and its angle from the %%%%%%%%%horizontal plane %% Calculating the regular problem %% tan(Alfa)=8/10 & tan(Beta)=6/10 then we can get Alfa & Beta; Alfa=atan(8/10)*57.3 Beta=atan(6/10)*57.3 Epsai = 180 - Alfa - Beta %%%%% Calculating the resultant from the Law of Cosines R = ((120^2) + (40^2) - 2*40*120*cos(Epsai/57.3))^0.5 %%%%% Calculating the angle Gama to obtain angle Fai as shown in the figure %%%%% sin(Gama)/40=sin(Epsai)/R consequently we can obtain sin(Gama) in %%%%% form of G G=(sin(Epsai/57.3)/R)*40 Gama=asin(G)*57.3 %%%%%%% in Degrees Fai=90-Alfa + Gama %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% Performing a loop calculation for P=[500 550 600 650 700 750 800 850 900] T1=[100 120 140 160 180 200 220 240 260]; T2=[30 40 50 60 70 80 90 100 110]; for I=0:1:8 I=I+1; %%%%% Calculating the resultant from the Law of Cosines R(I) = ((T1(I)^2) + (T2(I)^2) - 2*T1(I)*T2(I)*cos(Epsai/57.3))^0.5; %%%%% Calculating the angle Gama to obtain angle Fai as shown in the figure %%%%% sin(Gama)/40=sin(Epsai)/R consequently we can obtain sin(Gama) in %%%%% form of G G=(sin(Epsai/57.3)/R(I))*40; Gama=asin(G)*57.3; %%%%%%% in Degrees Fai(I)=90-Alfa + Gama;

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end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% Plotting Resultant R and the Corresponding Angle Fai for different T1 & T2 figure(1); plot3(T1,T2,R); xlabel('T1 in LB'), grid on;ylabel('T2 in LB'), grid on;zlabel('R in LB'), grid on; figure(2); plot(R,Fai); xlabel('R in LB'), grid on;ylabel('Fai in Degrees'), grid on;

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PROBLEM 2.16

Solve Problem 2.4 by trigonometry.

PROBLEM 2.4 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

Using the force triangle and the laws of cosines and sines:

We have: 180 (50 25 )

105

γ = ° − ° + °= °

Then 2 2 2

2

(4 kips) (6 kips) 2(4 kips)(6 kips)cos105

64.423 kips

8.0264 kips

R

R

= + − °

==

And 4 kips 8.0264 kips

sin(25 ) sin105

sin(25 ) 0.48137

25 28.775

3.775

αααα

=° + °° + =° + = °

= °

8.03 kips=R 3.8°

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PROBLEM 2.29

The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 720-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC.

SOLUTION

(a) 37

1237

(720 N)122220 N

=

=

=

xP P

2.22 kNP =

(b) 35

1235

(720 N)122100 N

y xP P=

=

=

2.10 kN=yP

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%%%%%%%%%Problem 2.29 How to calculate Resultant P by knowing one component %%%%%%%%%Px = 720 N as shown in figure Px = 720; %% Calculating the regular problem %% Taking into consideration that Px = P*sin(theta)where theta = the angle %% between Py and P -----> sin(theta)= 2.4/(2.4^2 + 7^2)^0.5 s= 2.4/(2.4^2 + 7^2)^0.5; P = Px/s %% Taking into consideration that Py = P*cos(theta)where theta = the angle %% between Py and P -----> cos(theta) = 7/(2.4^2 + 7^2)^0.5 Py = P*(7/(2.4^2 + 7^2)^0.5) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% Performing a loop calculation with Length L (from point D to point C %% varies from L=[1 2.4 4 6 8 10 12 14 16 18]) therefore calculating the %% corresponding P & Py L=[1 2.4 4 6 8 10 12 14 16 20 30 40]; for I=0:1:11 I=I+1; %% Taking into consideration that Px = P*sin(theta)where theta = the angle %% between Py and P -----> sin(theta)= L(I)/(L(I)^2 + 7^2)^0.5 s= L(I)/(L(I)^2 + 7^2)^0.5; P(I) = Px/s; %% Taking into consideration that Py = P*cos(theta)where theta = the angle %% between Py and P -----> cos(theta) = 7/(L(I)^2 + 7^2)^0.5 Py(I) = P(I)*(7/(L(I)^2 + 7^2)^0.5); end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% Plotting Resultant P & Px & Py with respect to length L

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Px = [720 720 720 720 720 720 720 720 720 720 720 720]; figure(1); plot(L,P,'o',L,Px,'*',L,Py,'+'); xlabel('L in m '), grid on;ylabel('P & Px & Py in N'), grid on;

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PROBLEM 2.31

Determine the resultant of the three forces of Problem 2.23.

PROBLEM 2.23 Determine the x and y components of each of the forces shown.

SOLUTION

Components of the forces were determined in Problem 2.23:

Force x Comp. (N) y Comp. (N)

800 lb +640 +480

424 lb –224 –360

408 lb +192 –360

608xR = + 240yR = −

(608 lb) ( 240 lb)

tan

240

60821.541

240 N

sin(21.541°)

653.65 N

x y

y

x

R R

R

R

R

α

α

= +

= + −

=

=

= °

=

=

R i j

i j

654 N=R 21.5°

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%%%%%%%%%Problem 2.31 How to calculate Resultant R by knowing all the %%%%%%%%%components of the applied forces as shown in figure %%%Calculating the regular problem as it is stated in the book %%Calculating the X & Y Components for each applied force as following:- %%%First force T1 = 800 N & x = 800 mm & y = 600 mm T1 = 800 T1x = 800*(800/(800^2 + 600^2)^0.5) T1y = 800*(600/(800^2 + 600^2)^0.5) %%%Second force T2 = 408 N & x = 480 mm & y = -900 mm T2 = 408 T2x = 408*(480/(480^2 + 900^2)^0.5) T2y = 408*(-900/(480^2 + 900^2)^0.5) %%%Third force T3 = 424 N & x = -560 mm & y = -900 mm T3 = 424 T3x = 424*(-560/(560^2 + 900^2)^0.5) T3y = 424*(-900/(560^2 + 900^2)^0.5) %%%%Calculating the Resultant R as a summation of X components and Y %%%%components as following:- Rx = T1x + T2x + T3x Ry = T1y + T2y + T3y R=(Rx^2 + Ry^2)^0.5 %%%%Calculating the angle of the resultant R as following:- theta = atan(Ry/Rx)*57.3 %%%% in Degrees %%In case of several acting forces or even enormous number of acting forces %%A Successive Loop should be constructed as following:- %%T = [T1 T2 T3 T4 T5 .........] %%x = [x1 x2 x3 x4 x5 .........] %%y = [y1 y2 y3 y4 y5 .........] T = [800 408 424] x = [800 480 -560] y = [600 -900 -900] Rx=0; Ry=0; for H=0:1:2 H=H+1 Tx(H)= T(H)*(x(H)/(x(H)^2 + y(H)^2)^0.5) Ty(H)= T(H)*(y(H)/(x(H)^2 + y(H)^2)^0.5) Rx = Rx + Tx(H) Ry = Ry + Ty(H) end R=(Rx^2 + Ry^2)^0.5

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%%%%Calculating the angle of the resultant R as following:- theta = atan(Ry/Rx)*57.3 %%%% in Degrees

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PROBLEM 2.32



SOLUTION


Force x Comp. (N) y Comp. (N)

80 N +61.3 +51.4

120 N +41.0 +112.8

150 N –122.9 +86.0

20.6xR = − 250.2yR = +

( 20.6 N) (250.2 N)

tan

250.2 Ntan

20.6 Ntan 12.1456

85.293

250.2 N

sin85.293

x y

y

x

R R

R

R

R

α

α

αα

= +

= − +

=

=

== °

=°

R i j

i j

251 N=R 85.3°

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PROBLEM 2.34



SOLUTION


Force x Comp. (lb) y Comp. (lb)

102 lb −48.0 +90.0

106 lb +56.0 +90.0

200 lb −160.0 −120.0

152.0xR = − 60.0yR =

( 152 lb) (60.0 lb)

tan

60.0 lbtan

152.0 lbtan 0.39474

21.541

α

α

αα

= +

= − +

=

=

== °

x y

y

x

R R

R

R

R i j

i j

60.0 lb

sin 21.541R =

° 163.4 lb=R 21.5°

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PROBLEM 2.36

Knowing that the tension in rope AC is 365 N, determine the resultant of the three forces exerted at point C of post BC.

SOLUTION

Determine force components:

Cable force AC: 960

(365 N) 240 N14601100

(365 N) 275 N1460

= − = −

= − = −

x

y

F

F

500-N Force: 24

(500 N) 480 N257

(500 N) 140 N25

x

y

F

F

= =

= =

200-N Force: 4

(200 N) 160 N5

3(200 N) 120 N

5

x

y

F

F

= =

= − = −

and

2 2

2 2

240 N 480 N 160 N 400 N

275 N 140 N 120 N 255 N

(400 N) ( 255 N)

474.37 N

= Σ = − + + == Σ = − + − = −

= +

= + −=

x x

y y

x y

R F

R F

R R R

Further: 255

tan40032.5

α

α

=

= °

474 N=R 32.5°

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%%%%%%%%%Problem 2.36 How to calculate Resultant P by knowing all the %%%%%%%%%components of the applied forces as shown in figure %%%Calculating the regular problem as it is stated in the book %%Calculating the X & Y Components for each applied force as following:- %%%First force T1 = 500 N & x = 24 m & y = 7 m T1 = 500 T1x = 500*(24/(24^2 + 7^2)^0.5) T1y = 500*(7/(24^2 + 7^2)^0.5) %%%Second force T2 = 200 N & x = 4 m & y = -3 m T2 = 200 T2x = 200*(4/(4^2 + 3^2)^0.5) T2y = 200*(-3/(4^2 + 3^2)^0.5) %%%Third force T3 = 365 N & x = -960 mm & y = -1100 mm T3 = 365 T3x = 365*(-960/(960^2 + 1100^2)^0.5) T3y = 365*(-1100/(960^2 + 1100^2)^0.5) %%%%Calculating the Resultant R as a summation of X components and Y %%%%components as following:- Rx = T1x + T2x + T3x Ry = T1y + T2y + T3y R=(Rx^2 + Ry^2)^0.5 %%%%Calculating the angle of the resultant R as following:- theta = atan(Ry/Rx)*57.3 %%%% in Degrees %%In case of several acting forces or even enormous number of acting forces %%A Successive Loop should be constructed as following:- %%T = [T1 T2 T3 T4 T5 .........] %%x = [x1 x2 x3 x4 x5 .........] %%y = [y1 y2 y3 y4 y5 .........] for I=1:1:50 %%P is the highest force of 500 N, a range of P will be proposed %%from 0 : 500 N in the following loop to discuss the effect on the %%resultant P(I)=I*10; T = [P(I) 200 365]; x = [24 4 -960]; y = [7 -3 -1100]; Rx=0; Ry=0; for H=0:1:2 H=H+1;

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Tx(H)= T(H)*(x(H)/(x(H)^2 + y(H)^2)^0.5); Ty(H)= T(H)*(y(H)/(x(H)^2 + y(H)^2)^0.5); Rx = Rx + Tx(H); Ry = Ry + Ty(H); end R(I)=(Rx^2 + Ry^2)^0.5; %%%%Calculating the angle of the resultant R as following:- theta(I) = atan(Ry/Rx)*57.3; %%%% in Degrees end figure(1); plot(P,theta,'o'); xlabel('Required P (The Highest Force)'), grid on;ylabel('theta in Degrees'), grid on; figure(2); plot(P,R,'+'); xlabel('Required P (The Highest Force)'), grid on;ylabel('Resultant in N'), grid on;

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39

PROBLEM 2.37

Knowing that α = 40°, determine the resultant of the three forces shown.

SOLUTION

60-lb Force: (60 lb)cos 20 56.382 lb

(60 lb)sin 20 20.521 lbx

y

F

F

= ° == ° =

80-lb Force: (80 lb)cos60 40.000 lb

(80 lb)sin 60 69.282 lbx

y

F

F

= ° == ° =

120-lb Force: (120 lb)cos30 103.923 lb

(120 lb)sin 30 60.000 lbx

y

F

F

= ° == − ° = −

and

2 2

200.305 lb

29.803 lb

(200.305 lb) (29.803 lb)

202.510 lb

x x

y y

R F

R F

R

= Σ == Σ =

= +=

Further: 29.803

tan200.305

α =

1 29.803tan

200.3058.46

α −=

= ° 203 lb=R 8.46°

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40

PROBLEM 2.38

Knowing that α = 75°, determine the resultant of the three forces shown.

SOLUTION

60-lb Force: (60 lb) cos 20 56.382 lb

(60 lb)sin 20 20.521 lbx

y

F

F

= ° == ° =

80-lb Force: (80 lb) cos 95 6.9725 lb

(80 lb)sin 95 79.696 lbx

y

F

F

= ° = −= ° =

120-lb Force: (120 lb) cos 5 119.543 lb

(120 lb)sin 5 10.459 lbx

y

F

F

= ° == ° =

Then 168.953 lb

110.676 lbx x

y y

R F

R F

= Σ == Σ =

and 2 2(168.953 lb) (110.676 lb)

201.976 lb

R = +=

110.676tan

168.953tan 0.65507

33.228

α

αα

=

== ° 202 lb=R 33.2°

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42

PROBLEM 2.40

For the post of Prob. 2.36, determine (a) the required tension in rope AC if the resultant of the three forces exerted at point C is to be horizontal, (b) the corresponding magnitude of the resultant.

SOLUTION

960 24 4(500 N) (200 N)

1460 25 548

640 N73

x x AC

x AC

R F T

R T

= Σ = − + +

= − + (1)

1100 7 3(500 N) (200 N)

1460 25 555

20 N73

y y AC

y AC

R F T

R T

= Σ = − + −

= − + (2)

(a) For R to be horizontal, we must have 0.yR =

Set 0yR = in Eq. (2): 55

20 N 073 ACT− + =

26.545 NACT = 26.5 NACT =

(b) Substituting for ACT into Eq. (1) gives

48(26.545 N) 640 N

73622.55 N

623 N

= − +

== =

x

x

x

R

R

R R 623 NR =

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45

PROBLEM 2.43

Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION

Free-Body Diagram

1100tan

96048.888

400tan

96022.620

α

α

β

β

=

= °

=

= °

Force Triangle

Law of sines:

15.696 kN

sin 22.620 sin 48.888 sin108.492AC BCT T

= =° ° °

(a) 15.696 kN

(sin 22.620 )sin108.492ACT = °

° 6.37 kNACT =

(b) 15.696 kN

(sin 48.888 )sin108.492BCT = °

° 12.47 kNBCT =

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46

PROBLEM 2.44

Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION

3

tan2.2553.130

1.4tan

2.2531.891

α

α

β

β

=

= °

=

= °

Free-Body Diagram

Law of sines: Force-Triangle

660 N

sin 31.891 sin 53.130 sin 94.979AC BCT T

= =° ° °

(a) 660 N

(sin 31.891 )sin 94.979ACT = °

° 350 NACT =

(b) 660 N

(sin 53.130 )sin 94.979BCT = °

° 530 NBCT =

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47

PROBLEM 2.45

Knowing that 20 ,α = ° determine the tension (a) in cable AC, (b) in rope BC.

SOLUTION

Free-Body Diagram Force Triangle

Law of sines: 1200 lb

sin 110 sin 5 sin 65AC BCT T

= =° ° °

(a) 1200 lb

sin 110sin 65ACT = °

° 1244 lbACT =

(b) 1200 lb

sin 5sin 65BCT = °

° 115.4 lbBCT =

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48

PROBLEM 2.46

Knowing that 55α = ° and that boom AC exerts on pin C a force directed along line AC, determine (a) the magnitude of that force, (b) the tension in cable BC.

SOLUTION


Law of sines: 300 lb

sin 35 sin 50 sin 95AC BCF T

= =° ° °

(a) 300 lb

sin 35sin 95ACF = °

° 172.7 lbACF =

(b) 300 lb

sin 50sin 95BCT = °

° 231 lbBCT =

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54

PROBLEM 2.52

Two cables are tied together at C and loaded as shown. Determine the range of values of P for which both cables remain taut.

SOLUTION

Free Body: C

12 4

0: 013 5x ACTΣ = − + =F P

13

15ACT P= (1)

5 3

0: 480 N 013 5y AC BCT T PΣ = + + − =F

Substitute for ACT from (1): 5 13 3

480 N 013 15 5BCP T P + + − =

14

480 N15BCT P= − (2)

From (1), 0ACT � requires 0.P �

From (2), 0BCT � requires 14

480 N, 514.29 N15

P P� �

Allowable range: 0 514 NP� �

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55

PROBLEM 2.53

A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that 30α = ° and 10β = ° and that the combined weight of the boatswain’s chair and the sailor is 900 N, determine the tension (a) in the support cable ACB, (b) in the traction cable CD.

SOLUTION

Free-Body Diagram

0: cos 10 cos 30 cos 30 0x ACB ACB CDF T T TΣ = ° − ° − ° =

0.137158CD ACBT T= (1)

0: sin 10 sin 30 sin 30 900 0y ACB ACB CDF T T TΣ = ° + ° + ° − =

0.67365 0.5 900ACB CDT T+ = (2)

(a) Substitute (1) into (2): 0.67365 0.5(0.137158 ) 900ACB ACBT T+ =

1212.56 NACBT = 1213 NACBT =

(b) From (1): 0.137158(1212.56 N)CDT = 166.3 NCDT =

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56

PROBLEM 2.54

A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that 25α = ° and 15β = ° and that the tension in cable CD is 80 N, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) in tension in the support cable ACB.

SOLUTION

Free-Body Diagram

0: cos 15 cos 25 (80 N)cos 25 0x ACB ACBF T TΣ = ° − ° − ° =

1216.15 NACBT =

0: (1216.15 N)sin 15 (1216.15 N)sin 25yFΣ = ° + °

(80 N)sin 25 0

862.54 N

W

W

+ ° − ==

(a) 863 NW =

(b) 1216 NACBT =

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57

PROBLEM 2.55

Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that 500P = lb and 650Q = lb, determine the magnitudes of the forces exerted on the rods A and B.

SOLUTION

Free-Body Diagram

Resolving the forces into x- and y-directions:

0A B= + + + =R P Q F F

Substituting components: (500 lb) [(650 lb)cos50 ]

[(650 lb)sin 50 ]

( cos50 ) ( sin 50 ) 0B A AF F F

= − + °− °+ − ° + ° =

R j i

j

i i j

In the y-direction (one unknown force):

500 lb (650 lb)sin 50 sin 50 0AF− − ° + ° =

Thus, 500 lb (650 lb)sin 50

sin 50AF+ °=

°

1302.70 lb= 1303 lbAF =

In the x-direction: (650 lb)cos50 cos50 0B AF F° + − ° =

Thus, cos50 (650 lb)cos50

(1302.70 lb)cos50 (650 lb)cos50B AF F= ° − °

= ° − °

419.55 lb= 420 lbBF =

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58

PROBLEM 2.56

Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the magnitudes of the forces exerted on rods A and B are 750AF = lb and 400BF = lb, determine the magnitudes of P and Q.

SOLUTION

Free-Body Diagram

Resolving the forces into x- and y-directions:

0A B= + + + =R P Q F F

Substituting components: cos 50 sin 50

[(750 lb)cos 50 ]

[(750 lb)sin 50 ] (400 lb)

P Q Q= − + ° − °− °+ ° +

R j i j

i

j i

In the x-direction (one unknown force):

cos 50 [(750 lb)cos 50 ] 400 lb 0Q ° − ° + =

(750 lb)cos 50 400 lb

cos 50

127.710 lb

Q° −=°

=

In the y-direction: sin 50 (750 lb)sin 50 0P Q− − ° + ° =

sin 50 (750 lb)sin 50

(127.710 lb)sin 50 (750 lb)sin 50

476.70 lb

P Q= − ° + °= − ° + °= 477 lb; 127.7 lbP Q= =

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61

PROBLEM 2.59

For the situation described in Figure P2.45, determine (a) the value of α for which the tension in rope BC is as small as possible, (b) the corresponding value of the tension.

PROBLEM 2.45 Knowing that 20 ,α = ° determine the tension (a) in cable AC, (b) in rope BC.

SOLUTION


To be smallest, BCT must be perpendicular to the direction of .ACT

(a) Thus, 5α = ° 5.00α = °

(b) (1200 lb)sin 5BCT = ° 104.6 lbBCT =

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%%%%%%%%%Problem 2.59 How to calculate value of alfa for which the tension %%%%%%%%%in rope BC is as small as possible. %%Measuring alfa from vertical line upward %%Considering the Law of Sines consequently (Tac/sin(90+90-alfa)) = %%(Tbc/sin(90+90-5)) = (1200/sin(180 -5-alfa). for I=1:1:180 alfa(I)=I Tbc(I) = (1200/sin((5+alfa(I))/57.3))*sin((180-5)/57.3); Tac(I) = (1200/sin((5+alfa(I))/57.3))*sin((180-alfa(I))/57.3); end figure(1); plot(alfa,Tbc,'o'); xlabel('alfa in Degrees '),grid on;ylabel('Tbc in LB'),grid on; xlim([0 180]) ylim([0 2000]) figure(2); plot(alfa,Tac,'+'); xlabel('alfa in Degrees '),grid on;ylabel('Tac in LB'),grid on; xlim([0 180]) ylim([0 2000]) figure(3); plot(alfa,Tac,'+',alfa,Tbc,'o'); xlabel('alfa in Degrees '),grid on;ylabel('Tac in LB & Tbc in LB'),grid on; xlim([0 180]) ylim([0 2000])

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62

PROBLEM 2.60

For the structure and loading of Problem 2.46, determine (a) the value of α for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension.

SOLUTION

BCT must be perpendicular to ACF to be as small as possible.

Free-Body Diagram: C Force Triangle is a right triangle

To be a minimum, BCT must be perpendicular to .ACF

(a) We observe: 90 30α = ° − ° 60.0α = °

(b) (300 lb)sin 50BCT = °

or 229.81 lbBCT = 230 lbBCT =

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63

PROBLEM 2.61

For the cables of Problem 2.48, it is known that the maximum allowable tension is 600 N in cable AC and 750 N in cable BC. Determine (a) the maximum force P that can be applied at C, (b) the corresponding value of α.

SOLUTION


(a) Law of cosines 2 2 2(600) (750) 2(600)(750)cos (25 45 )P = + − ° + °

784.02 NP = 784 NP =

(b) Law of sines sin sin (25 45 )

600 N 784.02 N

β ° + °=

46.0β = ° 46.0 25α∴ = ° + ° 71.0α = °

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64

PROBLEM 2.62

A movable bin and its contents have a combined weight of 2.8 kN. Determine the shortest chain sling ACB that can be used to lift the loaded bin if the tension in the chain is not to exceed 5 kN.

SOLUTION

Free-Body Diagram

tan0.6 m

α = h (1)

Isosceles Force Triangle

Law of sines: 12

12

(2.8 kN)sin

5 kN

(2.8 kN)sin

5 kN

16.2602

AC

AC

T

T

α

α

α

=

=

=

= °

From Eq. (1): tan16.2602 0.175000 m0.6 m

hh° = ∴ =

Half length of chain 2 2(0.6 m) (0.175 m)

0.625 m

AC= = +=

Total length: 2 0.625 m= × 1.250 m

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65

PROBLEM 2.63

Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) 4.5 in.,x = (b) 15 in.x =

SOLUTION

(a) Free Body: Collar A Force Triangle

50 lb

4.5 20.5

P = 10.98 lbP =

(b) Free Body: Collar A Force Triangle

50 lb

15 25

P = 30.0 lbP =

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66

PROBLEM 2.64

Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P = 48 lb.

SOLUTION

Free Body: Collar A Force Triangle

2 2 2(50) (48) 196

14.00 lb

N

N

= − ==

Similar Triangles

48 lb

20 in. 14 lb

x =

68.6 in.x =

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%%%%%%%%%Problem 2.64 How to calculate value of alfa while the tension in %%%%%%%%%rope AB is varing from T = 0 LB to 250 LB %%%%%%%%% %%Measuring alfa from horizontal line to the right at Point A %%Considering the Law of Sines consequently (T/sin(90)) = (48/sin(90-alfa)) %%= (N/sin(alfa) consequently Beta = sin(90-alfa)=48/T for I = 1:1:50 T(I)=I*5 Beta(I) = 48/T(I); alfa(I) = -asin(Beta(I))*57.3 + 90; X(I) = 20/tan(alfa(I)/57.3) N(I) = (T(I))*sin(alfa(I)/57.3) end figure(1); plot(T,X,'o'); xlabel('T in LB '),grid on;ylabel('X in in'),grid on; xlim([0 250]) ylim([0 100]) figure(2); plot(T,alfa,'*'); xlabel('T in LB '),grid on;ylabel('alfa in Degrees'),grid on; xlim([0 250]) ylim([0 100]) figure(3); plot(T,N,'+'); xlabel('T in LB '),grid on;ylabel('N in LB'),grid on; xlim([0 250]) ylim([0 200])

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67

PROBLEM 2.65

Three forces are applied to a bracket as shown. The directions of the two 150-N forces may vary, but the angle between these forces is always 50°. Determine the range of values of α for which the magnitude of the resultant of the forces acting at A is less than 600 N.

SOLUTION

Combine the two 150-N forces into a resultant force Q:

2(150 N)cos 25

271.89 N

Q = °=

Equivalent loading at A:

Using the law of cosines:

2 2 2(600 N) (500 N) (271.89 N) 2(500 N)(271.89 N)cos(55 )

cos(55 ) 0.132685

αα

= + + ° +° + =

Two values for :α 55 82.375

27.4

αα

° + == °

or 55 82.375

55 360 82.375

222.6

ααα

° + = − °° + = ° − °

= °

For 600 lb:R < 27.4 222.6α° < <

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68

PROBLEM 2.66

A 200-kg crate is to be supported by the rope-and-pulley arrangement shown. Determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Ch. 4.)

SOLUTION

Free-Body Diagram: Pulley A

50: 2 cos 0

281

cos 0.59655

53.377

xF P P α

αα

Σ = − + =

== ± °

For 53.377 :α = + °

160: 2 sin 53.377 1962 N 0

281yF P P

Σ = + ° − =

724 N=P 53.4°

For 53.377 :α = − °

160: 2 sin( 53.377 ) 1962 N 0

281yF P P

Σ = + − ° − =

1773=P 53.4°

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73

PROBLEM 2.71

Determine (a) the x, y, and z components of the 900-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION

cos 65

(900 N)cos 65

380.36 N

h

h

F F

F

= °= °=

(a) sin 20

(380.36 N)sin 20°x hF F= °

=

130.091 N,= −xF 130.1 NxF = −

sin 65

(900 N)sin 65°

815.68 N,

y

y

F F

F

= °

== + 816 NyF = +

cos 20

(380.36 N)cos 20

357.42 N

= °= °= +

z h

z

F F

F 357 NzF = +

(b) 130.091 N

cos900 N

xx

F

Fθ −= = 98.3xθ = °

815.68 N

cos900 N

yy

F

Fθ += = 25.0yθ = °

357.42 N

cos900 N

zz

F

Fθ += = 66.6zθ = °

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95

PROBLEM 2.93

Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLUTION

2 2 2

2 2 2

(40 in.) (45 in.) (60 in.)

(40 in.) (45 in.) (60 in.) 85 in.

(100 in.) (45 in.) (60 in.)

(100 in.) (45 in.) (60 in.) 125 in.

(40 in.) (45 in.) (60 in.)(425 lb)

85 in.AB AB AB AB

AB

AB

AC

AC

ABT T

AB

= − +

= + + =

= − +

= + + =

− += = =

i j k

i j k

i j kT λ

(200 lb) (225 lb) (300 lb)

(100 in.) (45 in.) (60 in.)(510 lb)

125 in.

(408 lb) (183.6 lb) (244.8 lb)

(608) (408.6 lb) (544.8 lb)

AB

AC AC AC AC

AC

AB AC

ACT T

AC

= − +

− += = =

= − += + = − +

T i j k

i j kT λ

T i j k

R T T i j k

Then: 912.92 lbR = 913 lbR =

and 608 lb

cos 0.66599912.92 lbxθ = = 48.2xθ = °

408.6 lb

cos 0.44757912.92 lbyθ = = − 116.6yθ = °

544.8 lb

cos 0.59677912.92 lbzθ = = 53.4zθ = °

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%%%%%%%%%Problem 2.93 How to calculate the resultant at point A from %%different acting forces using the unit vector & vector summation concept, %%considering the tension in cable AB = 425 LB and in cable AC = 510 LB %% Calculating each force in vector form by calculating the unit vector for %% each AB=[40 -45 60] %%% 40i -45j + 60k AC=[100 -45 60] %%% 100i - 45j + 60k lamdaAB=AB./(AB(1)^2 + AB(2)^2 + AB(3)^2)^0.5 Tab=lamdaAB.*425 lamdaAC=AC./(AC(1)^2 + AC(2)^2 + AC(3)^2)^0.5 Tac=lamdaAC.*510 R = Tab + Tac Rm=(R(1)^2 + R(2)^2 + R(3)^2)^0.5 %%Calculating theta angles with respect to X, Y, Z thetaX = acos(R(1)/Rm)*57.3 thetaY = acos(R(2)/Rm)*57.3 thetaZ = acos(R(3)/Rm)*57.3

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PROBLEM 2.107

Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that 0,Q = find the value of P for which the tension in cable AD is 305 N.

SOLUTION

0: 0A AB AC ADΣ = + + + =F T T T P where P=P i

(960 mm) (240 mm) (380 mm) 1060 mm

(960 mm) (240 mm) (320 mm) 1040 mm

(960 mm) (720 mm) (220 mm) 1220 mm

AB AB

AC AC

AD AD

= − − + =

= − − − =

= − + − =

i j k

i j k

i j k

48 12 19

53 53 53

12 3 4

13 13 13

305 N[( 960 mm) (720 mm) (220 mm) ]

1220 mm(240 N) (180 N) (55 N)

AB AB AB AB AB

AC AC AC AC AC

AD AD AD

ABT T T

AB

ACT T T

AC

T

= = = − − +

= = = − − −

= = − + −

= − + −

T λ i j k

T λ i j k

T λ i j k

i j k

Substituting into 0,AΣ =F factoring , , ,i j k and setting each coefficient equal to φ gives:

48 12

: 240 N53 13AB ACP T T= + +i (1)

:j 12 3

180 N53 13AB ACT T+ = (2)

:k 19 4

55 N53 13AB ACT T− = (3)

Solving the system of linear equations using conventional algorithms gives:

446.71 N

341.71 NAB

AC

T

T

== 960 NP =

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113

PROBLEM 2.108

Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that 1200 N,P = determine the values of Q for which cable AD is taut.

SOLUTION

We assume that 0ADT = and write 0: (1200 N) 0A AB AC QΣ = + + + =F T T j i

(960 mm) (240 mm) (380 mm) 1060 mm

(960 mm) (240 mm) (320 mm) 1040 mm

AB AB

AC AC

= − − + =

= − − − =

i j k

i j k

48 12 19

53 53 53

12 3 4

13 13 13

AB AB AB AB AB

AC AC AC AC AC

ABT T T

AB

ACT T T

AC

= = = − − +

= = = − − −

T λ i j k

T λ i j k

Substituting into 0,AΣ =F factoring , , ,i j k and setting each coefficient equal to φ gives:

48 12

: 1200 N 053 13AB ACT T− − + =i (1)

12 3

: 053 13AB ACT T Q− − + =j (2)

19 4

: 053 13AB ACT T− =k (3)

Solving the resulting system of linear equations using conventional algorithms gives:

605.71 N

705.71 N

300.00 N

AB

AC

T

T

Q

=== 0 300 NQ� �

Note: This solution assumes that Q is directed upward as shown ( 0),Q � if negative values of Q are considered, cable AD remains taut, but AC becomes slack for 460 N.Q = −

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114

PROBLEM 2.109

A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 60 N, determine the weight of the plate.

SOLUTION

We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A.

Free Body A:

0: 0AB AC ADF PΣ = + + + =T T T j

We have:

( )

(320 mm) (480 mm) (360 mm) 680 mm

(450 mm) (480 mm) (360 mm) 750 mm

(250 mm) (480 mm) 360 mm 650 mm

AB AB

AC AC

AD AD

= − − + =

= − + =

= − − =

i j k

i j k

i j k

Thus:

( )

8 12 9

17 17 17

0.6 0.64 0.48

5 9.6 7.2

13 13 13

AB AB AB AB AB

AC AC AC AC AC

AD AD AD AD AD

ABT T T

AB

ACT T T

AC

ADT T T

AD

= = = − − +

= = = − +

= = = − −

T λ i j k

T λ i j k

T λ i j k

Substituting into the Eq. 0FΣ = and factoring , , :i j k

8 5

0.617 13

12 9.60.64

17 13

9 7.20.48 0

17 13

AB AC AD

AB AC AD

AB AC AD

T T T

T T T P

T T T

− + +

+ − − − + + + − =

i

j

k

Dimensions in mm

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PROBLEM 2.109 (Continued)

Setting the coefficient of i, j, k equal to zero:

:i 8 5

0.6 017 13AB AC ADT T T− + + = (1)

:j 12 9.6

0.64 07 13AB AC ADT T T P− − − + = (2)

:k 9 7.2

0.48 017 13AB AC ADT T T+ − = (3)

Making 60 NACT = in (1) and (3):

8 5

36 N 017 13AB ADT T− + + = (1′)

9 7.2

28.8 N 017 13AB ADT T+ − = (3′)

Multiply (1′) by 9, (3′) by 8, and add:

12.6

554.4 N 0 572.0 N13 AD ADT T− = =

Substitute into (1′) and solve for :ABT

17 5

36 572 544.0 N8 13AB ABT T = + × =

Substitute for the tensions in Eq. (2) and solve for P :

12 9.6(544 N) 0.64(60 N) (572 N)

17 13844.8 N

P = + +

= Weight of plate 845 NP= =

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%%%%%%%%%Problem 2.109 How to calculate weight of a plate supported by %%tensions at point A in cables AD & AB & AC knowing that tension in AC is %%60 N AB=[-320 -480 360] %%% -320i -450j + 360k AC=[450 -480 360] %%% 450i - 450j + 360k AD=[250 -480 -360] %%% 250i -450j + 360k %%Tab=lamdaAB.*Tab lamdaAB=AB./(AB(1)^2 + AB(2)^2 + AB(3)^2)^0.5 %%Tac=lamdaAC.*Tac lamdaAC=AC./(AC(1)^2 + AC(2)^2 + AC(3)^2)^0.5 Tac=lamdaAC.*60 %%Tad=lamdaAD.*Tad lamdaAD=AD./(AD(1)^2 + AD(2)^2 + AD(3)^2)^0.5 %%%%Consider the weight force = Wj as it is upward %%%%As long the plate is in equilibrium state therefore the summation of %%%%forces is equal to zero in all directions X, Y, Z %%%Calculating the weight of the plate in matrix calculations %%%[lamdaAB(1) lamdaAD(1) 0] [Tab] [-Tac(1)] %%%[lamdaAB(2) lamdaAD(2) 1] * [Tad] = [-Tac(2)] %%%[lamdaAB(3) lamdaAD(3) 0] [ W ] [-Tac(3)] A = [lamdaAB(1) lamdaAD(1) 0;lamdaAB(2) lamdaAD(2) 1;lamdaAB(3) lamdaAD(3) 0] C = [-Tac(1) -Tac(2) -Tac(3)]' X = inv(A)*C Tab = X(1) Tad = X(2) W = X(3)

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PROBLEM 2.110

A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 520 N, determine the weight of the plate.

SOLUTION

See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below:

8 5

0.6 017 13AB AC ADT T T− + + = (1)

12 9.6

0.64 017 13AB AC ADT T T P− + − + = (2)

9 7.2

0.48 017 13AB AC ADT T T+ − = (3)

Making 520 NADT = in Eqs. (1) and (3):

8

0.6 200 N 017 AB ACT T− + + = (1′)

9

0.48 288 N 017 AB ACT T+ − = (3′)

Multiply (1′) by 9, (3′) by 8, and add:

9.24 504 N 0 54.5455 NAC ACT T− = =

Substitute into (1′) and solve for :ABT

17

(0.6 54.5455 200) 494.545 N8AB ABT T= × + =

Substitute for the tensions in Eq. (2) and solve for P:

12 9.6(494.545 N) 0.64(54.5455 N) (520 N)

17 13768.00 N

P = + +

= Weight of plate 768 NP= =

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%%%%%%%%%Problem 2.110 How to calculate weight of a plate supported by %%tensions at point A in cables AD & AB & AC knowing that tension in AD is %%520 N AB=[-320 -480 360] %%% -320i -450j + 360k AC=[450 -480 360] %%% 450i - 450j + 360k AD=[250 -480 -360] %%% 250i -450j + 360k lamdaAB=AB./(AB(1)^2 + AB(2)^2 + AB(3)^2)^0.5 %%Tab=lamdaAB.*Tab lamdaAC=AC./(AC(1)^2 + AC(2)^2 + AC(3)^2)^0.5 %%Tac=lamdaAC.*Tac lamdaAD=AD./(AD(1)^2 + AD(2)^2 + AD(3)^2)^0.5 Tad=lamdaAD.*520 %%%%Consider the weight force = -Wj as it is downward %%%%As long the plate is in equilibrium state therefore the summation of %%%%forces is equal to zero in all directions X, Y, Z %%%Calculating the weight of the plate in matrix calculations %%%[lamdaAB(1) lamdaAC(1) 0] [Tab] [-Tad(1)] %%%[lamdaAB(2) lamdaAC(2) -1] * [Tac] = [-Tad(2)] %%%[lamdaAB(3) lamdaAC(3) 0] [ W ] [-Tad(3)] A = [lamdaAB(1) lamdaAC(1) 0;lamdaAB(2) lamdaAC(2) 1;lamdaAB(3) lamdaAC(3) 0] C = [-Tad(1) -Tad(2) -Tad(3)]' X = inv(A)*C Tab = X(1) Tac = X(2) W = X(3)

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Chapter 3 Statics of Rigid Bodies

Equivalent Systems of Forces

3.1 INTRODUCTION In the preceding chapter it was assumed that each of the bodies considered could be treated as a single particle. Such a view, however, is not always possible, and a body, in general, should be treated as a combination of a large number of particles. The size of the body will have to be taken into consideration, as well as the fact that forces will act on different particles and thus will have different points of application. Most of the bodies considered in elementary mechanics are assumed to be rigid, a rigid body being defined as one which does not deform. Actual structures and machines, however, are never absolutely rigid and deform under the loads to which they are subjected. But these deformations are usually small and do not appreciably affect the conditions of equilibrium or motion of the structure under consideration. They are important, though, as far as the resistance of the structure to failure is concerned and are considered in the study of mechanics of materials. In this chapter you will study the effect of forces exerted on a rigid body, and you will learn how to replace a given system of forces by a simpler equivalent system. This analysis will rest on the fundamental assumption that the effect of a given force on a rigid body remains unchanged if that force is moved along its line of action (principle of transmissibility). The basic system is called a force-couple system. In the case of concurrent, coplanar, or parallel forces, the equivalent force-couple system can be further reduced to a single force, called the resultant of the system, or to a single couple, called the resultant couple of the system. 3.2 EXTERNAL AND INTERNAL FORCES Forces acting on rigid bodies can be separated into two groups: (1) external forces and (2) internal forces.

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1. The external forces represent the action of other bodies on the rigid body under consideration. They are entirely responsible for the external behavior of the rigid body. They will either cause it to move or ensure that it remains at rest. We shall be concerned only with external forces in this chapter and in Chaps. 4 and 5. 2 . The internal forces are the forces which hold together the particles forming the rigid body. If the rigid body is structurally composed of several parts, the forces holding the component parts together are also defined as internal forces. Internal forces will be considered in Chaps. 6 and 7. As an example of external forces, let us consider the forces acting on a disabled truck that three people are pulling forward by means of a rope attached to the front bumper (Fig. 3.1). The external forces acting on the truck are shown in a free-body diagram (Fig. 3.2). Let us first consider the weight of the truck. Although it embodies the effect of the earth’s pull on each of the particles forming the truck, the weight can be represented by the single force W. The point of application of this force, i.e., the point at which the force acts, is defined as the center of gravity of the truck. It will be seen in Chap. 5 how centers of gravity can be determined. The weight W tends to make the truck move vertically downward. In fact, it would actually cause the truck to move downward, i.e., to fall, if it were not for the presence of the ground. The ground opposes the downward motion of the truck by means of the reactions R 1 and R 2. These forces are exerted by the ground on the truck and must therefore be included among the external forces acting on the truck. The people pulling on the rope exert the force F. The point of application of F is on the front bumper. The force F tends to make the truck move forward in a straight line and does actually make it move, since no external force opposes this motion. (Rolling resistance has been neglected here for simplicity.) This forward motion of the truck, during which each straight line keeps its original orientation (the floor of the truck remains horizontal, and the walls remain vertical), is known as a translation. Other forces might cause the truck to move differently. For example, the force exerted by a jack placed under the front axle would cause the truck to pivot about its rear axle. Such a motion is a rotation. It can be concluded, therefore, that each of the external forces acting on a rigid body can, if unopposed, impart to the rigid body a motion of translation or rotation, or both.

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3.3 VECTOR PRODUCT OF TWO VECTORS In order to gain a better understanding of the effect of a force on a rigid body, a new concept, the concept of a moment of a force about a point, will be introduced at this time. This concept will be more clearly understood, and applied more effectively, if we first add to the mathematical tools at our disposal the vector product of two vectors. The vector product of two vectors P and Q is defined as the vector V which satisfies the following conditions. 1. The line of action of V is perpendicular to the plane containing P and Q (Fig. 3.3 a). 2. The magnitude of V is the product of the magnitudes of P and Q and of the sine of the angle u formed by P and Q (the measure of which will always be 180° or less); we thus have

V = PQ sin θ 3. The direction of V is obtained from the right-hand rule. Close your right hand and hold it so that your fingers are curled in the same sense as the rotation through u which brings the vector P in line with the vector Q ; your thumb will then indicate the direction of the vector V (Fig. 3.3 b ). Note that if P and Q do not have a common point of application, they should first be redrawn from the same point. The three vectors P , Q, and V —taken in that order—are said to form a right-handed rule. As stated above, the vector V satisfying these three conditions (which define it uniquely) is referred to as the vector product of P and Q ; it is represented by the mathematical expression in vector form as:

V = P x Q

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Fig. 3.3

We can now easily express the vector product V of two given vectors P and Q in terms of the rectangular components of these vectors. Resolving P and Q into components, we first write:

V = P x Q = (Pxi + Pyj + Pzk) x (Qxi + Qyj + Qz

V = (Pk)

yQz - PzQy)i + (PzQx - PxQz)j + (PxQy - PyQx

The rectangular components of the vector product V are thus found to be:

)k

Which is more easily memorized by the following determinate:

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PROBLEM 3.1

A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α = 25°, determine the moment of the force about Point B by resolving the force into horizontal and vertical components.

SOLUTION

Free-Body Diagram of Rod AB:

(9 in.)cos65

3.8036 in.

(9 in.)sin 65

8.1568 in.

x

y

= °== °=

(20 lb cos 25 ) ( 20 lb sin 25 )

(18.1262 lb) (8.4524 lb)

x yF F= +

= ° + − °= −

F i j

i j

i j

/ ( 3.8036 in.) (8.1568 in.)A B BA= = − +r i j

/

( 3.8036 8.1568 ) (18.1262 8.4524 )

32.150 147.852

115.702 lb-in.

B A B= ×= − + × −= −= −

M r F

i j i j

k k

115.7 lb-in.B =M

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%%%%%%%%%Problem 3.1 How to calculate the moment of force about point B as %%%%%%%%%shown in figure. %%Calulating the acting force of 20 LB in vector form as following:- %%F = (Fx)i + (Fy)j %%F = 20*cos(25/57.3)i - 20*sin(25/57.3)j F=[ 20*cos(25/57.3) -20*sin(25/57.3)] %%% F = 20*cos(25/57.3)i - 20*sin(25/57.3)j %%Calulating the moment arm BA in vector form as shown in figure. %% BA = (x)i + (y)j %% BA = -9*cos(65/57.3)i + 9*sin(65/57.3)j BA=[-9*cos(65/57.3) 9*sin(65/57.3)] %%% BA = -9*cos(65/57.3)i + 9*sin(65/57.3)j %%Calculating the moment acting about point B by force of 20 LB as %%Mb = BA x F by the cross product method as following:- %% | i j k | %%determinate of | BA(1) BA(2) 0 | %% | F(1) F(2) 0 | %% Mb = BA(1)*F(2) - F(1)*BA(2) %%%%Mb= (Mb)k %%Negative sign of the moment magnitude represents a clock wise acting %%moment

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PROBLEM 3.6

A 300-N force P is applied at Point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into horizontal and vertical components. (b) Using the result of part (a), determine the perpendicular distance from O to the line of action of P.

SOLUTION

(0.2 m)cos 40

0.153209 m

(0.2 m)sin 40

0.128558 m

x

y

= °== °=

/ (0.153209 m) (0.128558 m)A O∴ = +r i j

(a) (300 N)sin 30

150 N

(300 N)cos30

259.81 N

x

y

F

F

= °== °

=

(150 N) (259.81 N)= +F i j

/

(0.153209 0.128558 ) m (150 259.81 ) N

(39.805 19.2837 ) N m

(20.521 N m)

O A O= ×= + × += − ⋅= ⋅

M r F

i j i j

k k

k 20.5 N mO = ⋅M

(b) OM Fd=

20.521 N m (300 N)( )

0.068403 m

d

d

⋅ == 68.4 mmd =

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%%%%%%%%%Problem 3.6 How to calculate the moment of force about point O as %%%%%%%%%shown in figure. %%Calulating the acting force of 300 N in vector form as following:- %%F = (Fx)i + (Fy)j %%F = 300*sin(30/57.3)i + 300*cos(30/57.3)j F=[ 300*sin(30/57.3) 300*cos(30/57.3)] %%% F = 300*sin(30/57.3)i + 300*cos(30/57.3)j %%Calulating the moment arm OA in vector form as shown in figure. %% OA = (x)i + (y)j %% OA = 200*cos(40/57.3)i + 200*sin(40/57.3)j OA=[200*cos(40/57.3) 200*sin(40/57.3)] %%% OA = 200*cos(40/57.3)i + 200*sin(40/57.3)j %%Calculating the moment acting about point O by force of 300 N as %%Mo = OA x F by evaluating the cross product as following:- %% | i j k | %%Mo =determinate of | OA(1) OA(2) 0 | %% | F(1) F(2) 0 | %% Mo = OA(1)*F(2) - F(1)*OA(2) %%%%Mo= (Mo)k in N.mm Mo = Mo/1000 %%%%Mo in N.m %%Negative sign of the moment magnitude represents a clock wise acting %%Calculating the perpendicular distance from point O to the line of the %%acting force of the 300 N as following:- D = Mo/300 %%%% D in meter

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PROBLEM 3.11

A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) at Point C, (b) at Point E.

SOLUTION

(a) Slope of line: 0.875 m 5

1.90 m 0.2 m 12EC = =

+

Then 12

( )13ABx ABT T=

12(1040 N)

13960 N

=

= (a)

and 5

(1040 N)13400 N

AByT =

=

Then (0.875 m) (0.2 m)

(960 N)(0.875 m) (400 N)(0.2 m)

D ABx AByM T T= −

= −

760 N m= ⋅ or 760 N mD = ⋅M

(b) We have ( ) ( )D ABx ABxM T y T x= +

(960 N)(0) (400 N)(1.90 m)

760 N m

= += ⋅ (b)

or 760 N mD = ⋅M

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%%%%%%%%%Problem 3.11 How to calculate the moment of force about point D as %%%%%%%%%shown in figure using unit vector method to resolve the tension in %%%%%%%%%cable BC into the horizontal and vertical components %%unit vector of CB = unit vector of CE ---> lamdaCB = lamdaCE therefore %%lamdaCE can be easily calculated from the geometry %%CE = ((-d)i + 0j) - ((o.2)i + (0.857)j) d=1.9 %%%%d = 1.9 meter CE = [(-d - 0.2) (0 - 0.875)] lamdaCE =CE./(CE(1)^2 + CE(2)^2)^0.5 %%Calulating the acting tension of 1040 N in vector form as following:- %%T = (Fx)i + (Fy)j %%T = 1040*lamdaCE T = 1040*lamdaCE %%Calulating the moment arm DC in vector form as shown in figure. %% DC = (x)i + (y)j %% DC = (0.2)i + (0.875)j DC = [(0.2) (0.875)] %%Calculating the moment acting about point D by force of 1040 N as %%MD = DC x T by evaluating the cross product as following:- %% | i j k | %%MD =determinate of | DC(1) DC(2) 0 | %% | T(1) T(2) 0 | %% MD1 = DC(1)*T(2) - T(1)*DC(2) %%%%MD1 = (MD1)k in N.m %%Negative sign of the moment magnitude represents a clock wise acting %%Calulating the moment arm DE in vector form as shown in figure. %% DE = (x)i + (y)j %% DE = (-1.9)i (0)j DE = [(-1.9) (0)] %%Calculating the moment acting about point D by force of 1040 N as %%MD = DE x T by evaluating the cross product as following:- %% | i j k | %%MD =determinate of | DE(1) DE(2) 0 | %% | T(1) T(2) 0 | %% MD2 = DE(1)*T(2) - T(1)*DE(2) %%%%MD2= (MD2)k in N.m %%Negative sign of the moment magnitude represents a clock wise acting

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PROBLEM 3.12

It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the required moment about Point D.

SOLUTION

Slope of line: 0.875 m 7

2.80 m 0.2 m 24EC = =

+

Then 24

25ABx ABT T=

and 7

25ABy ABT T=

We have ( ) ( )D ABx AByM T y T x= +

24 7960 N m (0) (2.80 m)

25 251224 N

AB AB

AB

T T

T

⋅ = +

= or 1224 NABT =

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PROBLEM 3.13

It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If the capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified moment about Point D.

SOLUTION

The minimum value of d can be found based on the equation relating the moment of the force ABT about D:

max( ) ( )D AB yM T d=

where 960 N mDM = ⋅

max max( ) sin (2400 N)sinAB y ABT T θ θ= =

Now 2 2

2 2

0.875 msin

( 0.20) (0.875) m

0.875960 N m 2400 N ( )

( 0.20) (0.875)

d

dd

θ =+ +

⋅ = + +

or 2 2( 0.20) (0.875) 2.1875d d+ + =

or 2 2 2( 0.20) (0.875) 4.7852d d+ + =

or 23.7852 0.40 0.8056 0d d− − =

Using the quadratic equation, the minimum values of d are 0.51719 m and 0.41151 m.−

Since only the positive value applies here, 0.51719 md =

or 517 mmd =

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PROBLEM 3.15

Form the vector products B × C and B′ × C, where B = B′, and use the results obtained to prove the identity

1 1sin cos sin ( ) sin ( ).

2 2α β α β α β= + + −

SOLUTION

Note: (cos sin )

(cos sin )

(cos sin )

B

B

C

β ββ βα α

= +′ = −= +

B i j

B i j

C i j

By definition, | | sin ( )BC α β× = −B C (1)

| | sin ( )BC α β′× = +B C (2)

Now (cos sin ) (cos sin )B Cβ β α α× = + × +B C i j i j

(cos sin sin cos )BC β α β α= − k (3)

and (cos sin ) (cos sin )B Cβ β α α′× = − × +B C i j i j

(cos sin sin cos )BC β α β α= + k (4)

Equating the magnitudes of ×B C from Equations (1) and (3) yields:

sin( ) (cos sin sin cos )BC BCα β β α β α− = − (5)

Similarly, equating the magnitudes of ′×B C from Equations (2) and (4) yields:

sin( ) (cos sin sin cos )BC BCα β β α β α+ = + (6)

Adding Equations (5) and (6) gives:

sin( ) sin( ) 2cos sinα β α β β α− + + =

or 1 1

sin cos sin( ) sin( )2 2

α β α β α β= + + −

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176

PROBLEM 3.16

The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k.

SOLUTION

(a) We have | |A = ×P Q

where 7 3 3= − + −P i j k

2 2 5= + +Q i j k

Then 7 3 3

2 2 5

[(15 6) ( 6 35) ( 14 6) ]

(21) (29) ( 20)

× = − −

= + + − + + − −= + −

i j k

P Q

i j k

i j k

2 2 2(20) (29) ( 20)A = + + − or 41.0A =

(b) We have | |A = ×P Q

where 6 5 2= − −P i j k

2 5 1= − + −Q i j k

Then 6 5 2

2 5 1

[(5 10) (4 6) (30 10) ]

(15) (10) (20)

× = − −− −

= + + + + −= + +

i j k

P Q

i j k

i j k

2 2 2(15) (10) (20)A = + + or 26.9A =

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177

PROBLEM 3.17

A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) i + 2j − 5k and 4i − 7j − 5k, (b) 3i − 3j + 2k and −2i + 6j − 4k.

SOLUTION

(a) We have | |

×=×

A BλA B

where 1 2 5= + −A i j k

4 7 5= − −B i j k

Then 1 2 5

4 7 5

( 10 35) (20 5) ( 7 8)

15(3 1 1 )

× = + −− −

= − − + + + − −= − −

i j k

A B

i j k

i j k

and 2 2 2| | 15 ( 3) ( 1) ( 1) 15 11× = − + − + − =A B

15( 3 1 1 )

15 11

− − −= i j kλ or 1

( 3 )11

= − − −λ i j k

(b) We have | |

×=×

A BλA B

where 3 3 2= − +A i j k

2 6 4= − + −B i j k

Then 3 3 2

2 6 4

(12 12) ( 4 12) (18 6)

(8 12 )

× = −− −

= − + − + + −= +

i j k

A B

i j k

j k

and 2 2| | 4 (2) (3) 4 13× = + =A B

4(2 3 )

4 13

+= j kλ or 1

(2 3 )13

= +λ j k

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179

PROBLEM 3.19

Determine the moment about the origin O of the force F = 4i − 3j + 5k that acts at a Point A. Assume that the position vector of A is (a) r = 2i + 3j − 4k, (b) r = −8i + 6j − 10k, (c) r = 8i − 6j + 5k.

SOLUTION

O = ×M r F

(a) 2 3 4

4 3 5O = −

−

i j k

M

(15 12) ( 16 10) ( 6 12)= − + − − + − −i j k 3 26 18O = − −M i j k

(b) 8 6 10

4 3 5O = − −

−

i j k

M

(30 30) ( 40 40) (24 24)= − + − + + −i j k 0O =M

(c) 8 6 5

4 3 5O = −

−

i j k

M

( 30 15) (20 40) ( 24 24)= − + + − + − +i j k 15 20O = − −M i j

Note: The answer to Part (b) could have been anticipated since the elements of the last two rows of the determinant are proportional.

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180

PROBLEM 3.20

Determine the moment about the origin O of the force F = 2i + 3j − 4k that acts at a Point A. Assume that the position vector of A is (a) r = 3i − 6j + 5k, (b) r = i − 4j − 2k, (c) r = 4i + 6j − 8k.

SOLUTION

O = ×M r F

(a) 3 6 5

2 3 4O = −

−

i j k

M

(24 15) (10 12) (9 12)= − + + + +i j k 9 22 21O = + +M i j k

(b) 1 4 2

2 3 4O = − −

−

i j k

M

(16 6) ( 4 4) (3 8)= + + − + + +i j k 22 11O = +M i k

(c) 4 6 8

2 3 4O = −

−

i j k

M

( 24 24) ( 16 16) (12 12)= − + + − + + −i j k 0O =M

Note: The answer to Part (c) could have been anticipated since the elements of the last two rows of the determinant are proportional.

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213

PROBLEM 3.53

A single force P acts at C in a direction perpendicular to the handle BC of the crank shown. Knowing that Mx = +20 N ·

m and My = −8.75 N · m, and Mz = −30 N · m, determine the magnitude of P and the values of φ and θ.

SOLUTION

(0.25 m) (0.2 m)sin (0.2 m)cos

sin cos

0.25 0.2sin 0.2cos

0 sin cos

C

O C

P P

P P

θ θφ φ

θ θφ φ

= + += − +

= × =−

r i j k

P j k

i j k

M r P

Expanding the determinant, we find

(0.2) (sin cos cos sin )xM P θ φ θ φ= +

(0.2) sin( )xM P θ φ= + (1)

(0.25) cosyM P φ= − (2)

(0.25) sinzM P φ= − (3)

Dividing Eq. (3) by Eq. (2) gives: tan z

y

M

Mφ = (4)

30 N m

tan8.75 N m

φ − ⋅=− ⋅

73.740φ = 73.7φ = °

Squaring Eqs. (2) and (3) and adding gives:

2 2 2 2 2 2(0.25) or 4y z y zM M P P M M+ = = + (5)

2 24 (8.75) (30)

125.0 N

P = += 125.0 NP =

Substituting data into Eq. (1):

( 20 N m) 0.2 m(125.0 N)sin( )

( ) 53.130 and ( ) 126.87

20.6 and 53.1

θ φθ φ θ φ

θ θ

+ ⋅ = ++ = ° + = °

= − ° = °

53.1Q = °

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217

PROBLEM 3.57

The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 235-lb force P.

SOLUTION

(32 in.) (30 in.) (24 in.)AB = − −i j k

2 2 2(32) ( 30) ( 24) 50 in.AB = + − + − =

0.64 0.60 0.48ABAB

AB= = − −i k

λ

We shall apply the force P at Point G:

/ (5 in.) (30 in.)G B = +r i k

(21 in.) (38 in.) (18 in.)DG = − +i j k

2 2 2(21) ( 38) (18) 47 in.DG = + − + =

21 38 18

(235 lb)47

DGP

DG

− += = i j kP

(105 lb) (190 lb) (90 lb)= − +P i j k

The moment of P about AB is given by Eq. (3.46):

/

0.64 0.60 0.48

( ) 5 in. 0 30 in.

105 lb 190 lb 90 lbAB AB G B P

− −= ⋅ × =

−M rλ

0.64[0 (30 in.)( 190 lb)]

0.60[(30 in.)(105 lb) (5 in.)(90 lb)]

0.48[(5 in.)( 190 lb) 0]

2484 lb in.

AB = − −− −− − −

= + ⋅

M

207 lb ftAB = + ⋅M

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%%%%%%%%%Problem 3.57 How to calculate the moment of force %%%%%%%%%about a definite straight line AB %%%%%%%%%shown in figure using unit vector method to resolve the %%%%%%%%%force acting P into horizontal and vertical component and resolve %%%%%%%%%the rod AB into horizontal and vertical compnents of unity %%unit vector of AB where AB = 32i - 30j - 24j AB = [32 -30 -24] lamdaAB =AB./(AB(1)^2 + AB(2)^2 + AB(3)^2)^0.5 %%unit vector of DG where DG = 16i - 38j + 18j DG = [21 -38 18] lamdaDG =DG./(DG(1)^2 + DG(2)^2 + DG(3)^2)^0.5 %%Calulating the acting force P of 235 LB in vector form as following:- %%P = (Fx)i + (Fy)j + (Fz)K %%P = 235*lamdaDG P = 235*lamdaDG %%Calulating the moment arm CD in vector form as shown in figure. %% CD = (x)i + (y)j + (z)k %% CD= 0i + 23j + 0k CD = 23 %%%% DC = 23j %%Calculating the moment acting about point C by force of 235 LB as %%MC = CD x P by evaluating the cross product as following:- %% | i j k | %%MC =determinate of | 0 CD 0 | %% | P(1) P(2) P(3)| %% MC = [CD*P(3) -CD*P(1)] %%%%MC = (MCx)i + (MCy)k in LB.in %%Negative sign of the moment magnitude represents a clock wise acting %%Calculating the moment of force P about AB Rod Mab=lamdaAB(1)*MC(1) + lamdaAB(3)*MC(2)%%%%%%Scalar value of acting %%moment by force P about the rod AB in LB.in

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218

PROBLEM 3.58

The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q.

SOLUTION

(32 in.) (30 in.) (24 in.)AB = − −i j k

2 2 2(32) ( 30) ( 24) 50 in.AB = + − + − =

0.64 0.60 0.48ABAB

AB= = − −i j k

λ

We shall apply the force Q at Point H:

/ (32 in.) (17 in.)H B = − +r i j

(16 in.) (21 in.) (12 in.)DH = − − −i j k

2 2 2(16) ( 21) ( 12) 29 in.DH = + − + − =

16 21 12

(174 lb)29

DH

DH

− − −= = i j kQ

(96 lb) (126 lb) (72 lb)Q = − − −i j k

The moment of Q about AB is given by Eq. (3.46):

/

0.64 0.60 0.48

( ) 32 in. 17 in. 0

96 lb 126 lb 72 lbAB AB H B

− −= ⋅ × = −

− − −M r Qλ

0.64[(17 in.)( 72 lb) 0]

0.60[(0 ( 32 in.)( 72 lb)]

0.48[( 32 in.)( 126 lb) (17 in.)( 96 lb)]

2119.7 lb in.

AB = − −− − − −− − − − −

= − ⋅

M

176.6 lb ftAB = ⋅M

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%%%%%%%%%Problem 3.58 How to calculate the moment of force %%%%%%%%%about a definite straight line AB %%%%%%%%%shown in figure using unit vector method to resolve the %%%%%%%%%force acting Q into horizontal and vertical component and resolve %%%%%%%%%the rod AB into horizontal and vertical compnents of unity %%unit vector of AB where AB = 32i - 30j - 24j AB = [32 -30 -24] lamdaAB =AB./(AB(1)^2 + AB(2)^2 + AB(3)^2)^0.5 %%unit vector of DH where DH = -16i - 21j - 12j DH = [-16 -21 -12] lamdaDH =DH./(DH(1)^2 + DH(2)^2 + DH(3)^2)^0.5 %%Calulating the acting force Q of 174 LB in vector form as following:- %%Q = (Fx)i + (Fy)j + (Fz)K %%Q = 174*lamdaDH Q = 174*lamdaDH %%Calulating the moment arm CD in vector form as shown in figure. %% CD = (x)i + (y)j + (z)k %% CD= 0i + 23j + 0k CD = 23 %%%% DC = 23j %%Calculating the moment acting about point C by force of 174 LB as %%MC = CD x Q by evaluating the cross product as following:- %% | i j k | %%MC =determinate of | 0 CD 0 | %% | Q(1) Q(2) Q(3)| %% MC = [CD*Q(3) -CD*Q(1)] %%%%MC = (MCx)i + (MCy)k in LB.in %%Negative sign of the moment magnitude represents a clock wise acting %%Calculating the moment of force Q about AB Rod Mab=lamdaAB(1)*MC(1) + lamdaAB(3)*MC(2) %%%%%%Scalar value of acting %%moment by force Q about the rod AB in LB.in

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271

PROBLEM 3.107

The weights of two children sitting at ends A and B of a seesaw are 84 lb and 64 lb, respectively. Where should a third child sit so that the resultant of the weights of the three children will pass through C if she weighs (a) 60 lb, (b) 52 lb.

SOLUTION

(a) For the resultant weight to act at C, 0 60 lbC CM WΣ = =

Then (84 lb)(6 ft) 60 lb( ) 64 lb(6 ft) 0d− − =

2.00 ft to the right of d C=

(b) For the resultant weight to act at C, 0 52 lbC CM WΣ = =

Then (84 lb)(6 ft) 52 lb( ) 64 lb(6 ft) 0d− − =

2.31 ft to the right of d C=

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272

PROBLEM 3.108

A couple of magnitude M = 54 lb ⋅ in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC.

SOLUTION

(a) We have : ( 10 ) (30 cos 60 )

30 sin 60 ( 45 )

(30 lb) (15.9808 lb)

Σ = − + °+ ° + −

= − +

F R j i

j i

i j

or 34.0 lb=R 28.0°

(b) First reduce the given forces and couple to an equivalent force-couple system ( , )BR M at B.

We have : (54 lb in) (12 in.)(10 lb) (8 in.)(45 lb)

186 lb in.B BM MΣ = ⋅ + −

= − ⋅

Then with R at D, : 186 lb in (15.9808 lb)BM aΣ − ⋅ =

or 11.64 in.a =

and with R at E, : 186 lb in (30 lb)BM CΣ − ⋅ =

or 6.2 in.C =

The line of action of R intersects line AB 11.64 in. to the left of B and intersects line BC 6.20 in. below B.

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278

PROBLEM 3.114

Four ropes are attached to a crate and exert the forces shown. If the forces are to be replaced with a single equivalent force applied at a point on line AB, determine (a) the equivalent force and the distance from A to the point of application of the force when 30 ,α = ° (b) the value of α so that the single equivalent force is applied at Point B.

SOLUTION

We have

(a) For equivalence, : 100 cos30 400 cos65 90 cos65x xF RΣ − ° + ° + ° =

or 120.480 lbxR =

: 100 sin 160 400 sin 65 90 sin 65y yF RαΣ + + ° + ° =

or (604.09 100sin ) lbyR α= + (1)

With 30 ,α = ° 654.09 lbyR =

Then 2 2 654.09(120.480) (654.09) tan

120.480665 lb or 79.6

R θ

θ

= + =

= = °

Also

: (46 in.)(160 lb) (66 in.)(400 lb)sin 65

(26 in.)(400 lb)cos65 (66 in.)(90 lb)sin 65

(36 in.)(90 lb)cos65 (654.09 lb)

AM

d

Σ + °+ ° + °+ ° =

or 42,435 lb in. and 64.9 in.AM dΣ = ⋅ = 665 lbR = 79.6°

and R is applied 64.9 in. to the right of A.

(b) We have 66 in.d =

Then : 42,435 lb in (66 in.)A yM RΣ ⋅ =

or 642.95 lbyR =

Using Eq. (1): 642.95 604.09 100sinα= + or 22.9α = °

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%%%%%%%%%Problem 3.114 How to calculate the equivalent force of a proposed %%%%%%%%%system of forces %%Considering an equivalent force R has two components Rx & Ry for the %%proposed system of forces ie. Summation of all forces in X direction is %%equal to Rx and Summation of all forces in Y direction is equal to Ry Rx=90*cos(65/57.3) + 400*cos(65/57.3) - 100*cos(30/57.3) Ry=90*sin(65/57.3) + 400*sin(65/57.3) + 100*sin(30/57.3) + 160 %%Calculating equivalent force angle by using tan(theta)=Ry/Rx theta = atan(Ry/Rx)*57.3 %%%%theta in Degrees %%Calculating the application point of the R over the edge AB %%Summation of the system moments = the equivalent moment by the equivalent %%force R we can take the moments about any arbitrary point ex. point A %%Assume D is the distance between the equivalent force and point A D =(160*46 + 90*sin(65/57.3)*66 + 400*sin(65/57.3)*66 + 90*cos(65/57.3)*36 + 400*cos(65/57.3)*26)/Ry %%Considering D = 66 in -----> the application angle of the equivalent %%force and alfa angle could be calculated as following:- Ry =(160*46 + 90*sin(65/57.3)*66 + 400*sin(65/57.3)*66 + 90*cos(65/57.3)*36 + 400*cos(65/57.3)*26)/66 sinalfa=(90*sin(65/57.3) + 400*sin(65/57.3) + 160 - Ry)/(-100) alfa=asin(sinalfa)*57.3 Rx=90*cos(65/57.3) + 400*cos(65/57.3) - 100*cos(alfa/57.3) R = (Rx^2 + Ry^2)^0.5

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306

PROBLEM 3.137*

Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz-plane.

SOLUTION

First, reduce the given force system to a force-couple at the origin.

We have : (84 N) (80 N) 116 NRΣ − − = =F j k R

and : ( ) RO O C OΣ Σ × + Σ =M r F M M

0.6 0 0.1 0.4 0.3 0 ( 30 32 ) N m

0 84 0 0 0 80

RO+ + − − ⋅ =

i j k i j k

j k M

(15.6 N m) (2 N m) (82.4 N m)RO = − ⋅ + ⋅ − ⋅M i j k

(a) (84.0 N) (80.0 N)= − −R j k

(b) We have 1

84 80[ (15.6 N m) (2 N m) (82.4 N m) ]

11655.379 N m

RR O RM

R= ⋅ =

− −= − ⋅ − ⋅ + ⋅ − ⋅

= ⋅

Rλ M λ

j ki j k

and 1 1 (40.102 N m) (38.192 N m)RM λ= = − ⋅ − ⋅M j k

Then pitch 1 55.379 N m0.47741 m

116 N

Mp

R

⋅= = = or 0.477 mp =

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307

PROBLEM 3.137* (Continued)

(c) We have 1 2

2 1 [( 15.6 2 82.4 ) (40.102 38.192 )] N m

(15.6 N m) (42.102 N m) (44.208 N m)

RO

RO

= +

= − = − + − − − ⋅= − ⋅ + ⋅ − ⋅

M M M

M M M i j k j k

i j k

We require 2 /

( 15.6 42.102 44.208 ) ( ) (84 80 )

(84 ) (80 ) (84 )

Q O

x z

z x x

= ×

− + − = + × −= + −

M r R

i j k i k j k

i j k

From i: 15.6 84

0.185714 m

z

z

− == −

or 0.1857 mz = −

From k: 44.208 84

0.52629 m

x

x

− = −=

or 0.526 mx =

The axis of the wrench intersects the xz-plane at

0.526 m 0 0.1857 mx y z= = = −

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%%%%%%%%%Problem 3.137 How to calculate the equivalent wrench %%(Force - Couple) %%of a proposed system of forces and moments %%Calculating the equivalent force at the orgin Rx = 0 Ry = -84 Rz = -80 R = (Rx^2 + Ry^2 + Rz^2)^0.5 %% ie. R = 0i + (Ry)j + (Rz)k %%%Calculating the unit vector of R lamdaR=[Rx Ry Rz]./(Rx^2 + Ry^2 + Rz^2)^0.5 %%Calculating the equivalent moment at the orgin in form of two components %%in Y & Z directions Mox = 84*0.1 - 80*0.3 Moy = -30 + 80*0.4 Moz = -32 - 84*0.6 MoMAG = (Mox^2 + Moy^2 + Moz^2)^0.5 Mo = [Mox Moy Moz] %% ie. Mo = (Mox)i + (Moy)j + (Moz)k %%%Calculating the component of Mo in the direction of R M1MAG = lamdaR(1)*Mo(1) + lamdaR(2)*Mo(2) + lamdaR(3)*Mo(3) %%%Scalar value of M1 on N.m M1 = M1MAG*lamdaR %%%%Vector form in N.m M2 = Mo - M1 %%%% in N.m Pitch = M1/R %%%% in meters %%%Calculating the shift of the wrench in xz plane through M2 = rxz X R %%% | i j k | %%% M2 = Determinate | rx ry rz| = (M2x)i + (M2y)j + (M2z)k %%% | Rx Ry Rz| rx = M2(3)/Ry %%% in meters rz = -M2(1)/Ry %%% in meters

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308

PROBLEM 3.138*

Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz-plane.

SOLUTION

First, reduce the given force system to a force-couple at the origin at B.

(a) We have 8 15

: (26.4 lb) (17 lb)17 17 Σ − − + =

F k i j R

(8.00 lb) (15.00 lb) (26.4 lb)= − − −R i j k

and 31.4 lbR =

We have /: RB A B A A B BΣ × + + =M r F M M M

8 150 10 0 220 238 264 220 14(8 15 )

17 170 0 26.4

(152 lb in.) (210 lb in.) (220 lb in.)

RB

RB

= − − − + = − − + −

= ⋅ − ⋅ − ⋅

i j k

M k i j i k i j

M i j k

(b) We have 1

8.00 15.00 26.4[(152 lb in.) (210 lb in.) (220 lb in.) ]

31.4246.56 lb in.

RR O RM

R= ⋅ =

− − −= ⋅ ⋅ − ⋅ − ⋅

= ⋅

Rλ M λ

i j ki j k

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309

PROBLEM 3.138* (Continued)

and 1 1 (62.818 lb in.) (117.783 lb in.) (207.30 lb in.)RM λ= = − ⋅ − ⋅ − ⋅M i j k

Then pitch 1 246.56 lb in.7.8522 in.

31.4 lb

Mp

R

⋅= = = or 7.85 in.p =

(c) We have 1 2

2 1 (152 210 220 ) ( 62.818 117.783 207.30 )

(214.82 lb in.) (92.217 lb in.) (12.7000 lb in.)

RB

RB

= +

= − = − − − − − −= ⋅ − ⋅ − ⋅

M M M

M M M i j k i j k

i j k

We require 2 /Q B= ×M r R

214.82 92.217 12.7000 0

8 15 26.4

(15 ) (8 ) (26.4 ) (15 )

x z

z z x x

− − =− − −

= − + −

i j k

i j k

i j j k

From i: 214.82 15 14.3213 in.z z= =

From k: 12.7000 15 0.84667 in.x x− = − =

The axis of the wrench intersects the xz-plane at 0.847 in. 0 14.32 in.x y z= = =

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PROBLEM 3.158

A concrete foundation mat in the shape of a regular hexagon of side 12 ft supports four column loads as shown. Determine the magnitudes of the additional loads that must be applied at B and F if the resultant of all six loads is to pass through the center of the mat.

SOLUTION

From the statement of the problem, it can be concluded that the six applied loads are equivalent to the resultant R at O. It then follows that

0 or 0 0O x zM MΣ = Σ = Σ =M

For the applied loads:

Then 0: (6 3 ft) (6 3 ft)(10 kips) (6 3 ft)(20 kips)

(6 3 ft) 0

x B

F

F

F

Σ = + −

− =

M

or 10B FF F− = (1)

0: (12 ft)(15 kips) (6 ft) (6 ft)(10 kips)

(12 ft)(30 kips) (6 ft)(20 kips) (6 ft) 0z B

F

F

F

Σ = + −− − + =

M

or 60B FF F+ = (2)

Then Eqs. (1) (2)+ 35.0 kipsB =F

and 25.0 kipsF =F

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Chapter 4 Equilibrium of Rigid Bodies

4.1 INTRODUCTION We saw in the preceding chapter that the external forces acting on a rigid body can be reduced to a force-couple system at some arbitrary point O. When the force and the couple are both equal to zero, the external forces form a system equivalent to zero, and the rigid body is said to be in equilibrium.

Resolving each force and each moment into its rectangular components, we can express the necessary and sufficient conditions for the equilibrium of a rigid body with the following six scalar equations:

4.2 REACTIONS AT SUPPORTS AND CONNECTIONS FOR A 2 DIMENSIONAL STRUCTURE In the first part of this chapter, the equilibrium of a two-dimensional structure is considered; i.e., it is assumed that the structure being analyzed and the forces applied to it are contained in the same plane. Clearly, the reactions needed to maintain the structure in the same position will also be contained in this plane. The reactions exerted on a two-dimensional structure can be divided into three groups corresponding to three types of supports, or connections as following:

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Table 4.1

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4.3 STATICALLY INDETERMINATE REACTIONS. PARTIAL CONSTRAINTS In the two examples considered in the preceding section (Figs. 4.2 and 4.3), the types of supports used were such that the rigid body could not possibly move under the given loads or under any other loading conditions. In such cases, the rigid body is said to be completely constrained. We also recall that the reactions corresponding to these supports involved three unknowns and could be determined by solving the three equations of equilibrium. When such a situation exists, the reactions are said to be statically determinate. Consider Fig. 4.4a, in which the truss shown is held by pins at A and B. These supports provide more constraints than are necessary to keep the truss from moving under the given loads or under any other loading conditions. We also note from the free-body diagram of Fig. 4.4b that the corresponding reactions involve four unknowns. Since, only three independent equilibrium equations are available, there are more unknowns than equations; thus, all of the unknowns cannot be determined. While the equations ∑𝑀𝑀𝐴𝐴 = 0 and ∑𝑀𝑀𝐵𝐵 = 0 yield the vertical components By and Ay, respectively, the equation ∑𝐹𝐹𝑥𝑥 = 0 gives only the sum Ax + Bx of the horizontal components of the reactions at A and B. The components Ax and Bx

are said to be statically indeterminate. They could be determined by considering the deformations produced in the truss by the given loading, but this method is beyond the scope of statics and belongs to the study of mechanics of materials.

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The supports used to hold the truss shown in Fig. 4.5a,b consist of rollers at A and B. Clearly, the constraints provided by these supports are not sufficient to keep the truss from moving. While any vertical motion is prevented, the truss is free to move horizontally. The truss is said to be partially constrained. The examples of Figs. 4.6 and 4.7 lead us to conclude that a rigid body is improperly constrained whenever the supports, even though they may provide a sufficient number of reactions, that are arranged in such a way that the reactions must be either concurrent or parallel.

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345

PROBLEM 4.1

Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION

Free-Body Diagram:

2

2

(350 kg)(9.81 m/s ) 3.4335 kN

(1400 kg)(9.81 m/s ) 13.7340 kNt

W

W

= =

= =

(a) Rear wheels: 0: (1.7 m 2.05 m) (2.05 m) (1.2 m) 2 (3 m) 0B tM W W W AΣ = + + + − =

(3.4335 kN)(3.75 m) (3.4335 kN)(2.05 m)

(13.7340 kN)(1.2 m) 2 (3 m) 0A

++ − =

6.0659 kNA = + 6.07 kN=A

(b) Front wheels: 0: 2 2 0y tF W W W A BΣ = − − − + + =

3.4335 kN 3.4335 kN 13.7340 kN 2(6.0659 kN) 2 0B− − − + + =

4.2346 kNB = + 4.23 kN=B

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%%Problem 4.1 How to calculate the ground reactions on a truck of 1400 kg %%As long as the case is in equilibrium state it is available to use the %%relevant equilibrium equations ---> Summation of Forces = 0 & Summation %%of Moment at any point = 0 %%Calculating the moments about point A %% Rb*3 - 1400*1.8 - 350*(3.75 - 2.8) + 350*(4.5 - 3.75) = 0 Rb = ((1400*1.8 + 350*(3.75 - 2.8) - 350*(4.5 - 3.75))/3) %%% in kg Ra = (2*350 + 1400 -Rb) %%% in kg Rb = (Rb/2)*9.8 %%%As we have 2 wheels in front in N Ra = (Ra/2)*9.8 %%%As we have 2 wheels in rear in N %%Calculating the maximum available distance between C & D %%Rb should be estimated to equal 0 to calculate the maximum available %%distance CD keeping weight in D location. CD = (1400*1.8 + 350*(3.75 - 2.8))/350 - 2.8 + 3.75 %%%% in meters for I=1:1:91 CD1(I) = I*0.1; Rb1(I) = ((1400*1.8 + 350*(3.75 - 2.8) - 350*(2.8 + CD1(I) - 3.75))/3); %%% in kg Ra1(I) = (2*350 + 1400 -Rb1(I)); %%% in kg Rb1(I) = (Rb1(I)/2)*9.8; %%%As we have 2 wheels in front in N Ra1(I) = (Ra1(I)/2)*9.8; %%%As we have 2 wheels in rear in N end figure(1); plot(CD1,Rb1,'o',CD1,Ra1,'*'); xlabel('CD Distance in meters '),grid on;ylabel('Reaction on one wheel in front in N'),grid on; xlim([0 10]) ylim([0 11000])

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346

PROBLEM 4.2

Solve Problem 4.1, assuming that crate D is removed and that the position of crate C is unchanged.

PROBLEM 4.1 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION

Free-Body Diagram:

2

2

(350 kg)(9.81 m/s ) 3.4335 kN

(1400 kg)(9.81 m/s ) 13.7340 kNt

W

W

= =

= =

(a) Rear wheels: 0: (1.7 m 2.05 m) (1.2 m) 2 (3 m) 0B tM W W AΣ = + + − =

(3.4335 kN)(3.75 m) (13.7340 kN)(1.2 m) 2 (3 m) 0A+ − =

4.8927 kNA = + 4.89 kN=A

(b) Front wheels: 0: 2 2 0y tM W W A BΣ = − − + + =

3.4335 kN 13.7340 kN 2(4.8927 kN) 2 0B− − + + =

3.6911 kNB = + 3.69 kN=B

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347

PROBLEM 4.3

A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if 10 in.,a = (b) if 7 in.a =

SOLUTION

Free-Body Diagram: 0: 0x xF BΣ = =

0: (40 lb)(6 in.) (30 lb) (10 lb)( 8 in.) (12 in.) 0BM a a AΣ = − − + + =

(40 160)

12

aA

−= (1)

0: (40 lb)(6 in.) (50 lb)(12 in.) (30 lb)( 12 in.)

(10 lb)( 20 in.) (12 in.) 0A

y

M a

a B

Σ = − − − +− + + =

(1400 40 )

12ya

B+=

Since (1400 40 )

0,12x

aB B

+= = (2)

(a) For 10 in.,a =

Eq. (1): (40 10 160)

20.0 lb12

A× −= = + 20.0 lb=A

Eq. (2): (1400 40 10)

150.0 lb12

B+ ×= = + 150.0 lb=B

(b) For 7 in.,a =

Eq. (1): (40 7 160)

10.00 lb12

A× −= = + 10.00 lb=A

Eq. (2): (1400 40 7)

140.0 lb12

B+ ×= = + 140.0 lb=B

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348

PROBLEM 4.4

For the bracket and loading of Problem 4.3, determine the smallest distance a if the bracket is not to move.

PROBLEM 4.3 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if 10 in.,a = (b) if 7 in.a =

SOLUTION

Free-Body Diagram:

For no motion, reaction at A must be downward or zero; smallest distance a for no motion corresponds to 0.=A

0: (40 lb)(6 in.) (30 lb) (10 lb)( 8 in.) (12 in.) 0BM a a AΣ = − − + + =

(40 160)

12

aA

−=

0: (40 160) 0A a= − = 4.00 in.a =

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Resolve the problem above after replacing the 10 lb force with 20 lb force.

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349

PROBLEM 4.5

A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when 35 ,α = ° (b) the corresponding reaction at each of the two wheels.

SOLUTION

Free-Body Diagram:

2

1

2

(40 kg)(9.81 m/s ) 392.40 N

(300 mm)sin (80 mm)cos

(430 mm)cos (300 mm)sin

(930 mm)cos

W mg

a

a

b

α αα αα

= = == −= −=

From free-body diagram of hand truck,

Dimensions in mm

2 10: ( ) ( ) ( ) 0BM P b W a W aΣ = − + = (1)

0: 2 2 0yF P W BΣ = − + = (2)

For 35α = °

1

2

300sin 35 80cos35 106.541 mm

430cos35 300sin 35 180.162 mm

930cos35 761.81 mm

a

a

b

= ° − ° == ° − ° == ° =

(a) From Equation (1):

(761.81 mm) 392.40 N(180.162 mm) 392.40 N(106.54 mm) 0P − + =

37.921 NP = or 37.9 N=P

(b) From Equation (2):

37.921 N 2(392.40 N) 2 0B− + = or 373 N=B

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350

PROBLEM 4.6

Solve Problem 4.5 when α = 40°.

PROBLEM 4.5 A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when α = 35°, (b) the corresponding reaction at each of the two wheels.

SOLUTION Free-Body Diagram:

2

1

2

(40 kg)(9.81 m/s )

392.40 N

(300 mm)sin (80 mm)cos

(430 mm)cos (300 mm)sin

(930 mm)cos

W mg

W

a

a

b

α αα αα

= === −= −=

From F.B.D.:

2 10: ( ) ( ) ( ) 0BM P b W a W aΣ = − + =

2 1( )/P W a a b= − (1)

0: 2 0yF W W P BΣ = − − + + =

1

2B W P= − (2)

For 40 :α = °

1

2

300sin 40 80cos 40 131.553 mm

430cos 40 300sin 40 136.563 mm

930cos 40 712.42 mm

a

a

b

= ° − ° == ° − ° == ° =

(a) From Equation (1): 392.40 N (0.136563 m 0.131553 m)

0.71242 mP

−=

2.7595 NP = 2.76 N=P

(b) From Equation (2): 1

392.40 N (2.7595 N)2

B = − 391 N=B

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351

PROBLEM 4.7

A 3200-lb forklift truck is used to lift a 1700-lb crate. Determine the reaction at each of the two (a) front wheels A, (b) rear wheels B.

SOLUTION

Free-Body Diagram:

(a) Front wheels: 0: (1700 lb)(52 in.) (3200 lb)(12 in.) 2 (36 in.) 0BM AΣ = + − =

1761.11 lbA = + 1761 lb=A

(b) Rear wheels: 0: 1700 lb 3200 lb 2(1761.11 lb) 2 0yF BΣ = − − + + =

688.89 lbB = + 689 lb=B

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352

PROBLEM 4.8

For the beam and loading shown, determine (a) the reaction at A, (b) the tension in cable BC.

SOLUTION

Free-Body Diagram:

(a) Reaction at A: 0: 0x xF AΣ = =

0: (15 lb)(28 in.) (20 lb)(22 in.) (35 lb)(14 in.)

(20 lb)(6 in.) (6 in.) 0B

y

M

A

Σ = + ++ − =

245 lbyA = + 245 lb=A

(b) Tension in BC: 0: (15 lb)(22 in.) (20 lb)(16 in.) (35 lb)(8 in.)

(15 lb)(6 in.) (6 in.) 0A

BC

M

F

Σ = + +− − =

140.0 lbBCF = + 140.0 lbBCF =

Check: 0: 15 lb 20 lb 35 lb 20 lb 0

105 lb 245 lb 140.0 0

y BCF A FΣ = − − = − + − =

− + − =

0 0 (Checks)=

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353

PROBLEM 4.9

For the beam and loading shown, determine the range of the distance a for which the reaction at B does not exceed 100 lb downward or 200 lb upward.

SOLUTION

Assume B is positive when directed .

Sketch showing distance from D to forces.

0: (300 lb)(8 in. ) (300 lb)( 2 in.) (50 lb)(4 in.) 16 0DM a a BΣ = − − − − + =

600 2800 16 0a B− + + =

(2800 16 )

600

Ba

+= (1)

For 100 lbB = 100 lb,= − Eq. (1) yields:

[2800 16( 100)] 1200

2 in.600 600

a+ −≥ = = 2.00 in.a ≥

For 200B = 200 lb,= + Eq. (1) yields:

[2800 16(200)] 6000

10 in.600 600

a+≤ = = 10.00 in.a ≤

Required range: 2.00 in. 10.00 in.a≤ ≤

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354

PROBLEM 4.10

The maximum allowable value of each of the reactions is 180 N. Neglecting the weight of the beam, determine the range of the distance d for which the beam is safe.

SOLUTION

0: 0x xF BΣ = =

yB B=

0: (50 N) (100 N)(0.45 m ) (150 N)(0.9 m ) (0.9 m ) 0AM d d d B dΣ = − − − − + − =

50 45 100 135 150 0.9 0d d d B Bd− + − + + − =

180 N m (0.9 m)

300

Bd

A B

⋅ −=−

(1)

0: (50 N)(0.9 m) (0.9 m ) (100 N)(0.45 m) 0BM A dΣ = − − + =

45 0.9 45 0A Ad− + + =

(0.9 m) 90 N mA

dA

− ⋅= (2)

Since 180 N,B ≤ Eq. (1) yields

180 (0.9)180 18

0.15 m300 180 120

d−≥ = =

− 150.0 mmd ≥

Since 180 N,A ≤ Eq. (2) yields

(0.9)180 90 72

0.40 m180 180

d−≤ = = 400 mmd ≤

Range: 150.0 mm 400 mmd≤ ≤

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355

PROBLEM 4.11

Three loads are applied as shown to a light beam supported by cables attached at B and D. Neglecting the weight of the beam, determine the range of values of Q for which neither cable becomes slack when P = 0.

SOLUTION

0: (3.00 kN)(0.500 m) (2.25 m) (3.00 m) 0B DM T QΣ = + − =

0.500 kN (0.750) DQ T= + (1)

0: (3.00 kN)(2.75 m) (2.25 m) (0.750 m) 0D BM T QΣ = − − =

11.00 kN (3.00) BQ T= − (2)

For cable B not to be slack, 0,BT ≥ and from Eq. (2),

11.00 kNQ ≤

For cable D not to be slack, 0,DT ≥ and from Eq. (1),

0.500 kNQ ≥

For neither cable to be slack,

0.500 kN 11.00 kNQ≤ ≤

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356

PROBLEM 4.12

Three loads are applied as shown to a light beam supported by cables attached at B and D. Knowing that the maximum allowable tension in each cable is 4 kN and neglecting the weight of the beam, determine the range of values of Q for which the loading is safe when P = 0.

SOLUTION

0: (3.00 kN)(0.500 m) (2.25 m) (3.00 m) 0B DM T QΣ = + − =

0.500 kN (0.750) DQ T= + (1)

0: (3.00 kN)(2.75 m) (2.25 m) (0.750 m) 0D BM T QΣ = − − =

11.00 kN (3.00) BQ T= − (2)

For 4.00 kN,BT ≤ Eq. (2) yields

11.00 kN 3.00(4.00 kN)Q ≥ − 1.000 kNQ ≥ −

For 4.00 kN,DT ≤ Eq. (1) yields

0.500 kN 0.750(4.00 kN)Q ≤ + 3.50 kNQ ≤

For loading to be safe, cables must also not be slack. Combining with the conditions obtained in Problem 4.11,

0.500 kN 3.50 kNQ≤ ≤

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357

PROBLEM 4.13

For the beam of Problem 4.12, determine the range of values of Q for which the loading is safe when P = 1 kN.

PROBLEM 4.12 Three loads are applied as shown to a light beam supported by cables attached at B and D. Knowing that the maximum allowable tension in each cable is 4 kN and neglecting the weight of the beam, determine the range of values of Q for which the loading is safe when P = 0.

SOLUTION

0: (3.00 kN)(0.500 m) (1.000 kN)(0.750 m) (2.25 m) (3.00 m) 0B DM T QΣ = − + − =

0.250 kN 0.75 DQ T= + (1)

0: (3.00 kN)(2.75 m) (1.000 kN)(1.50 m)

(2.25 m) (0.750 m) 0D

B

M

T Q

Σ = +− − =

13.00 kN 3.00 BQ T= − (2)

For the loading to be safe, cables must not be slack and tension must not exceed 4.00 kN.

Making 0 4.00 kNBT≤ ≤ in Eq. (2), we have

13.00 kN 3.00(4.00 kN) 13.00 kN 3.00(0)Q− ≤ ≤ −

1.000 kN 13.00 kNQ≤ ≤ (3)

Making 0 4.00 kNDT≤ ≤ in Eq. (1), we have

0.250 kN 0.750(0) 0.250 kN 0.750(4.00 kN)Q+ ≤ ≤ +

0.250 kN 3.25 kNQ≤ ≤ (4)

Combining Eqs. (3) and (4), 1.000 kN 3.25 kNQ≤ ≤

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358

PROBLEM 4.14

For the beam of Sample Problem 4.2, determine the range of values of P for which the beam will be safe, knowing that the maximum allowable value of each of the reactions is 30 kips and that the reaction at A must be directed upward.

SOLUTION

0: 0x xF BΣ = =

yB=B

0: (3 ft) (9 ft) (6 kips)(11 ft) (6 kips)(13 ft) 0AM P BΣ = − + − − =

3 48 kipsP B= − (1)

0: (9 ft) (6 ft) (6 kips)(2 ft) (6 kips)(4 ft) 0BM A PΣ = − + − − =

1.5 6 kipsP A= + (2)

Since 30 kips,B ≤ Eq. (1) yields

(3)(30 kips) 48 kipsP ≤ − 42.0 kipsP ≤

Since 0 30 kips,A≤ ≤ Eq. (2) yields

0 6 kips (1.5)(30 kips)1.6 kipsP+ ≤ ≤

6.00 kips 51.0 kipsP≤ ≤

Range of values of P for which beam will be safe:

6.00 kips 42.0 kipsP≤ ≤

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359

PROBLEM 4.15

The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

At B: 0.18 m

0.24 my

x

T

T=

3

4y xT T= (1)

(a) 0: (0.18 m) (240 N)(0.4 m) (240 N)(0.8 m) 0C xM TΣ = − − =

1600 NxT = +

From Eq. (1): 3

(1600 N) 1200 N4yT = =

2 2 2 21600 1200 2000 Nx yT T T= + = + = 2.00 kNT =

(b) 0: 0x x xF C TΣ = − =

1600 N 0 1600 Nx xC C− = = + 1600 Nx =C

0: 240 N 240 N 0y y yF C TΣ = − − − =

1200 N 480 N 0yC − − =

1680 NyC = + 1680 Ny =C

46.4

2320 NC

α = °= 2.32 kN=C 46.4°

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360

PROBLEM 4.16

Solve Problem 4.15, assuming that 0.32 m.a =

PROBLEM 4.15 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

At B: 0.32 m

0.24 m

4

3

y

x

y x

T

T

T T

=

=

0: (0.32 m) (240 N)(0.4 m) (240 N)(0.8 m) 0C xM TΣ = − − =

900 NxT =

From Eq. (1): 4

(900 N) 1200 N3yT = =

2 2 2 2900 1200 1500 Nx yT T T= + = + = 1.500 kNT =

0: 0x x xF C TΣ = − =

900 N 0 900 Nx xC C− = = + 900 Nx =C

0: 240 N 240 N 0y y yF C TΣ = − − − =

1200 N 480 N 0yC − − =

1680 NyC = + 1680 Ny =C

61.8

1906 NC

α = °= 1.906 kN=C 61.8°

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361

PROBLEM 4.17

The lever BCD is hinged at C and attached to a control rod at B. If P = 100 lb, determine (a) the tension in rod AB, (b) the reaction at C.

SOLUTION

Free-Body Diagram:

(a) 0: (5 in.) (100 lb)(7.5 in.) 0CM TΣ = − =

150.0 lbT =

(b) 3

0: 100 lb (150.0 lb) 05x xF CΣ = + + =

190 lbxC = − 190 lbx =C

4

0: lb) 05y yF CΣ = + (150.0 =

120 lbyC = − 120 lby =C

32.3α = ° 225 lbC = 225 lb=C 32.3°

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362

PROBLEM 4.18

The lever BCD is hinged at C and attached to a control rod at B. Determine the maximum force P that can be safely applied at D if the maximum allowable value of the reaction at C is 250 lb.

SOLUTION

Free-Body Diagram:

0: (5 in.) (7.5 in.) 0CM T PΣ = − =

1.5T P=

3

0: (1.5 ) 05x xF P C PΣ = + + =

1.9xC P= − 1.9x P=C

4

0: (1.5 ) 05y yF C PΣ = + =

1.2= −yC P 1.2=y PC

2 2

2 2(1.9 ) (1.2 )

x yC C C

P P

= +

= +

2.2472C P=

For 250 lb,C =

250 lb 2.2472P=

111.2 lbP = 111.2 lb=P

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363

PROBLEM 4.19

Two links AB and DE are connected by a bell crank as shown. Knowing that the tension in link AB is 720 N, determine (a) the tension in link DE, (b) the reaction at C.

SOLUTION

Free-Body Diagram:

0: (100 mm) (120 mm) 0C AB DEM F FΣ = − =

5

6DE ABF F= (1)

(a) For 720 NABF =

5

(720 N)6DEF = 600 NDEF =

(b) 3

0: (720 N) 05x xF CΣ = − + =

432 NxC = +

4

0: (720 N) 600 N 05

1176 N

y y

y

F C

C

Σ = − + − =

= +

1252.84 N

69.829

C

α== °

1253 N=C 69.8°

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364

PROBLEM 4.20

Two links AB and DE are connected by a bell crank as shown. Determine the maximum force that may be safely exerted by link AB on the bell crank if the maximum allowable value for the reaction at C is 1600 N.

SOLUTION

See solution to Problem 4.15 for F.B.D. and derivation of Eq. (1).

5

6DE ABF F= (1)

3 3

0: 05 5x AB x x ABF F C C FΣ = − + = =

4

0: 05y AB y DEF F C FΣ = − + − =

4 50

5 649

30

AB y AB

y AB

F C F

C F

− + − =

=

2 2

2 21(49) (18)

301.74005

x y

AB

AB

C C C

F

C F

= +

= +

=

For 1600 N, 1600 N 1.74005 ABC F= = 920 NABF =

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PROBLEM 4.38

A light rod AD is supported by frictionless pegs at B and C and rests against a frictionless wall at A. A vertical 120-lb force is applied at D. Determine the reactions at A, B, and C.

SOLUTION

Free-Body Diagram:

0: cos30 (120 lb)cos60 0xF AΣ = ° − ° =

69.28 lbA = 69.3 lb=A

0: (8 in.) (120 lb)(16 in.)cos30

(69.28 lb)(8 in.)sin 30 0BM CΣ = − °

+ ° =

173.2 lbC = 173.2 lb=C 60.0°

0: (8 in.) (120 lb)(8 in.)cos30

(69.28 lb)(16 in.)sin 30 0CM BΣ = − °

+ ° =

34.6 lbB = 34.6 lb=B 60.0°

Check: 0: 173.2 34.6 (69.28)sin 30 (120)sin 60 0yFΣ = − − ° − ° =

0 0 (check)=

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389

PROBLEM 4.41

The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C, D, and E when 30 .θ = °

SOLUTION

Free-Body Diagram:

0: cos30 20 40 0yF EΣ = ° − − =

60 lb

69.282 lbcos 30°

E = = 69.3 lb=E 60.0°

0: (20 lb)(4 in.) (40 lb)(4 in.)

(3 in.) sin 30 (3 in.) 0

Σ = −− + ° =

DM

C E

80 3 69.282(0.5)(3) 0C− − + =

7.9743 lbC = 7.97 lb=C

0: sin 30 0xF E C DΣ = ° + − =

(69.282 lb)(0.5) 7.9743 lb 0D+ − =

42.615 lbD = 42.6 lb=D

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390

PROBLEM 4.42

The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine (a) the smallest value of θ for which the equilibrium of the bracket is maintained, (b) the corresponding reactions at C, D, and E.

SOLUTION

Free-Body Diagram:

0: cos 20 40 0yF E θΣ = − − =

60

cosE

θ= (1)

0: (20 lb)(4 in.) (40 lb)(4 in.) (3 in.)

60+ sin 3 in. 0

cos

DM C

θθ

Σ = − −

=

1

(180 tan 80)3

C θ= −

(a) For 0,C = 180 tan 80θ =

4

tan 23.9629

θ θ= = ° 24.0θ = °

From Eq. (1): 60

65.659cos 23.962

E = =°

0: sin 0xF D C E θΣ = − + + =

(65.659) sin 23.962 26.666 lb= =D

(b) 0 26.7 lb= =C D 65.7 lb=E 66.0°

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391

PROBLEM 4.43

Beam AD carries the two 40-lb loads shown. The beam is held by a fixed support at D and by the cable BE that is attached to the counterweight W. Determine the reaction at D when (a) 100W = lb, (b) 90 lb.W =

SOLUTION

(a) 100 lbW =

From F.B.D. of beam AD:

0: 0x xF DΣ = =

0: 40 lb 40 lb 100 lb 0y yF DΣ = − − + =

20.0 lbyD = − or 20.0 lb=D

0: (100 lb)(5 ft) (40 lb)(8 ft)

(40 lb)(4 ft) 0D DM MΣ = − +

+ =

20.0 lb ftDM = ⋅ or 20.0 lb ftD = ⋅M

(b) 90 lbW =

From F.B.D. of beam AD:

0: 0x xF DΣ = =

0: 90 lb 40 lb 40 lb 0y yF DΣ = + − − =

10.00 lbyD = − or 10.00 lb=D

0: (90 lb)(5 ft) (40 lb)(8 ft)

(40 lb)(4 ft) 0D DM MΣ = − +

+ =

30.0 lb ftDM = − ⋅ or 30.0 lb ftD = ⋅M

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392

PROBLEM 4.44

For the beam and loading shown, determine the range of values of W for which the magnitude of the couple at D does not exceed 40 lb ⋅ ft.

SOLUTION

For min ,W 40 lb ftDM = − ⋅

From F.B.D. of beam AD: min0: (40 lb)(8 ft) (5 ft)

(40 lb)(4 ft) 40 lb ft 0DM WΣ = −

+ − ⋅ =

min 88.0 lbW =

For max,W 40 lb ftDM = ⋅

From F.B.D. of beam AD: max0: (40 lb)(8 ft) (5 ft)

(40 lb)(4 ft) 40 lb ft 0DM WΣ = −

+ + ⋅ =

max 104.0 lbW = or 88.0 lb 104.0 lbW≤ ≤

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393

PROBLEM 4.45

An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a 100-mm radius, determine the reaction at A in each case.

SOLUTION

2(8 kg)(9.81 m/s ) 78.480 NW mg= = =

(a) 0: 0x xF AΣ = =

0: 0y yF A WΣ = − = 78.480 Ny =A

0: (1.6 m) 0A AM M WΣ = − =

(78.480 N)(1.6 m)AM = + 125.568 N mA = ⋅M

78.5 N=A 125.6 N mA = ⋅M

(b) 0: 0x xF A WΣ = − = 78.480x =A

0: 0y yF A WΣ = − = 78.480y =A

(78.480 N) 2 110.987 N= =A 45°

0: (1.6 m) 0A AM M WΣ = − =

(78.480 N)(1.6 m) 125.568 N mA AM = + = ⋅M

111.0 N=A 45° 125.6 N mA = ⋅M

(c) 0: 0x xF AΣ = =

0: 2 0y yF A WΣ = − =

2 2(78.480 N) 156.960 NyA W= = =

0: 2 (1.6 m) 0A AM M WΣ = − =

2(78.480 N)(1.6 m)AM = + 251.14 N mA = ⋅M

157.0 N=A 251 N mA = ⋅M

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394

PROBLEM 4.46

A tension of 20 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C.

SOLUTION

Free-Body Diagram:

0: (20 N) 0x xF CΣ = + = 20 NxC = −

0: (20 N) 0y yF CΣ = − = 20 NyC = +

28.3 N=C 45.0°

0: (20 N)(0.160 m) (20 N)(0.055 m) 0C CM MΣ = + + =

4.30 N mCM = − ⋅ 4.30 N mC = ⋅M

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396

PROBLEM 4.48

The rig shown consists of a 1200-lb horizontal member ABC and a vertical member DBE welded together at B. The rig is being used to raise a 3600-lb crate at a distance x = 12 ft from the vertical member DBE. If the tension in the cable is 4 kips, determine the reaction at E, assuming that the cable is (a) anchored at F as shown in the figure, (b) attached to the vertical member at a point located 1 ft above E.

SOLUTION

Free-Body Diagram:

0: (3600 lb) (1200 lb) (6.5 ft) (3.75 ft) 0E EM M x T= + + − =

3.75 3600 7800EM T x= − − (1)

(a) For 12 ft and 4000 lbs,x T= =

3.75(4000) 3600(12) 7800

36,000 lb ftEM = − −

= ⋅

0 0x xF EΣ = ∴ =

0: 3600 lb 1200 lb 4000 0y yF EΣ = − − − =

8800 lbyE =

8.80 kips=E ; 36.0 kip ftE = ⋅M

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(b) 0: (3600 lb)(12 ft) (1200 lb)(6.5 ft) 0E EM MΣ = + + =

51,000 lb ftEM = − ⋅

0 0x xF EΣ = ∴ =

0: 3600 lb 1200 lb 0y yF EΣ = − − =

4800 lbyE =

4.80 kips=E ; 51.0 kip ftE = ⋅M

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%%Problem 4.48 how to calculate the reactions at point E for a balanced %%system as shown in figure %%The is balanced, therefore the summation of forces in both directions X, %%Y and the summation of moment at any point = 0. %%Calculating the moment about point E as following:- %%Assuming the reaction torque at E is in counterclock wise direction. for I=1:1:150 T(I) = I*100 %%% T in lbs Me(I) = T(I)*3.75 - 3600*12 - 1200*6.5; %%Calculating Ry at point E from the summation of forces in Y direction, %%considering Rx = 0 from the FBD. Ry(I) = T(I) + 3600 + 1200; end figure(1); plot(T, Me,'o'); xlabel('Tension in Cable ADCF '),grid on;ylabel('Reaction Couple in lb.ft'),grid on; xlim([0 14000]) ylim([-60000 0]) figure(2); plot(T, Ry,'*'); xlabel('Tension in Cable ADCF '),grid on;ylabel('Reaction Ry in lb'),grid on; xlim([0 6000]) ylim([4000 10000]) %%In case the cable is anchored at a point located 1 ft above point E. %%in this case tension T is eliminated from the equations directly. Me0 = -3600*12 - 1200*6.5 %%Calculating Ry at point E from the summation of forces in Y direction, %%considering Rx = 0 from the FBD. Ry0 = 3600 + 1200

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398

PROBLEM 4.49

For the rig and crate of Prob. 4.48, and assuming that cable is anchored at F as shown, determine (a) the required tension in cable ADCF if the maximum value of the couple at E as x varies from 1.5 to 17.5 ft is to be as small as possible, (b) the corresponding maximum value of the couple.

SOLUTION

Free-Body Diagram:

0: (3600 lb) (1200 lb)(6.5 ft) (3.75 ft) 0E EM M x T= + + − =

3.75 3600 7800EM T x= − − (1)

For 1.5 ft, Eq. (1) becomesx =

1( ) 3.75 3600(1.5) 7800EM T= − − (2)

For 17.5 ft, Eq. (1) becomesx =

2( ) 3.75 3600(17.5) 7800EM T= − −

(a) For smallest max value of | |,EM we set

1 2( ) ( )E EM M−=

3.75 13,200 3.75 70,800T T− = − + 11.20 kips=T

(b) From Equation (2), then

3.75(11.20) 13.20EM = − | | 28.8 kip ftEM = ⋅

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%%Problem 4.49 how to calculate the reactions at point E for a balanced %%system as shown in figure %%The is balanced, therefore the summation of forces in both directions X, %%Y and the summation of moment at any point = 0. %%Calculating the moment about point E as following:- %%Assume that Me at x=1.5ft is equal to -Me at x=17.5 ft %% i.e. T*3.75 - 3600*1.5 - 1200*6.5 = -(T*3.75 - 3600*17.5 - 1200*6.5) T = (3600*(17.5 + 1.5) + 1200*(6.5 + 6.5))/(3.75 + 3.75) Me = T*3.75 - 3600*1.5 - 1200*6.5 Me0 = T*3.75 - 3600*17.5 - 1200*6.5 Memin = T*3.75 - 3600*((17.5+1.5)/2) - 1200*6.5 %%Calculating Ry at point E from the summation of forces in Y direction, %%considering Rx = 0 from the FBD. Ry = T + 3600 + 1200 %%In Case that we Assume that Me at x=1.5ft is equal to 1.5*Me at x=17.5 ft T1 = (3600*(-17.5*1.5 + 1.5) + 1200*(-6.5*1.5 + 6.5))/(1*3.75 - 1.5*3.75) Me1 = T1*3.75 - 3600*1.5 - 1200*6.5 Me10 = T1*3.75 - 3600*17.5 - 1200*6.5 %%In Case that we Assume that Me at x=1.5ft is equal to -1.5*Me at x=17.5 ft T2 = (3600*(17.5*1.5 + 1.5) + 1200*(6.5*1.5 + 6.5))/(1*3.75 + 1.5*3.75) Me2 = T2*3.75 - 3600*1.5 - 1200*6.5 Me20 = T2*3.75 - 3600*17.5 - 1200*6.5

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440

PROBLEM 4.83

A thin ring of mass 2 kg and radius r = 140 mm is held against a frictionless wall by a 125-mm string AB. Determine (a) the distance d, (b) the tension in the string, (c) the reaction at C.

SOLUTION

Free-Body Diagram:

(Three-force body)

The force T exerted at B must pass through the center G of the ring, since C and W intersect at that point.

Thus, points A, B, and G are in a straight line.

(a) From triangle ACG: 2 2

2 2

( ) ( )

(265 mm) (140 mm)

225.00 mm

d AG CG= −

= −=

225 mmd =

Force Triangle

2(2 kg)(9.81 m/s ) 19.6200 NW = =

Law of sines: 19.6200 N

265 mm 140 mm 225.00 mm

T C= =

(b) 23.1 NT =

(c) 12.21 N=C

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441

PROBLEM 4.84

A uniform rod AB of length 2R rests inside a hemispherical bowl of radius R as shown. Neglecting friction, determine the angle θ corresponding to equilibrium.

SOLUTION

Based on the F.B.D., the uniform rod AB is a three-force body. Point E is the point of intersection of the three forces. Since force A passes through O, the center of the circle, and since force C is perpendicular to the rod, triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle.

Note that the angle α of triangle DOA is the central angle corresponding to the inscribed angleθ of triangle DCA.

2α θ=

The horizontal projections of , ( ),AEAE x and , ( ),AGAG x are equal.

AE AG Ax x x= =

or ( )cos 2 ( )cosAE AGθ θ=

and (2 )cos 2 cosR Rθ θ=

Now 2cos 2 2cos 1θ θ= −

then 24cos 2 cosθ θ− =

or 24cos cos 2 0θ θ− − =

Applying the quadratic equation,

cos 0.84307 and cos 0.59307θ θ= = −

32.534 and 126.375 (Discard)θ θ= ° = ° or 32.5θ = °

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448

PROBLEM 4.91

A 200-mm lever and a 240-mm-diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

SOLUTION

Dimensions in mm

We have six unknowns and six equations of equilibrium. —OK

0: ( 120 ) ( ) (120 160 ) (80 200 ) ( 720 ) 0C x yD D TΣ = − × + + − × + − × − =M k i j j k i k i j

3 3120 120 120 160 57.6 10 144 10 0x yD D T T− + − − + × + × =j i k j i k

Equating to zero the coefficients of the unit vectors:

:k 3120 144 10 0T− + × = (a) 1200 NT =

:i 3120 57.6 10 0 480 Ny yD D+ × = = −

: 120 160(1200 N) 0xD− − =j 1600 NxD = −

0:xFΣ = 0x xC D T+ + = 1600 1200 400 NxC = − =

0:yFΣ = 720 0y yC D+ − = 480 720 1200 NyC = + =

0:zFΣ = 0zC =

(b) (400 N) (1200 N) ; (1600 N) (480 N)= + = − −C i j D i j

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465

PROBLEM 4.105

A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C.

SOLUTION

Free-Body Diagram: Five unknowns and six equations of equilibrium, but equilibrium is maintained ( 0).ACMΣ =

1.2

2.4B

A

==

r k

r k

0.8 0.6 2.4 2.6 m

0.8 1.2 2.4 2.8 m

AD AD

AE AE

= − + − =

= + − =

i j k

i j k

( 0.8 0.6 2.4 )

2.6

(0.8 1.2 2.4 )2.8

ADAD

AEAE

TADT

AD

TAET

AE

= = − + −

= = + −

i j k

i j k

0: ( 3 kN) 0C A AD A AE BMΣ = × + × + × − =r T r T r j

0 0 2.4 0 0 2.4 1.2 ( 3.6 kN) 02.6 2.8

0.8 0.6 2.4 0.8 1.2 2.4

AD AET T+ + × − =

− − −

i j k i j k

k j

Equate coefficients of unit vectors to zero:

: 0.55385 1.02857 4.32 0

: 0.73846 0.68671 0AD AE

AD AE

T T

T T

− − + =− + =

i

j

(1)

0.92857AD AET T= (2)

From Eq. (1): 0.55385(0.92857) 1.02857 4.32 0AE AET T− − + =

1.54286 4.32

2.800 kNAE

AE

T

T

== 2.80 kNAET =

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466


From Eq. (2): 0.92857(2.80) 2.600 kNADT = = 2.60 kNADT =

0.8 0.80: (2.6 kN) (2.8 kN) 0 0

2.6 2.80.6 1.2

0: (2.6 kN) (2.8 kN) (3.6 kN) 0 1.800 kN2.6 2.82.4 2.4

0: (2.6 kN) (2.8 kN) 0 4.80 kN2.6 2.8

x x x

y y y

z z z

F C C

F C C

F C C

Σ = − + = =

Σ = + + − = =

Σ = − − = =

(1.800 kN) (4.80 kN)= +C j k

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469

PROBLEM 4.107

A 10-ft boom is acted upon by the 840-lb force shown. Determine the tension in each cable and the reaction at the ball-and-socket joint at A.

SOLUTION

We have five unknowns and six equations of equilibrium, but equilibrium is maintained ( 0).xMΣ =

Free-Body Diagram:

( 6 ft) (7 ft) (6 ft) 11 ft

( 6 ft) (7 ft) (6 ft) 11 ft

( 6 7 6 )11

( 6 7 6 )11

BDBD BD

BEBE BE

BD BD

BE BE

TBDT T

BD

TBET T

BE

= − + + =

= − + − =

= = − + +

= = − + −

i j k

i j k

i j k

i j k

0: ( 840 ) 0A B BD B BE CM T TΣ = × + × + × − =r r r j

6 ( 6 7 6 ) 6 ( 6 7 6 ) 10 ( 840 ) 011 11BD BET T× − + + + × − + − + × − =i i j k i i j k i j

42 36 42 36

8400 011 11 11 11BD BD BE BET T T T− + + − =k j k j k

Equate coefficients of unit vectors to zero:

36 36

: 011 11BD BE BE BDT T T T− + = =i

42 42

: 8400 011 11BD BET T+ − =k

42

2 840011 BDT

=

1100 lbBDT =

1100 lbBET =

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6 6

0: (1100 lb) (1100 lb) 011 11x xF AΣ = − − =

1200 lbxA =

7 7

0: (1100 lb) (1100 lb) 840 lb 011 11y yF AΣ = + + − =

560 lbyA = −

6 6

0: (1100 lb) (1100 lb) 011 11z zF AΣ = + − =

0zA = (1200 lb) (560 lb)= −A i j

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Chapter 5 Distributed Forces

Centroids and Center of Gravity

5.1 INTRODUCTION We have assumed so far that the attraction exerted by the earth on a rigid body could be represented by a single force W. This force, called the force of gravity or the weight of the body, was to be applied at the center of gravity of the body. Actually, the earth exerts a force on each of the particles forming the body. The action of the earth on a rigid body should thus be represented by a large number of small forces distributed over the entire body. You will learn in this chapter, however, that all of these small forces can be replaced by a single equivalent force W. You will also learn how to determine the center of gravity, i.e., the point of application of the resultant W, for bodies of various shapes. In the first part of the chapter, two-dimensional bodies, such as flat plates and wires contained in a given plane, are considered. Two concepts closely associated with the determination of the center of gravity of a plate or a wire are introduced: the concept of the centroid of an area or a line and the concept of the first moment of an area or a line with respect to a given axis. You will also learn that the computation of the area of a surface of revolution or of the volume of a body of revolution is directly related to the determination of the centroid of the line or area used to generate that surface or body of revolution (theorems of Pappus-Guldinus). The determination of the centroid of an area simplifies the analysis of beams subjected to distributed loads and the computation of the forces exerted on submerged rectangular surfaces, such as hydraulic gates and portions of dams. In the last part of the chapter, you will learn how to determine the center of gravity of a three-dimensional body as well as the centroid of a volume and the first moments of that volume with respect to the coordinate planes.

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5.2 CENTER OF GRAVITY OF A TWO-DIMENSIONAL BODY Let us first consider a flat horizontal plate (Fig. 5.1). We can divide the plate into n

small elements. The coordinates of the first element are denoted by x1 and y1, those

of the second element by x2 and y2, etc. The forces exerted by the earth on the

elements of the plate will be denoted, respectively, by ΔW1, ΔW2, . . . , ΔWn

�𝑭𝑭𝒛𝒛 = 𝑾𝑾 = ∆𝑾𝑾𝟏𝟏 + ∆𝑾𝑾𝟐𝟐 … … . .∆𝑾𝑾𝒏𝒏

.

These forces or weights are directed toward the center of the earth; however, for all

practical purposes they can be assumed to be parallel. Their resultant is therefore a

single force in the same direction. The magnitude W of this force is obtained by

adding the magnitudes of the elemental weights.

To obtain the coordinates 𝐗𝐗� and 𝐘𝐘� of the point G where the resultant W should be applied, we write that the moments of W about the y and x axes are equal to the sum of the corresponding moments of the elemental weights,

�𝑴𝑴𝒙𝒙 = 𝑿𝑿�𝑾𝑾 = 𝒙𝒙𝟏𝟏∆𝑾𝑾𝟏𝟏 + 𝒙𝒙𝟐𝟐∆𝑾𝑾𝟐𝟐 … … . .𝒙𝒙𝒏𝒏∆𝑾𝑾𝒏𝒏

�𝑴𝑴𝒚𝒚 = 𝒀𝒀�𝑾𝑾 = 𝒚𝒚𝟏𝟏∆𝑾𝑾𝟏𝟏 + 𝒚𝒚𝟐𝟐∆𝑾𝑾𝟐𝟐 … … . .𝒚𝒚𝒏𝒏∆𝑾𝑾𝒏𝒏

Fig. 5.1 Center of Gravity of the Plate

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553

PROBLEM 5.1

Locate the centroid of the plane area shown.

SOLUTION

2, inA , inx , iny 3, inxA 3, inyA

1 8 0.5 4 4 32

2 3 2.5 2.5 7.5 7.5

Σ 11 11.5 39.5

X A x AΣ =

2 3(11in ) 11.5 inX = 1.045 in.X =

Y A y AΣ = Σ

(11) 39.5Y = 3.59 in.Y =

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554

PROBLEM 5.2


SOLUTION

For the area as a whole, it can be concluded by observation that

2

(72 mm)3

Y = or 48.0 mmY =

Dimensions in mm

2, mmA , mmx 3, mmxA

1 1

30 72 10802

× × = 20 21,600

2 1

48 72 17282

× × = 46 79,488

Σ 2808 101,088

Then X A x A= Σ (2808) 101,088X = or 36.0 mmX =

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555

PROBLEM 5.3


SOLUTION

2, mmA , mmx , mmy 3, mmxA 3, mmyA

1 126 54 6804× = 9 27 61,236 183,708

2 1

126 30 18902

× × = 30 64 56,700 120,960

3 1

72 48 17282

× × = 48 16− 82,944 27,648−

Σ 10,422 200,880 277,020

Then X A xAΣ = Σ

2 2(10,422 m ) 200,880 mmX = or 19.27 mmX =

and Y A yAΣ = Σ

2 3(10,422 m ) 270,020 mmY = or 26.6 mmY =

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556

PROBLEM 5.4


SOLUTION

2, inA , inx , iny 3, inxA 3, inyA

1 1

(12)(6) 362

= 4 4 144 144

2 (6)(3) 18= 9 7.5 162 135

Σ 54 306 279

Then XA xA= Σ

(54) 306X = 5.67 in.X =

(54) 279

YA yA

Y

= Σ= 5.17 in.Y =

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557

PROBLEM 5.5


SOLUTION

By symmetry, X Y=

Component 2, inA , in.x 3, inxA

I Quarter circle 2(10) 78.544

π = 4.2441 333.33

II Square 2(5) 25− = − 2.5 62.5−

Σ 53.54 270.83

2 3: (53.54 in ) 270.83 inX A x A XΣ = Σ =

5.0585 in.X = 5.06 in.X Y= =

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558

PROBLEM 5.6


SOLUTION

2, inA , in.x , in.y 3, inxA 3, inyA

1 14 20 280× = 7 10 1960 2800

2 2(4) 16π π− = − 6 12 –301.59 –603.19

Σ 229.73 1658.41 2196.8

Then 1658.41

229.73

xAX

A

Σ= =Σ

7.22 in.X =

2196.8

229.73

yAY

A

Σ= =Σ

9.56 in.Y =

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559

PROBLEM 5.7


SOLUTION

By symmetry, 0X =

Component 2, inA , in.y 3, inyA

I Rectangle (3)(6) 18= 1.5 27.0

II Semicircle 2(2) 6.282

π− = − 2.151 13.51−

Σ 11.72 13.49

2 3(11.72 in. ) 13.49 in

1.151 in.

Y A y A

Y

Y

Σ = Σ

==

0X =

1.151in.Y =

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560

PROBLEM 5.8


SOLUTION

2, mmA , mmx , mmy 3, mmx A 3, mmyA

1 (60)(120) 7200= –30 60 3216 10− × 3432 10×

2 2(60) 2827.44

π = 25.465 95.435 372.000 10× 3269.83 10×

3 2(60) 2827.44

π− = − –25.465 25.465 372.000 10× 372.000 10− ×

Σ 7200 372.000 10− × 3629.83 10×

Then 3(7200) 72.000 10XA x A X= Σ = − × 10.00 mmX = −

3(7200) 629.83 10YA y A Y= Σ = × 87.5 mmY =

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561

PROBLEM 5.9


SOLUTION


1 1

(120)(75) 45002

= 80 25 3360 10× 3112.5 10×

2 (75)(75) 5625= 157.5 37.5 3885.94 10× 3210.94 10×

3 2(75) 4417.94

π− = − 163.169 43.169 3720.86 10− × 3190.716 10− ×

Σ 5707.1 3525.08 10× 3132.724 10×

Then 3(5707.1) 525.08 10XA x A X= Σ = × 92.0 mmX =

3(5707.1) 132.724 10YA y A Y= Σ = × 23.3 mmY =

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562

PROBLEM 5.10


SOLUTION

First note that symmetry implies 0X =

2, inA , in.y 3, inyA

1 2(8)

100.5312

π− = − 3.3953 –341.33

2 2(12)

226.192

π = 5.0930 1151.99

Σ 125.659 810.66

Then 3

2

810.66 in

125.66 in

y AY

A

Σ= =Σ

or 6.45 in.Y =

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563

PROBLEM 5.11


SOLUTION

First note that symmetry implies 0X =

2, mA , my 3, myA

1 4

4.5 3 183

× × = 1.2 21.6

2 2(1.8) 5.08942

π− = − 0.76394 −3.8880

Σ 12.9106 17.7120

Then 3

2

17.7120 m

12.9106 m

y AY

A

Σ= =Σ

or 1.372 mY =

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564

PROBLEM 5.12


SOLUTION

Area mm2 , mmx , mmy 3, mmxA 3, mmyA

1 31(200)(480) 32 10

3= × 360 60

611.52 10× 61.92 10×

2 31(50)(240) 4 10

3− = × 180 15

60.72 10− × 60.06 10− ×

Σ 328 10× 610.80 10× 61.86 10×

:X A x AΣ = Σ 3 2 6 3(28 10 mm ) 10.80 10 mmX × = ×

385.7 mmX = 386 mmX =

:Y A y AΣ = Σ 3 2 6 3(28 10 mm ) 1.86 10 mmY × = ×

66.43 mmY = 66.4 mmY =

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565

PROBLEM 5.13


SOLUTION


1 (15)(80) 1200= 40 7.5 348 10× 39 10×

2 1

(50)(80) 1333.333

= 60 30 380 10× 340 10×

Σ 2533.3 3128 10× 349 10×

Then XA xA= Σ

3(2533.3) 128 10X = × 50.5 mmX =

3(2533.3) 49 10

YA yA

Y

= Σ

= × 19.34 mmY =

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566

PROBLEM 5.14


SOLUTION

Dimensions in in.

2, inA , in.x , in.y 3, inxA 3, inyA

1 2

(4)(8) 21.3333

= 4.8 1.5 102.398 32.000

2 1

(4)(8) 16.00002

− = − 5.3333 1.33333 85.333 −21.333

Σ 5.3333 17.0650 10.6670

Then X A xAΣ = Σ

2 3(5.3333 in ) 17.0650 inX = or 3.20 in.X =

and Y A yAΣ = Σ

2 3(5.3333 in ) 10.6670 inY = or 2.00 in.Y =

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567

PROBLEM 5.15


SOLUTION

Dimensions in mm


1 47 26 1919.512

π × × = 0 11.0347 0 21,181

2 1

94 70 32902

× × = −15.6667 −23.333 −51,543 −76,766

Σ 5209.5 −51,543 −55,584

Then 51,543

5209.5

x AX

A

Σ −= =Σ

9.89 mmX = −

55,584

5209.5

y AY

A

Σ −= =Σ

10.67 mmY = −

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568

PROBLEM 5.16

Determine the x coordinate of the centroid of the trapezoid shown in terms of h1, h2, and a.

SOLUTION

A x xA

1 11

2h a

1

3a 2

11

6h a

2 21

2h a

2

3a 2

22

6h a

Σ 1 21

( )2

a h h+ 2

1 21

( 2 )6

a h h+

21

1 261

1 22

( 2 )

( )

a h hx AX

A a h h

+Σ= =Σ +

1 2

1 2

21

3

h hX a

h h

+=

+

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631

PROBLEM 5.67

For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports.

SOLUTION

I

II

I II

1(150lb/ft)(9 ft) 675 lb

21

(120 lb/ft)(9ft) 540lb2

675 540 1215 lb

: (1215) (3)(675) (6)(540) 4.3333 ft

R

R

R R R

XR x R X X

= =

= =

= + = + =

= Σ = + =

(a) 1215 lb=R 4.33 ftX =

(b) Reactions: 0: (9 ft) (1215 lb) (4.3333 ft) 0AM BΣ = − =

585.00 lbB = 585 lb=B

0: 585.00 lb 1215 lb 0yF AΣ = + − =

630.00 lbA = 630 lb=A

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632

PROBLEM 5.68

Determine the reactions at the beam supports for the given loading.

SOLUTION

First replace the given loading by the loadings shown below. Both loading are equivalent since they are both defined by a linear relation between load and distance and have the same values at the end points.

1

2

1(900 N/m)(1.5 m) 675 N

21

(400 N/m)(1.5 m) 300 N2

R

R

= =

= =

0: (675 N)(1.4 m) (300 N)(0.9 m) (2.5 m)AM B CΣ = − + + =

270 N=B 270 N=B

0: 675 N 300 N 270 N 0yF AΣ = − + + =

105.0 N=A 105.0 N=A

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633

PROBLEM 5.69


SOLUTION

I

I

II

II

(200 lb/ft)(15 ft)

3000 lb

1(200 lb/ft)(6 ft)

2600 lb

R

R

R

R

==

=

=

0: (3000 lb)(1.5 ft) (600 lb)(9 ft 2ft) (15 ft) 0AM BΣ = − − + + =

740 lbB = 740 lb=B

0: 740 lb 3000 lb 600 lb 0yF AΣ = + − − =

2860 lbA = 2860 lb=A

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634

PROBLEM 5.70


SOLUTION

I

II

(200 lb/ft)(4 ft) 800 lb

1(150 lb/ft)(3 ft) 225 lb

2

R

R

= =

= =

0: 800 lb 225 lb 0yF AΣ = − + =

575 lb=A

0: (800 lb)(2 ft) (225 lb)(5 ft) 0A AM MΣ = − + =

475 lb ftA = ⋅M

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635

PROBLEM 5.71


SOLUTION

I

II

1(4 kN/m)(6 m)

212 kN

(2 kN/m)(10m)

20 kN

R

R

=

===

0: 12 kN 20 kN 0yF AΣ = − − =

32.0 kN=A

0: (12 kN)(2 m) (20 kN)(5 m) 0A AM MΣ = − − =

124.0 kN mA = ⋅M

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636

PROBLEM 5.72


SOLUTION

First replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a parabolic relation between load and distance and the values at the end points are the same.

We have I

II

(6 m)(300 N/m) 1800 N

2(6 m)(1200 N/m) 4800 N

3

R

R

= =

= =

Then 0: 0x xF AΣ = =

0: 1800 N 4800 N 0y yF AΣ = + − =

or 3000 NyA = 3.00 kN=A

15

0: (3 m)(1800 N) m (4800 N) 04A AM M

Σ = + − =

or 12.6 kN mAM = ⋅ 12.60 kN m= ⋅AM

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637

PROBLEM 5.73


SOLUTION

We have I

II

1(12 ft)(200 lb/ft) 800 lb

31

(6 ft)(100 lb/ft) 200 lb3

R

R

= =

= =

Then 0: 0x xF AΣ = =

0: 800 lb 200 lb 0y yF AΣ = − − =

or 1000 lbyA = 1000 lb=A

0: (3 ft)(800 lb) (16.5 ft)(200 lb) 0A AM MΣ = − − =

or 5700 lb ftAM = ⋅ 5700 lb ftA = ⋅M

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638

PROBLEM 5.74

Determine the reactions at the beam supports for the given loading when wO = 400 lb/ft.

SOLUTION

I

II

1 1(12 ft) (400 lb/ft)(12 ft) 2400 lb

2 21

(300 lb/ft)(12 ft) 1800 lb2

OR w

R

= = =

= =

0: (2400 lb)(1 ft) (1800 lb)(3 ft) (7 ft) 0Σ = − + =BM C

428.57 lb=C 429 lb=C

0: 428.57 lb 2400 lb 1800 lb 0Σ = + − − =yF B

3771 lb=B 3770 lb=B

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639

PROBLEM 5.75

Determine (a) the distributed load wO at the end A of the beam ABC for which the reaction at C is zero, (b) the corresponding reaction at B.

SOLUTION

For ,Ow I

II

1(12 ft) 6

21

(300 lb/ft)(12 ft) 1800 lb2

O OR w w

R

= =

= =

(a) For 0,C = 0: (6 )(1 ft) (1800 lb)(3 ft) 0B OM wΣ = − = 900 lb/ftOw =

(b) Corresponding value of I :R

I 6(900) 5400 lb= =R

0: 5400 lb 1800 lb 0Σ = − − =yF B 7200 lb=B

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640

PROBLEM 5.76

Determine (a) the distance a so that the vertical reactions at supports A and B are equal, (b) the corresponding reactions at the supports.

SOLUTION

(a)

We have I

II

1( m)(1800 N/m) 900 N

21

[(4 ) m](600 N/m) 300(4 ) N2

R a a

R a a

= =

= − = −

Then 0: 900 300(4 ) 0y y yF A a a BΣ = − − − + =

or 1200 600y yA B a+ = +

Now 600 300 (N)y y y yA B A B a= = = + (1)

Also, 0: (4 m) 4 m [(900 ) N]3B ya

M A a Σ = − + −

1

(4 ) m [300(4 ) N] 03

a a + − − =

or 2400 700 50yA a a= + − (2)

Equating Eqs. (1) and (2), 2600 300 400 700 50a a a+ = + −

or 2 8 4 0a a− + =

Then 28 ( 8) 4(1)(4)

2a

± − −=

or 0.53590 ma = 7.4641 ma =

Now 4 ma ≤ 0.536 ma =

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641


(b) We have 0: 0x xF AΣ = =

From Eq. (1):

600 300(0.53590)

y yA B=

= +

761 N= 761 N= =A B

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642

PROBLEM 5.77

Determine (a) the distance a so that the reaction at support B is minimum, (b) the corresponding reactions at the supports.

SOLUTION

(a)

We have I

II

1( m)(1800 N/m) 900 N

21

[(4 )m](600 N/m) 300(4 ) N2

R a a

R a a

= =

= − = −

Then 8

0: m (900 N) m [300(4 )N] (4 m) 03 3A ya a

M a a B+ Σ = − − − + =

or 250 100 800yB a a= − + (1)

Then 100 100 0ydBa

da= − = or 1.000 ma =

(b) From Eq. (1): 250(1) 100(1) 800 750 NyB = − + = 750 N=B

and 0: 0x xF AΣ = =

0: 900(1) N 300(4 1) N 750 N 0y yF AΣ = − − − + =

or 1050 NyA = 1050 N=A

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643

PROBLEM 5.78

A beam is subjected to a linearly distributed downward load and rests on two wide supports BC and DE, which exert uniformly distributed upward loads as shown. Determine the values of wBC and wDE corresponding to equilibrium when 600Aw = N/m.

SOLUTION

We have I

II

1(6 m)(600 N/m) 1800 N

21

(6 m)(1200 N/m) 3600 N2(0.8 m)( N/m) (0.8 ) N

(1.0 m)( N/m) ( ) NBC BC BC

DE DE DE

R

R

R w w

R w w

= =

= =

= == =

Then 0: (1 m)(1800 N) (3 m)(3600 N) (4 m)( N) 0G DEM wΣ = − − + =

or 3150 N/mDEw =

and 0: (0.8 ) N 1800 N 3600 N 3150 N 0y BCF wΣ = − − + =

or 2812.5 N/mBCw = 2810 N/mBCw =

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644

PROBLEM 5.79

A beam is subjected to a linearly distributed downward load and rests on two wide supports BC and DE, which exert uniformly distributed upward loads as shown. Determine (a) the value of wA so that wBC = wDE, (b) the corresponding values of wBC and wDE.

SOLUTION

(a)

We have I

II

1(6 m)( N/m) (3 ) N

21

(6 m)(1200 N/m) 3600 N2(0.8 m)( N/m) (0.8 ) N

(1 m)( N/m) ( ) N

A A

BC BC BC

DE DE DE

R w w

R

R w w

R w w

= ⋅

= =

= == =

Then 0: (0.8 ) N (3 ) N 3600 N ( ) N 0y BC A DEF w w wΣ = − − + =

or 0.8 3600 3BC DE Aw w w+ = +

Now 5

20003BC DE BC DE Aw w w w w= = = + (1)

Also, 0: (1 m)(3 N) (3 m)(3600 N) (4 m)( N) 0G A DEM w wΣ = − − + =

or 3

27004DE Aw w= + (2)

Equating Eqs. (1) and (2),

5 3

2000 27003 4A Aw w+ = +

or 8400

N/m11Aw = 764 N/mAw =

(b) Eq. (1) 5 8400

20003 11BC DEw w = = +

or 3270 N/mBC DEw w= =

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645

PROBLEM 5.80

The cross section of a concrete dam is as shown. For a 1-m-wide dam section, determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam.

SOLUTION

(a) Consider free body made of dam and section BDE of water. (Thickness = 1 m)

3 2(3 m)(10 kg/m )(9.81 m/s )p =

3 3 21

3 3 22

3 3 23

3 3 2

(1.5 m)(4 m)(1 m)(2.4 10 kg/m )(9.81 m/s ) 144.26 kN

1(2 m)(3 m)(1 m)(2.4 10 kg/m )(9.81 m/s ) 47.09 kN

32

(2 m)(3 m)(1 m)(10 kg/m )(9.81 m/s ) 39.24 kN31 1

(3 m)(1 m)(3 m)(10 kg/m )(9.81 m/s ) 44.145 kN2 2

= × =

= × =

= =

= = =

W

W

W

P Ap

0: 44.145 kN 0Σ = − =xF H

44.145 kN=H 44.1 kN=H

0: 141.26 47.09 39.24 0Σ = − − − =yF V

227.6 kN=V 228 kN=V

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646


1

2

3

1(1.5 m) 0.75 m

21

1.5 m (2 m) 2 m45

1.5 m (2 m) 2.75 m8

x

x

x

= =

= + =

= + =

0: (1 m) 0Σ = − Σ + =AM xV xW P

(227.6 kN) (141.26 kN)(0.75 m) (47.09 kN)(2 m)

(39.24 kN)(2.75 m) (44.145 kN)(1 m) 0

(227.6 kN) 105.9 94.2 107.9 44.145 0

(227.6) 263.9 0

x

x

x

− −− + =− − − + =− =

1.159 mx = (to right of A)

(b) Resultant of face BC:

Consider free body of section BDE of water.

59.1 kN− =R 41.6°

59.1 kN=R 41.6°

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647

PROBLEM 5.81

The cross section of a concrete dam is as shown. For a 1-m-wide dam section, determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam.

SOLUTION

(a) Consider free body made of dam and triangular section of water shown. (Thickness = 1 m.)

3 3 2(7.2m)(10 kg/m )(9.81m/s )=p

3 3 21

3 3 22

3 3 23

3 3 2

2(4.8 m)(7.2 m)(1 m)(2.4 10 kg/m )(9.81 m/s )

3542.5 kN

1(2.4 m)(7.2 m)(1 m)(2.4 10 kg/m )(9.81 m/s )

2203.4 kN1

(2.4 m)(7.2 m)(1 m)(10 kg/m )(9.81 m/s )284.8 kN

1 1(7.2 m)(1 m)(7.2 m)(10 kg/m )(9.81 m/s )

2 2

= ×

=

= ×

=

=

=

= =

=

W

W

W

P Ap

254.3 kN

0: 254.3 kN 0xF HΣ = − = 254 kN=H

0: 542.5 203.4 84.8 0yF VΣ = − − − =

830.7 kNV = 831 kN=V

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648


(b) 1

2

3

5(4.8 m) 3 m

81

4.8 (2.4) 5.6 m32

4.8 (2.4) 6.4 m3

x

x

x

= =

= + =

= + =

0: (2.4 m) 0AM xV xW PΣ = − Σ + =

(830.7 kN) (3 m)(542.5 kN) (5.6 m)(203.4 kN)

(6.4 m)(84.8 kN) (2.4 m)(254.3 kN) 0

(830.7) 1627.5 1139.0 542.7 610.3 0

(830.7) 2698.9 0

x

x

x

− −− + =

− − − + =− =

3.25 mx = (to right of A)

(c) Resultant on face BC:

Direct computation:

3 3 2

2

2 2

2

(10 kg/m )(9.81 m/s )(7.2 m)

70.63 kN/m

(2.4) (7.2)

7.589 m

18.43

1

21

(70.63 kN/m )(7.589 m)(1 m)2

P gh

P

BC

R PA

ρ

θ

= =

=

= +== °

=

= 268 kN=R 18.43°

Alternate computation: Use free body of water section BCD.

268 kN− =R 18.43°

268 kN=R 18.43°

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Chapter 6 Analysis of Structures

Simple Truss (Pin Joints & Two-Forces Members)

6.1 INTRODUCTION The problems considered in the preceding chapters concerned the equilibrium of a single rigid body, and all forces involved were external to the rigid body. We now consider problems dealing with the equilibrium of structures made of several connected parts. These problems call for the determination not only of the external forces acting on the structure but also of the forces which hold together the various parts of the structure. From the point of view of the structure as a whole, these forces are internal forces. Consider, for example, the crane shown in Fig. 6.1a, which carries a load W. The crane consists of three beams AD, CF, and BE connected by frictionless pins; it is supported by a pin at A and by a cable DG. The free-body diagram of the crane has been drawn in Fig. 6.1b. The external forces, which are shown in the diagram, include the weight W, the two components Ax and Ay

of the reaction at A, and the force T exerted by the cable at D. The internal forces holding the various parts of the crane together do not appear in the diagram. If, however, the crane is dismembered and if a free-body diagram is drawn for each of its component parts, the forces holding the three beams together will also be represented, since these forces are external forces from the point of view of each component part (Fig. 6.1c).

Fig. 6.1 Simple Truss

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6.2 ANALYSIS OF TRUSSES BY THE METHOD OF JOINTS The simple truss can be considered as a group of pins and two-force members. The truss shown, whose free-body diagram is shown in Fig. 6.2a, can thus be dismembered, and a free-body diagram can be drawn for each pin and each member (Fig. 6.2b). Each member is acted upon by two forces, one at each end; these forces have the same magnitude, same line of action, and opposite sense. Furthermore, Newton’s third law indicates that the forces of action and reaction between a member and a pin are equal and opposite. Therefore, the forces exerted by a member on the two pins it connects must be directed along that member and be equal and opposite. The common magnitude of the forces exerted by a member on the two pins it connects is commonly referred to as the force in the member considered, even though this quantity is actually a scalar. Since the lines of action of all the internal forces in a truss are known, the analysis of a truss reduces to computing the forces in its various members and to determining whether each of its members is in tension or in compression.

Fig. 6.2 Free Diagram of a Simple Truss

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Forces on each joint can be illustrated through the Free-Body Diagram as it is shown in the following figure Fig. 6.3, consequently the unknown forces can be obtained from equalizing the summation of all forces on a specified joint in both direction X, Y to zero.

Fig. 6.3 Free Body Diagram of Truss Joints

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PROBLEM 6.1

Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.

SOLUTION

Free body: Entire truss:

0: 0 0y y yF BΣ = = =B

0: (3.2 m) (48 kN)(7.2 m) 0C xM BΣ = − − =

108 kN 108 kNx xB = − =B

0: 108 kN 48 kN 0xF CΣ = − + =

60 kNC = 60 kN=C

Free body: Joint B:

108 kN

5 4 3BCAB FF = =

180.0 kNABF T=

144.0 kNBCF T=

Free body: Joint C:

60 kN

13 12 5AC BCF F

= =

156.0 kNACF C=

144 kN (checks)BCF =

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%%Problem 6.1 How to calculate all the forces in each member as shown in %%figure using the technique of FUNCTIONS as the following:- %%Calculating the reactions acting on the entire truss from FBD %%As the summation of moments about any point = Zero, then %%Taking the moment about point B to calculate the reaction acting on %%joint C as the following:- Cx = (48*4)/3.2 %%As the summation of forces = Zero in any direction,thus By = 0, %%calculating Bx as following:- Bx = Cx + 48 %%Calculating the FBD at joint B %%Three Balanced Forces theta1 = atan(4/3) theta2 = (pi/2) theta3 = atan(3/4) %%F the force between theta1 & theta2 F1 = abs(Bx) [F2,F3] = TBF(theta1, theta2, theta3, F1); Fbc = F2 Fab = F3 %%Calculating the FBD at joint C %%Three Balanced Forces theta1 = atan(7.2/3) theta2 = (pi/2) theta3 = atan(3/7.2) %%F the force between theta1 & theta2 F1 = abs(Cx) [F2,F3] = TBF(theta1, theta2, theta3, F1); Fcb = F2 Fca = F3

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%%Problem 6.1 How to calculate all the forces in each member as shown in %%figure using the technique of FUNCTIONS as the following:- %%Three Balanced Forces function [F2,F3] = TBF(theta1, theta2, theta3, F1) F2 = (F1/sin(theta3))*sin(theta1); F3 = (F1/sin(theta3))*sin(theta2);

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744

PROBLEM 6.2


SOLUTION

Reactions:

0: 1260 lbAMΣ = =C

0: 0x xFΣ = =A

0: 960 lby yFΣ = =A

Joint B:

300 lb

12 13 5BCAB FF = =

720 lbABF T=

780 lbBCF C=

Joint A:

4

0: 960 lb 05y ACF FΣ = − − =

1200 lbACF = 1200 lbACF C=

3

0: 720 lb (1200 lb) 0 (checks)5xFΣ = − =

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%%Problem 6.2 How to calculate all the forces in each member as shown in %%figure using the technique of FUNCTIONS as the following:- %%Calculating the reactions acting on the entire truss from FBD %%As the summation of moments about any point = Zero, then %%Taking the moment about point A to calculate the reaction acting on %%joint C as the following:- Cy = (300*63)/15 %%As the summation of forces = Zero in any direction,thus Ax = 0, %%calculating Ay as following:- Ay = 300 - Cy %%Calculating the FBD at joint A %%Three Balanced Forces theta1 = (pi/2) theta2 = atan(15/20) theta3 = atan(20/15) %%F the force between theta1 & theta2 F1 = abs(Ay) [F2,F3] = TBF(theta1, theta2, theta3, F1); Fac = F2 Fab = F3 %%Calculating the FBD at joint C %%Three Balanced Forces theta1 = atan(48/20) theta2 = atan(15/20) theta3 = pi - theta1 - theta2 %%F the force between theta1 & theta2 F1 = abs(Cy) [F2,F3] = TBF(theta1, theta2, theta3, F1); Fac = F2 Fcb = F3

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%%Problem 6.1 How to calculate all the forces in each member as shown in %%figure using the technique of FUNCTIONS as the following:- %%Three Balanced Forces function [F2,F3] = TBF(theta1, theta2, theta3, F1) F2 = (F1/sin(theta3))*sin(theta1); F3 = (F1/sin(theta3))*sin(theta2);

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745

PROBLEM 6.3


SOLUTION

2 2

2 2

3 1.25 3.25 m

3 4 5 m

AB

BC

= + =

= + =

Reactions:

0: (84 kN)(3 m) (5.25 m) 0AM CΣ = − =

48 kN=C

0: 0x xF A CΣ = − =

48 kNx =A

0: 84 kN 0y yF AΣ = = =

84 kNy =A

Joint A:

12

0: 48 kN 013x ABF FΣ = − =

52 kNABF = + 52.0 kNABF T=

5

0: 84 kN (52 kN) 013y ACF FΣ = − − =

64.0 kNACF = + 64.0 kNACF T=

Joint C:

48 kN

5 3BCF

= 80.0 kNBCF C=

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746

PROBLEM 6.4


SOLUTION

Free body: Truss:

From the symmetry of the truss and loading, we find 600 lb= =C D

Free body: Joint B: 300 lb

2 15BCAB FF

= =

671 lbABF T= 600 lbBCF C=

Free body: Joint C:

3

0: 600 lb 05y ACF FΣ = + =

1000 lbACF = − 1000 lbACF C=

4

0: ( 1000 lb) 600 lb 05x CDF FΣ = − + + = 200 lbCDF T=

From symmetry:

1000 lb , 671 lb ,AD AC AE ABF F C F F T= = = = 600 lbDE BCF F C= =

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747

PROBLEM 6.5


SOLUTION

Reactions:

0: (24) (4 2.4)(12) (1)(24) 0D yM FΣ = − + − =

4.2 kipsy =F

0: 0x xFΣ = =F

0: (1 4 1 2.4) 4.2 0yF DΣ = − + + + + =

4.2 kips=D

Joint A:

0: 0x ABF FΣ = = 0ABF =

0 : 1 0y ADF FΣ = − − =

1 kipADF = − 1.000 kipADF C=

Joint D: 8

0: 1 4.2 017y BDF FΣ = − + + =

6.8 kipsBDF = − 6.80 kipsBDF C=

15

0: ( 6.8) 017x DEF FΣ = − + =

6 kipsDEF = + 6.00 kipsDEF T=

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Joint E:

0 : 2.4 0y BEF FΣ = − =

2.4 kipsBEF = + 2.40 kipsBEF T=

Truss and loading symmetrical about c .L

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%%Problem 6.5 How to calculate all the forces in each member as shown in %%figure using the technique of FUNCTIONS as the following:- %%Calculating the reactions acting on the entire truss from FBD %%As the summation of moments about any point = Zero, then %%Taking the moment about point F to calculate the reaction acting on %%joint F as the following:- Dy = (1*24 + 4*12 + 2.4*12)/24 %%As the summation of forces = Zero in any direction,thus Fx = 0, %%calculating Fy as following:- Fy = 1 + 4 +1 +2.4 - Dy %%Calculating the FBD at joint D %%Three Balanced Forces theta1 = (pi/2) theta2 = (pi*3/2) theta3 = atan(6.4/12) theta4 = (0) %%theta1, theta2, theta3, theta4 are related to F1, F2, F3, F4 %%respectively. F1 = Dy F2 = 1 [F3,F4] = FBF(theta1, theta2, theta3, theta4, F1, F2); Fdb = F3 Fde = F4 %%Calculating the FBD at joint B %%Three Balanced Forces theta1 = atan(6.4/12) theta2 = (pi*3/2) theta3 = 2*pi - atan(6.4/12) theta4 = pi/2 %%theta1, theta2, theta3, theta4 are related to F1, F2, F3, F4 %%respectively. F1 = -Fdb F2 = 4 [F3,F4] = FBF(theta1, theta2, theta3, theta4, F1, F2); Fbf = F3 Fbe = F4

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%%Problem 6.5 How to calculate all the forces in each member as shown in %%figure using the technique of FUNCTIONS as the following:- %%Four Balanced Forces %%F1 F2 F3 F4 are four forces acting on a single joint %%theta1, theta2, theta3, theta4 are related to F1, F2, F3, F4 %%respectively. function [F3,F4] = FBF(theta1, theta2, theta3, theta4, F1, F2) A = [cos(theta3) cos(theta4);sin(theta3) sin(theta4)]; C = [-(F1*cos(theta1)+F2*cos(theta2)) -(F1*sin(theta1)+F2*sin(theta2))]' F = inv(A)*C F3 = F(1) F4 = F(2)

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749

PROBLEM 6.6


SOLUTION

2 2

2 2

5 12 13 ft

12 16 20 ft

AD

BCD

= + =

= + =

Reactions: 0: 0x xF DΣ = =

0: (21 ft) (693 lb)(5 ft) 0E yM DΣ = − = 165 lby =D

0: 165 lb 693 lb 0yF EΣ = − + = 528 lb=E

Joint D: 5 4

0: 013 5x AD DCF F FΣ = + = (1)

12 3

0: 165 lb 013 5y AD DCF F FΣ = + + = (2)

Solving Eqs. (1) and (2) simultaneously,

260 lbADF = − 260 lbADF C=

125 lbDCF = + 125 lbDCF T=

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750


Joint E:

5 4

0: 013 5x BE CEF F FΣ = + = (3)

12 3

0: 528 lb 013 5y BE CEF F FΣ = + + = (4)

Solving Eqs. (3) and (4) simultaneously,

832 lbBEF = − 832 lbBEF C=

400 lbCEF = + 400 lbCEF T=

Joint C:

Force polygon is a parallelogram (see Fig. 6.11, p. 209).

400 lbACF T=

125.0 lbBCF T=

Joint A: 5 4

0: (260 lb) (400 lb) 013 5x ABF FΣ = + + =

420 lbABF = − 420 lbABF C=

12 3

0: (260 lb) (400 lb) 013 5

0 0 (Checks)

yFΣ = − =

=

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751

PROBLEM 6.7


SOLUTION


0: 2(5 kN) 0x xF CΣ = + =

10 kN 10 kNxxC = − =C

0: (2 m) (5 kN)(8 m) (5 kN)(4 m) 0CM DΣ = − − =

30 kN 30 kND = + =D

0: 30 kN 0 30 kN 30 kNyy y yF C CΣ = + = = − =C

Free body: Joint A:

5 kN

4 117AB ADF F

= =

20.0 kNABF T=

20.6 kNADF C=

Free body: Joint B:

1

0: 5 kN 05

x BDF FΣ = + =

5 5 kNBDF = − 11.18 kNBDF C=

2

0: 20 kN ( 5 5 kN) 05

y BCF FΣ = − − − =

30 kNBCF = + 30.0 kNBCF T=

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Free body: Joint C:

0: 10 kN 0x CDF FΣ = − =

10 kNCDF = + 10.00 kNCDF T=

0: 30 kN 30 kN 0 (checks)yFΣ = − =

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753

PROBLEM 6.8


SOLUTION

Reactions:

0: 16 kNC xMΣ = =A

0: 9 kNy yFΣ = =A

0: 16 kNxFΣ = =C

Joint E:

3 kN

5 4 3BE DEF F= =

5.00 kNBEF T=

4.00 kNDEF C= Joint B:

4

0: (5 kN) 05x ABF FΣ = − =

4 kNABF = + 4.00 kNABF T=

3

0: 6 kN (5 kN) 05y BDF FΣ = − − − =

9 kNBDF = − 9.00 kNBDF C=

Joint D:

3

0: 9 kN 05y ADF FΣ = − + =

15 kNADF = + 15.00 kNADF T=

4

0: 4 kN (15 kN) 05x CDF FΣ = − − − =

16 kNCDF = − 16.00 kNCDF C=

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754

PROBLEM 6.9

Determine the force in each member of the Gambrel roof truss shown. State whether each member is in tension or compression.

SOLUTION

Free body: Truss:

0: 0x xFΣ = =H

Because of the symmetry of the truss and loading,

1

total load2y= =A H

1200 lby= =A H

Free body: Joint A:

900 lb

5 4 3ACAB FF = = 1500 lbAB C=F

1200 lbAC T=F

Free body: Joint C:

BC is a zero-force member.

0BC =F 1200 lbCEF T=

Free body: Joint B:

24 4 4

0: (1500 lb) 025 5 5x BD BEF F FΣ = + + =

or 24 20 30,000 lbBD BEF F+ = − (1)

7 3 3

0: (1500) 600 025 5 5y BD BEF F FΣ = − + − =

or 7 15 7,500 lbBD BEF F− = − (2)

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755


Multiply Eq. (1) by 3, Eq. (2) by 4, and add:

100 120,000 lbBDF = − 1200 lbBDF C=

Multiply Eq. (1) by 7, Eq. (2) by –24, and add:

500 30,000 lbBEF = − 60.0 lbBEF C=

Free body: Joint D:

24 24

0: (1200 lb) 025 25x DFF FΣ = + =

1200 lbDFF = − 1200 lbDFF C=

7 7

0: (1200 lb) ( 1200 lb) 600 lb 025 25y DEF FΣ = − − − − =

72.0 lbDEF = 72.0 lbDEF T=

Because of the symmetry of the truss and loading, we deduce that

=EF BEF F 60.0 lbEFF C=

EG CEF F= 1200 lbEGF T=

FG BCF F= 0FGF =

FH ABF F= 1500 lbFHF C=

GH ACF F= 1200 lbGHF T=

Note: Compare results with those of Problem 6.11.

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756

PROBLEM 6.10

Determine the force in each member of the Howe roof truss shown. State whether each member is in tension or compression.

SOLUTION

Free body: Truss:

0: 0x xFΣ = =H


1

total load2yA H= =

1200 lby= =A H

Free body: Joint A:

900 lb

5 4 3ACAB FF = = 1500 lbAB C=F

1200 lbAC T=F

Free body: Joint C:

BC is a zero-force member.

0BC =F 1200 lbCE T=F

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Free body: Joint B:

4 4 4

0: (1500 lb) 05 5 5x BD BCF F FΣ = + + =

or 1500 lbBD BEF F+ = − (1)

3 3 3

0: (1500 lb) 600 lb 05 5 5y BD BEF F FΣ = − + − =

or 500 lbBD BEF F− = − (2)

Add Eqs. (1) and (2): 2 2000 lbBDF = − 1000 lbBDF C=

Subtract Eq. (2) from Eq. (1): 2 1000 lbBEF = − 500 lbBEF C=

Free Body: Joint D:

4 4

0: (1000 lb) 05 5x DFF FΣ = + =

1000 lbDFF = − 1000 lbDFF C=

3 3

0: (1000 lb) ( 1000 lb) 600 lb 05 5y DEF FΣ = − − − − =

600 lbDEF = + 600 lbDEF T=

Because of the symmetry of the truss and loading, we deduce that

=EF BEF F 500 lbEFF C=

EG CEF F= 1200 lbEGF T=

FG BCF F= 0FGF =

FH ABF F= 1500 lbFHF C=

GH ACF F= 1200 lbGHF T=

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758

PROBLEM 6.11

Determine the force in each member of the Pratt roof truss shown. State whether each member is in tension or compression.

SOLUTION

Free body: Truss:

0: 0x xF AΣ = =

Due to symmetry of truss and load,

1

total load 21 kN2yA H= = =

Free body: Joint A:

15.3 kN

37 35 12ACAB FF = =

47.175 kN 44.625 kNAB ACF F= = 47.2 kNABF C=

44.6 kNACF T=

Free body: Joint B:

From force polygon: 47.175 kN, 10.5 kNBD BCF F= = 10.50 kNBCF C=

47.2 kNBDF C=

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Free body: Joint C: 3

0: 10.5 05y CDF FΣ = − = 17.50 kNCDF T=

4

0: (17.50) 44.625 05x CEF FΣ = + − =

30.625 kNCEF = 30.6 kNCEF T=

Free body: Joint E: DE is a zero-force member. 0DEF =

Truss and loading symmetrical about .cL

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760

PROBLEM 6.12

Determine the force in each member of the Fink roof truss shown. State whether each member is in tension or compression.

SOLUTION

Free body: Truss:

0: 0x xFΣ = =A


1

total load2y = =A G

6.00 kNy = =A G

Free body: Joint A:

4.50 kN

2.462 2.25 1ACAB FF = =

11.08 kNABF C=

10.125 kNACF = 10.13 kNACF T=

Free body: Joint B:

3 2.25 2.25

0: (11.08 kN) 05 2.462 2.462x BC BDF F FΣ = + + = (1)

4 11.08 kN

0: 3 kN 05 2.462 2.462

BDy BC

FF FΣ = − + + − = (2)

Multiply Eq. (2) by –2.25 and add to Eq. (1):

12

6.75 kN 0 2.81255 BC BCF F+ = = − 2.81 kNBCF C=

Multiply Eq. (1) by 4, Eq. (2) by 3, and add:

12 12

(11.08 kN) 9 kN 02.462 2.462BDF + − =

9.2335 kNBDF = − 9.23 kNBDF C=

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761


Free body: Joint C: 4 4

0: (2.8125 kN) 05 5y CDF FΣ = − =

2.8125 kN,CDF = 2.81 kNCDF T=

3 3

0: 10.125 kN (2.8125 kN) (2.8125 kN) 05 5x CEF FΣ = − + + =

6.7500 kNCEF = + 6.75 kNCEF T=

Because of the symmetry of the truss and loading, we deduce that DE CDF F= 2.81 kNCDF T=

DF BDF F= 9.23 kNDFF C=

EF BCF F= 2.81 kNEFF C=

EG ACF F= 10.13 kNEGF T=

FG ABF F= 11.08 kNFGF C=

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762

PROBLEM 6.13

Using the method of joints, determine the force in each member of the double-pitch roof truss shown. State whether each member is in tension or compression.

SOLUTION

Free body: Truss:

0: (18 m) (2 kN)(4 m) (2 kN)(8 m) (1.75 kN)(12 m)

(1.5 kN)(15 m) (0.75 kN)(18 m) 0AM HΣ = − − −

− − =

4.50 kN=H

0: 0

0: 9 0

9 4.50

x x

y y

y

F A

F A H

A

Σ = =Σ = + − =

= − 4.50 kNy =A

Free body: Joint A:

3.50 kN

2 157.8262 kN

ACAB

AB

FF

F C

= =

= 7.83 kNABF C=

7.00 kNACF T=

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Free body: Joint B:

2 2 1

0: (7.8262 kN) 05 5 2

x BD BCF F FΣ = + + =

or 0.79057 7.8262 kNBD BCF F+ = − (1)

1 1 1

0: (7.8262 kN) 2 kN 05 5 2

y BD BCF F FΣ = + − − =

or 1.58114 3.3541BD BCF F− = − (2)

Multiply Eq. (1) by 2 and add Eq. (2):

3 19.0065

6.3355 kNBD

BD

F

F

= −= − 6.34 kNBDF C=

Subtract Eq. (2) from Eq. (1):

2.37111 4.4721

1.8861 kNBC

BC

F

F

= −= − 1.886 kNBCF C=

Free body: Joint C:

2 1

0: (1.8861 kN) 05 2

y CDF FΣ = − =

1.4911 kNCDF = + 1.491 kNCDF T=

1 1

0: 7.00 kN (1.8861 kN) (1.4911 kN) 02 5

x CEF FΣ = − + + =

5.000 kNCEF = + 5.00 kNCEF T=

Free body: Joint D:

2 1 2 1

0: (6.3355 kN) (1.4911 kN) 05 2 5 5

x DF DEF F FΣ = + + − =

or 0.79057 5.5900 kNDF DEF F+ = − (1)

1 1 1 2

0: (6.3355 kN) (1.4911 kN) 2 kN 05 2 5 5

y DF DEF F FΣ = − + − − =

or 0.79057 1.1188 kNDF DEF F− = − (2)

Add Eqs. (1) and (2): 2 6.7088 kNDFF = −

3.3544 kNDFF = − 3.35 kNDFF C=

Subtract Eq. (2) from Eq. (1): 1.58114 4.4712 kNDEF = −

2.8278 kNDEF = − 2.83 kNDEF C=

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Free body: Joint F:

1 2

0: (3.3544 kN) 02 5

x FGF FΣ = + =

4.243 kNFGF = − 4.24 kNFGF C=

1 1

0: 1.75 kN (3.3544 kN) ( 4.243 kN) 05 2

y EFF FΣ = − − + − − =

2.750 kNEFF = 2.75 kNEFF T=

Free body: Joint G:

1 1 1

0: (4.243 kN) 02 2 2

x GH EGF F FΣ = − + =

or 4.243 kNGH EGF F− = − (1)

1 1 1

0: (4.243 kN) 1.5 kN 02 2 2

y GH EGF F FΣ = − − − − =

or 6.364 kNGH EGF F+ = − (2)

Add Eqs. (1) and (2): 2 10.607GHF = −

5.303GHF = − 5.30 kNGHF C=

Subtract Eq. (1) from Eq. (2): 2 2.121 kNEGF = −

1.0605 kNEGF = − 1.061 kNEGF C=

Free body: Joint H:

3.75 kN

1 1EHF = 3.75 kNEHF T=

We can also write 3.75 kN

12GHF

= 5.30 kN (Checks)GHF C=

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765

PROBLEM 6.14

The truss shown is one of several supporting an advertising panel. Determine the force in each member of the truss for a wind load equivalent to the two forces shown. State whether each member is in tension or compression.

SOLUTION


0: (800 N)(7.5 m) (800 N)(3.75 m) (2 m) 0FM AΣ = + − =

2250 NA = + 2250 N=A

0: 2250 N 0y yF FΣ = + =

2250 N 2250 Ny yF = − =F

0: 800 N 800 N 0x xF FΣ = − − + = 1600 N 1600 Nx xF = + =F

Joint D:

800 N

8 15 17DE BDF F= =

1700 NBDF C=

1500 NDEF T=

Joint A:

2250 N

15 17 8ACAB FF= =

2250 NABF C=

1200 NACF T=

Joint F:

0: 1600 N 0x CFF FΣ = − =

1600 NCFF = + 1600 NCFF T=

0: 2250 N 0y EFF FΣ = − =

2250 NEFF = + 2250 NEFF T=

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766


Joint C: 8

0: 1200 N 1600 N 017x CEF FΣ = − + =

850 NCEF = − 850 NCEF C=

15

0: 017y BC CEF F FΣ = + =

15 15

( 850 N)17 17BC CEF F= − = − −

750 NBCF = + 750 NBCF T=

Joint E: 8

0: 800 N (850 N) 017x BEF FΣ = − − + =

400 NBEF = − 400 NBEF C=

15

0: 1500 N 2250 N (850 N) 017

0 0 (checks)

yFΣ = − + =

=

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Chapter 7 Shear and Bending Moment of Beams

7.1 SHEAR AND BENDING MOMENT IN A BEAM Consider a beam AB subjected to various concentrated and distributed loads (Fig. 7.8a). We propose to determine the shearing force and bending moment at any point of the beam. In the example considered here, the beam is simply supported, but the method used could be applied to any type of statically determinate beam. First we determine the reactions at A and B by choosing the entire beam as a free body (Fig. 7.8b); writing MA = 0 and MB = 0, we obtain, respectively, RB and RA

.

Fig. 7.1 Shear Force and Bending Moment in a Beam Fig. 7.2 Positive Directions of Shear Force and Bending Moment

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7.2 SHEAR AND BENDING-MOMENT DIAGRAMS Now that shear and bending moment have been clearly defined in sense as well as in magnitude, we can easily record their values at any point of a beam by plotting these values against the distance x measured from one end of the beam. The graphs obtained in this way are called, respectively, the shear diagram and the bending moment diagram. As an example, consider a simply supported beam AB of span L subjected to a single concentrated load P applied at its midpoint D (Fig. 7.3a). We first determine the reactions at the supports from the free-body diagram of the entire beam (Fig. 7.3b); we find that the magnitude of each reaction is equal to P/2. Next we cut the beam at a point C between A and D and draw the free-body diagrams of AC and CB (Fig. 7.3c). Assuming that shear and bending moment are positive, we direct the internal forces V and V’ and the internal couples M and M’ as indicated in Fig. 7.2a. Considering the free body AC and writing that the sum of the vertical components and the sum of the moments about C of the forces acting on the free body are zero, we find V = +P/2 and M = +Px/2. Both shear and bending moment are therefore positive; this can be checked by observing that the reaction at A tends to shear off and to bend the beam at C as indicated in Fig. 7.2b and c. We can plot V and M between A and D (Fig. 7.3e and f); the shear has a constant value V = P/2, while the bending moment increases linearly from M = 0 at x = 0 to M = PL/4 at x = L/2. Cutting, now, the beam at a point E between D and B and considering the free body EB (Fig. 7.3d), we write that the sum of the vertical components and the sum of the moments about E of the forces acting on the free body are zero. We obtain V = -P/2 and M = P(L - x)/2. The shear is therefore negative and the bending moment positive; this can be checked by observing that the reaction at B bends the beam at E as indicated in Fig. 7.2c but tends to shear it off in a manner opposite to that shown in Fig. 7.2b. We can complete, now, the shear and bending-moment diagrams of (Fig. 7.3e and f); the shear has a constant value V =-P/2 between D and B, while the bending moment decreases linearly from M = PL/4 at x = L/2 to M = 0 at x = L. It should be noted that when a beam is subjected to concentrated loads only, the shear is of constant value between loads and the bending moment varies linearly between loads, but when a beam is subjected to distributed loads, the shear and bending moment vary quite differently.

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Fig. 7.3 Shear Force and Bending Moment Diagrams

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1041

PROBLEM 7.29

For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION

0: 0yF wx VΣ = − − =

V wx= −

1 0: 02

xM wx M

Σ = + =

21

2M wx= −

max| |V wL=

2max

1| |

2M wL=

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1042

PROBLEM 7.30


SOLUTION

2

0 01 1 1

2 2 2

x xwx w x w

L L = =

By similar D’s

2

01

0: 02y

xF w V

LΣ = − − =

2

01

2

xV w

L= −

2

01

0: 02 3y

x xF w M

L

Σ = + =

3

01

6

xM w

L= −

max 01

| |2

V w L=

2max 0

1| |

6M w L=

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1043

PROBLEM 7.31


SOLUTION

Free body: entire beam

2

0: 03 3DL L

M P P AL Σ = − − =

/3P=A

0: 0x xF DΣ = =

0: 03y yP

F P P DΣ = − + + =

/3yD P= − /3P=D

(a) Shear and bending moment. Since the loading consists of concentrated loads, the shear diagram is made of horizontal straight-line segments and the B. M. diagram is made of oblique straight-line segments.

Just to the right of A:

10: 03yP

F VΣ = − + = 1 /3V P= + �

1 10: (0) 03

PM MΣ = − = 1 0M = �

Just to the right of B:

20: 0,3yP

F V PΣ = − + − = 2 2 /3V P= − �

2 20: (0) 03 3

P LM M P

Σ = − + =

2 /9M PL= + �

Just to the right of C:

30: 03yP

F P P VΣ = − + − = 3 /3V P= + �

3 32

0: (0) 03 3 3

P L LM M P P

Σ = − + − =

3 /9M PL= − �

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1044


Just to the left of D:

40: 03yP

F VΣ = − = 4 3

PV = + �

4 40: (0) 03

PM MΣ = − − = 4 0M = �

(b) max| | 2 /3;V P= max| | /9M PL=

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1045

PROBLEM 7.32


SOLUTION

(a) Shear and bending moment.


1 ;V P= + 1 0M = �

Just to the right of B:

20: 0;yF P P VΣ = − − = 2 0V = �

2 20: 0;2

LM M P

Σ = − =

2 /2M PL= + �

Just to the left of C:

30: 0,yF P P VΣ = − − = 3 0V = �

3 30: 0,2

LM M P PL

Σ = + − =

3 /2M PL= + �

(b) max| |V P=

max| | /2M PL=

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1046

PROBLEM 7.33


SOLUTION

(a) FBD Beam: 00: 0C yM LA MΣ = − =

0y

M

L=A

0: 0y yF A CΣ = − + =

0M

L=C

Along AB:

0

0

0: 0 yM

F VL

MV

L

Σ = − − =

= −

0

0

0: 0JM

M x ML

MM x

L

Σ = + =

= −

Straight with 0 at 2

MM B= −

Along BC:

0 00: 0 y

M MF V V

L LΣ = − − = = −

0

0 00: 0 1KM x

M M x M M ML L

Σ = + − = = −

Straight with 0 at 0 at 2

MM B M C= =

(b) From diagrams: max 0| | /V M L=

0max| | at

2

MM B=

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1047

PROBLEM 7.34


SOLUTION

Free body: Portion AJ

0: 0yF P VΣ = − − = V P= −

0: 0Σ = + − =J xM M P PL ( )M P L x= −

(a) The V and M diagrams are obtained by plotting the functions V and M.

(b) max| |V P=

max| |M PL=

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1048

PROBLEM 7.35


SOLUTION

(a) Just to the right of A:

1 10 15 kN 0yF V MΣ = = + =

Just to the left of C:

2 215 kN 15 kN mV M= + = + ⋅


3 315 kN 5 kN mV M= + = + ⋅

Just to the right of D:

4 415 kN 12.5 kN mV M= − = + ⋅

Just to the right of E:

5 535 kN 5 kN mV M= − = + ⋅

At B: 12.5 kN mBM = − ⋅

(b) max| | 35.0 kNV = max| | 12.50 kN mM = ⋅

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1049

PROBLEM 7.36


SOLUTION

Free body: Entire beam

0: (3.2 m) (40 kN)(0.6 m) (32 kN)(1.5 m) (16 kN)(3 m) 0AM BΣ = − − − =

37.5 kNB = + 37.5 kN=B

0: 0x xF AΣ = =

0: 37.5 kN 40 kN 32 kN 16 kN 0y yF AΣ = + − − − =

50.5 kNyA = + 50.5 kN=A

(a) Shear and bending moment

Just to the right of A: 1 50.5 kNV = 1 0M =


20: 50.5 kN 40 kN 0yF VΣ = − − = 2 10.5 kNV = +

2 20: (50.5 kN)(0.6 m) 0M MΣ = − = 2 30.3 kN mM = + ⋅


30: 50.5 40 32 0yF VΣ = − − − = 3 21.5 kNV = −

3 30: (50.5)(1.5) (40)(0.9) 0M MΣ = − + = 3 39.8 kN mM = + ⋅

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1050


Just to the right of E:

40: 37.5 0yF VΣ = + = 4 37.5 kNV = −

4 40: (37.5)(0.2) 0M MΣ = − + = 4 7.50 kN mM = + ⋅

At B: 0B BV M= =

(b) max| | 50.5 kNV =

max| | 39.8 kN mM = ⋅

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1051

PROBLEM 7.37


SOLUTION

Free body: Entire beam

0: (6 ft) (6 kips)(2 ft) (12 kips)(4 ft) (4.5 kips)(8 ft) 0AM EΣ = − − − =

16 kipsE = + 16 kips=E

0: 0x xF AΣ = =

0: 16 kips 6 kips 12 kips 4.5 kips 0y yF AΣ = + − − − =

6.50 kipsyA = + 6.50 kips=A

(a) Shear and bending moment


1 16.50 kips 0V M= + =


20: 6.50 kips 6 kips 0yF VΣ = − − = 2 0.50 kipsV = +

2 20: (6.50 kips)(2 ft) 0M MΣ = − = 2 13 kip ftM = + ⋅


30: 6.50 6 12 0yF VΣ = − − − = 3 11.5 kipsV = +

3 30: (6.50)(4) (6)(2) 0M MΣ = − − = 3 14 kip ftM = + ⋅

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1052


Just to the right of E: 40: 4.5 0yF VΣ = − = 4 4.5 kipsV = +

4 40: (4.5)2 0M MΣ = − − = 4 9 kip ftM = − ⋅

At B: 0B BV M= =

(b) max| | 11.50 kipsV =

max| | 14.00 kip ftM = ⋅

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Chapter 8 Friction

Pulleys, Tension Belts and Journal Bearings

8.1 INTRODUCTION In the preceding chapters, it was assumed that surfaces in contact were either frictionless or rough. If they were frictionless, the force each surface exerted on the other was normal to the surfaces and the two surfaces could move freely with respect to each other. If they were rough, it was assumed that tangential forces could develop to prevent the motion of one surface with respect to the other. This view was a simplified one. Actually, no perfectly frictionless surface exists. When two surfaces are in contact, tangential forces, called friction forces, will always develop if one attempts to move one surface with respect to the other. On the other hand, these friction forces are limited in magnitude and will not prevent motion if sufficiently large forces are applied. The distinction between frictionless and rough surfaces is thus a matter of degree. This will be seen more clearly in the present chapter, which is devoted to the study of friction and of its applications to common engineering situations. There are two types of friction: dry friction, sometimes called Coulomb friction, and fluid friction. Fluid friction develops between layers of fluid moving at different velocities. Fluid friction is of great importance in problems involving the flow of fluids through pipes and orifices or dealing with bodies immersed in moving fluids. It is also basic in the analysis of the motion of lubricated mechanisms. Such problems are considered in texts on fluid mechanics. The present study is limited to dry friction, i.e., to problems involving rigid bodies which are in contact along nonlubricated surfaces. In the first part of this chapter, the equilibrium of various rigid bodies and structures, assuming dry friction at the surfaces of contact, is analyzed. Later a number of specific engineering applications where dry friction plays an important role are considered: wedges, square-threaded screws, journal bearings, thrust bearings, rolling resistance, and belt friction.

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Experimental evidence shows that the maximum value Fm

of the static-friction force is proportional to the normal component N of the reaction of the surface (Fig.8.1), we have:

𝑭𝑭𝒎𝒎 = 𝒎𝒎𝒔𝒔𝑵𝑵

Where ms is a constant called the coefficient of static friction. Similarly, the magnitude Fk

𝑭𝑭𝒌𝒌 = 𝒎𝒎𝒌𝒌𝑵𝑵

of the kinetic-friction force may be put in the form:

Where mk is a constant called the coefficient of kinetic friction. The coefficients of friction ms and mk

do not depend upon the area of the surfaces in contact.

Fig. 8.1 Friction Force with respect to the Acting one

8.2 ANGLES OF FRICTION It is sometimes convenient to replace the normal force N and the friction force F by their resultant R. Let us consider again a block of weight W resting on a horizontal plane surface. If no horizontal force is applied to the block, the resultant R reduces to the normal force N (Fig. 8.2a). However, if the applied force P has a horizontal component Px which tends to move the block, the force R will have a horizontal component F and, thus, will form an angle f with the normal to the surface (Fig. 8.2b). If Px is increased until motion becomes impending, the angle between R and the vertical grows and reaches a maximum value (Fig. 8.2c). This value is called the angle of static friction and is denoted by ∅Rs. From the geometry of Fig. 8.2c, we note that:

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𝐭𝐭𝐭𝐭𝐭𝐭(∅𝒔𝒔) =𝑭𝑭𝒎𝒎𝑵𝑵 =

𝒎𝒎𝒔𝒔𝑵𝑵𝑵𝑵 = 𝒎𝒎𝒔𝒔

If motion actually takes place, the magnitude of the friction force drops to Fk; similarly, the angle f between R and N drops to a lower value ∅Rk

𝐭𝐭𝐭𝐭𝐭𝐭(∅𝒌𝒌) =𝑭𝑭𝒌𝒌𝑵𝑵 =

𝒎𝒎𝒌𝒌𝑵𝑵𝑵𝑵 = 𝒎𝒎𝒌𝒌

, called the angle of kinetic friction (Fig. 8.2d). From the geometry of Fig. 8.2d, we write:

Fig. 8.2

Fig. 8.3

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8.3 BELT FRICTION Fig. 8.4 shows the relation between the Minimum T1 and larger T2 considering the friction coefficient between the belt and the pulley of ms

as noted by the following equations.

Fig. 8.4

𝒍𝒍𝒍𝒍 �𝑻𝑻𝟐𝟐𝑻𝑻𝟏𝟏� = 𝒎𝒎𝒔𝒔𝜷𝜷

Or

𝑻𝑻𝟐𝟐𝑻𝑻𝟏𝟏

= 𝒆𝒆𝒎𝒎𝒔𝒔𝜷𝜷

8.4 JOURNAL BEARINGS. AXLE FRICTION Journal bearings are used to provide lateral support to rotating shafts and axles. Thrust bearings, which will be studied in the next section, are used to provide axial support to shafts and axles. If the journal bearing is fully lubricated, the frictional resistance depends upon the speed of rotation, the clearance between axle and bearing, and the viscosity of the lubricant. As indicated in Sec. 8.1, such problems are studied in fluid mechanics. The methods of this chapter, however, can be applied to the study of axle friction when the bearing is not lubricated or only partially lubricated. It can then be assumed that the axle and the bearing are in direct contact along a single straight line (Fig. 8.5).

𝑴𝑴 = 𝑹𝑹𝑹𝑹𝒔𝒔𝑹𝑹𝒍𝒍(∅𝐤𝐤)

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Fig. 8.5

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1221

PROBLEM 8.1

Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = 25° and P = 750 N.

SOLUTION

Assume equilibrium:

0: (1200 N)sin 25° (750 N)cos 25° 0xF FΣ = + − =

172.6 NF = + 172.6 N=F

0: (1200 N)cos 25° (750 N)sin 25° 0yF NΣ = − − =

1404.5 NN =

Maximum friction force: 0.35(1404.5 N) 491.6 Nm sF Nμ= = =

Since ,mF F< block is in equilibrium

Friction force: 172.6 N=F 25.0°

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1222

PROBLEM 8.2

Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = 30° and P = 150 N.

SOLUTION

Assume equilibrium:

0: (1200 N)sin 30° (150 N)cos 30° 0xF FΣ = + − =

470.1 NF = − 470.1 N=F

0: (1200 N)cos 30 (150 N)sin 30 0 yF NΣ = − ° − ° =

1114.2 NN =

(a) Maximum friction force:

0.35(1114.2 N)

390.0 N

m sF Nμ===

Since F is and ,mF F> block moves down

(b) Actual friction force: 0.25(1114.2 N) 279 Nk kF F Nμ= = = = 279 N=F 30.0°

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1223

PROBLEM 8.3

Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when P = 100 lb.

SOLUTION

Assume equilibrium:

0: (45 lb)sin 30 (100 lb)cos 40 0xF FΣ = + ° − ° =

54.0 lbF = +

0: (45 lb)cos 30 (100 lb)sin 40 0yF NΣ = − ° − ° =

103.2 lbN =


0.40(103.2 lb)

41.30 lb

m sF Nμ===

We note that .mF F> Thus, block moves up

(b) Actual friction force:

0.30(103.2 lb) 30.97 lb,k kF F Nμ= = = = 31.0 lb=F 30.0°

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1224

PROBLEM 8.4

Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when P = 60 lb.

SOLUTION

Assume equilibrium:

0: (45 lb)sin 30 (60 lb)cos 40 0xF FΣ = + ° − ° =

23.46 lbF = +

0: (45 lb)cos 30 (60 lb)sin 40 0yF NΣ = − ° − ° =

77.54 lbN =


0.40(77.54 lb)

31.02 lb

m sF Nμ===

We check that mF F< . Thus, block is in equilibrium

(b) Thus 23.46 lbF = + 23.5 lb=F 30.0°

Note: We have 0.30(77.54) 23.26 lb.k kF Nμ= = = Thus .kF F> If block originally in motion, it will keep

moving with 23.26 lb.kF =

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1225

PROBLEM 8.5

Determine the smallest value of P required to (a) start the block up the incline, (b) keep it moving up, (c) prevent it from moving down.

SOLUTION

(a) To start block up the incline:

1

0.40

tan 0.40 21.80

s

s

μ

φ −

=

= = °

From force triangle:

45 lb

sin 51.80 sin 28.20

P =° °

74.8 lbP =

(b) To keep block moving up:

1

0.30

tan 0.30 16.70

k

k

μ

φ −

=

= = °


45 lb

sin 46.70 sin 33.30°

P =°

59.7 lbP =

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1226


(c) To prevent block from moving down:


45 lb

sin 8.20 sin 71.80°

P =°

6.76 lbP =

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1231

PROBLEM 8.10

Determine the range of values of P for which equilibrium of the block shown is maintained.

SOLUTION

FBD block:

(Impending motion down):

1 1

1

tan tan 0.25

(500 N) tan (30° tan 0.25)

143.03 N

s s

P

φ μ− −

−

= =

= −=

(Impending motion up):

1(500 N) tan (30° tan 0.25)

483.46 N

P −= +=

Equilibrium is maintained for 143.0 N 483 NP≤ ≤

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1232

PROBLEM 8.11

The 20-lb block A and the 30-lb block B are supported by an incline that is held in the position shown. Knowing that the coefficient of static friction is 0.15 between the two blocks and zero between block B and the incline, determine the value of θ for which motion is impending.

SOLUTION

Since motion impends, sF Nμ= between A + B

Free body: Block A

Impending motion:

10: 20cosyF N θΣ = =

10: 20sin 0θ μΣ = − − =x sF T N

20sin 0.15(20cos )T θ θ= +

20sin 3cosT θ θ= + (1)

Free body: Block B

Impending motion:

10: 30sin 0x sF T Nθ μΣ = − + =

130sin sT Nθ μ= −

30sin 0.15(20cos )θ θ= −

30sin 3cosT θ θ= − (2)

Eq. (1)-Eq. (2): 20sin 3cos 30sin 3cos 0θ θ θ θ+ − + =

6

6cos 10sin : tan ;10

θ θ θ= = 30.96θ = ° 31.0θ = °

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1233

PROBLEM 8.12

The 20-lb block A and the 30-lb block B are supported by an incline that is held in the position shown. Knowing that the coefficient of static friction is 0.15 between all surfaces of contact, determine the value of θ for which motion is impending.

SOLUTION

Since motion impends, sF Nμ= at all surfaces.

Free body: Block A

Impending motion:

10: 20cosyF N θΣ = =

10: 20sin 0θ μΣ = − − =x sF T N

20sin 0.15(20cos )T θ θ= +

20sin 3cosT θ θ= + (1)

Free body: Block B

Impending motion:

2 10: 30cos 0yF N NθΣ = − − =

2

2 2

30cos 20cos 50cos

0.15(50cos ) 6cos

θ θ θ

μ θ θ

= + =

= = =s

N

F N

1 20: 30sin 0x s sF T N Nθ μ μΣ = − + + =

30sin 0.15(20cos ) 0.15(50cos )T θ θ θ= − −

30sin 3cos 7.5cosT θ θ θ= − − (2)

Eq. (1) subtracted by Eq. (2): 20sin 3cos 30sin 3cos 7.5cos 0θ θ θ θ θ+ − + + =

13.5

13.5cos 10sin , tan10

θ θ θ °= = 53.5θ = °

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1236

PROBLEM 8.15

A 40-kg packing crate must be moved to the left along the floor without tipping. Knowing that the coefficient of static friction between the crate and the floor is 0.35, determine (a) the largest allowable value of α, (b) the corresponding magnitude of the force P.

SOLUTION

(a) Free-body diagram

If the crate is about to tip about C, contact between crate and ground is only at C and the reaction R is applied at C. As the crate is about to slide, R must form with the vertical an angle

1 1tan tan

0.35 19.29s sφ μ− −= =

= °

Since the crate is a 3-force body. P must pass through E where R and W intersect.

0.4 m1.1429 m

tan 0.35

1.1429 0.5 0.6429 m

0.6429 mtan

0.4 m

s

CFEF

EH EF HF

EH

HB

θ

α

= = =

= − = − =

= =

58.11α = ° 58.1α = °

(b) Force Triangle

0.424sin19.29 sin128.82

P WP W= =

° °

20.424(40 kg)(9.81 m/s ),P = 166.4 NP =

Note: After the crate starts moving, sμ should be replaced by the lower value .kμ This will

yield a larger value of .α

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1237

PROBLEM 8.16

A 40-kg packing crate is pulled by a rope as shown. The coefficient of static friction between the crate and the floor is 0.35. If α = 40°, determine (a) the magnitude of the force P required to move the crate, (b) whether the crate will slide or tip.

SOLUTION

Force P for which sliding is impending

(We assume that crate does not tip)

2(40 kg)(9.81 m/s ) 392.4 NW = =

0: sin 40 0yF N W PΣ = − + ° =

sin 40N W P= − ° (1)

0: 0.35 cos 40 0xF N PΣ = − ° =

Substitute for N from Eq. (1): 0.35( sin 40 ) cos 40 0W P P− ° − ° =

0.35

0.35sin 40 cos 40

WP =

° + ° 0.3532P W=

Force P for which crate rotates about C

(We assume that crate does not slide)

0: ( sin 40 )(0.8 m) ( cos 40 )(0.5 m)CM P PΣ = ° + °

(0.4 m) 0W− =

0.4

0.44580.8sin 40 0.5cos 40

WP W= =

° + °

Crate will first slide

0.3532(392.4 N)P = 138.6 NP =

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1337

PROBLEM 8.103

A 300-lb block is supported by a rope that is wrapped 1 12 times around a horizontal

rod. Knowing that the coefficient of static friction between the rope and the rod is 0.15, determine the range of values of P for which equilibrium is maintained.

SOLUTION

1.5 turns 3 radβ π= =

For impending motion of W up,

(0.15)3(300 lb)

1233.36 lb

sP We eμ β π= ==

For impending motion of W down,

(0.15)3(300 lb)

72.971 lb

sP We eμ β π− −= ==

For equilibrium, 73.0 lb 1233 lbP≤ ≤

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1339

PROBLEM 8.105

A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of static friction is 0.25, determine (a) the smallest value of the mass m for which equilibrium is possible, (b) the corresponding tension in portion BC of the rope.

SOLUTION

We apply Eq. (8.14) to pipe B and pipe C.

2

1

μ β= sT

eT

(8.14)

Pipe B: 2 1,A BCT W T T= =

2

0.25,3sπμ β= =

0.25(2 /3) /6A

BC

We e

Tπ π= = (1)

Pipe C: 2 1, , 0.25,3BC D sT T T Wπμ β= = = =

0.025( /3) /12BC

D

Te e

Wπ π= = (2)

(a) Multiplying Eq. (1) by Eq. (2):

/6 /12 /6 /12 /4 2.193π π π π π+= ⋅ = = =A

D

We e e e

W

50 kg

2.193 2.193 2.193 2.193

AWgA D A

DW W m

W mg

= = = = =

22.8 kgm =

(b) From Eq. (1): 2

/6

(50 kg)(9.81 m/s )291 N

1.688A

BCW

Teπ= = =

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1340

PROBLEM 8.106

A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of static friction is 0.25, determine (a) the largest value of the mass m for which equilibrium is possible, (b) the corresponding tension in portion BC of the rope.

SOLUTION

See FB diagrams of Problem 8.105. We apply Eq. (8.14) to pipes B and C.

Pipe B: 1 22

, , 0.25,3A BC sT W T Tπμ β= = = =

0.25(2 /3) /62

1

:μ β π π= = =s BC

A

TTe e e

T W (1)

Pipe C: 1 2, , 0.25,3BC D sT T T Wπμ β= = = =

0.25( /3) /122

1

:s D

BC

T We e e

T Tμ β π π= = = (2)

(a) Multiply Eq. (1) by Eq. (2):

/6 /12 /6 /12 /4 2.193D

A

We e e e

Wπ π π π π+= ⋅ = = =

0 2.193 2.193 2.193(50 kg)A AW W m m= = = 109.7 kgm =

(b) From Eq. (1): /6 2(50 kg)(9.81 m/s )(1.688) 828 NBC AT W eπ= = =

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1346

PROBLEM 8.112

A flat belt is used to transmit a couple from drum B to drum A. Knowing that the coefficient of static friction is 0.40 and that the allowable belt tension is 450 N, determine the largest couple that can be exerted on drum A.

SOLUTION

FBD’s drums:

7

180 306 6

5180 30

6 6

A

B

π πβ π

π πβ π

= ° + ° = + =

= ° − ° = − =

Since ,B Aβ β< slipping will impend first on B (friction coefficients being equal)

So 2 max 1

(0.4)5 /61 1450 N or 157.914 N

s BT T T e

T e T

μ β

π

= =

= =

1 20: (0.12 m)( ) 0A AM M T TΣ = + − =

(0.12 m)(450 N 157.914 N) 35.05 N mAM = − = ⋅

35.1 N mAM = ⋅

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1347

PROBLEM 8.113

A flat belt is used to transmit a couple from pulley A to pulley B. The radius of each pulley is 60 mm, and a force of magnitude P = 900 N is applied as shown to the axle of pulley A. Knowing that the coefficient of static friction is 0.35, determine (a) the largest couple that can be transmitted, (b) the corresponding maximum value of the tension in the belt.

SOLUTION

Drum A:

(0.35)2

1

2 13.0028

sT

e eT

T T

μ π π= =

=

180 radiansβ π= ° =

(a) Torque: 0: (675.15 N)(0.06 m) (224.84 N)(0.06 m)AM MΣ = − +

27.0 N mM = ⋅

(b) 1 20: 900 N 0xF T TΣ = + − =

1 1

1

1

2

3.0028 900 N 0

4.00282 900

224.841 N

3.0028(224.841 N) 675.15 N

T T

T

T

T

+ − ==== =

max 675 NT =

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1349

PROBLEM 8.115

The speed of the brake drum shown is controlled by a belt attached to the control bar AD. A force P of magnitude 25 lb is applied to the control bar at A. Determine the magnitude of the couple being applied to the drum, knowing that the coefficient of kinetic friction between the belt and the drum is 0.25, that a = 4 in., and that the drum is rotating at a constant speed (a) counterclockwise, (b) clockwise.

SOLUTION

(a) Counterclockwise rotation

Free body: Drum

0.252

1

2 1

8 in. 180 radians

2.1933

2.1933

k

r

Te e

T

T T

μ β π

β π= = ° =

= = =

=

Free body: Control bar

1 20: (12 in.) (4 in.) (25 lb)(28 in.) 0CM T TΣ = − − =

1 1

1

2

(12) 2.1933 (4) 700 0

216.93 lb

2.1933(216.93 lb) 475.80 lb

T T

T

T

− − === =

Return to free body of drum

1 20: (8 in.) (8 in.) 0EM M T TΣ = + − =

(216.96 lb)(8 in.) (475.80 lb)(8 in.) 0M + − =

2070.9 lb in.M = ⋅ 2070 lb in.M = ⋅

(b) Clockwise rotation

0.252

1

2 1

8 in. rad

2.1933

2.1933

k

r

Te e

T

T T

μ β π

β π= =

= = =

=

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1350


Free body: Control rod

2 10: (12 in.) (4 in.) (25 lb)(28 in.) 0CM T TΣ = − − =

1 1

1

2

2

2.1933 (12) (4) 700 0

31.363 lb

2.1933(31.363 lb)

68.788 lb

T T

T

T

T

− − ====

Return to free body of drum

1 20: (8 in.) (8 in.) 0EM M T TΣ = + − =

(31.363 lb)(8 in.) (68.788 lb)(8 in.) 0M + − =

299.4 lb in.M = ⋅ 299 lb in.M = ⋅

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1351

PROBLEM 8.116

The speed of the brake drum shown is controlled by a belt attached to the control bar AD. Knowing that a = 4 in., determine the maximum value of the coefficient of static friction for which the brake is not self-locking when the drum rotates counterclockwise.

SOLUTION

2

1

2 1

8 in., 180 radians

s s

s

r

Te e

T

T e T

μ β μ π

μ π

β π= = ° =

= =

=

Free body: Control rod

1 20: (28 in.) (12 in.) (4 in.) 0CM P T TΣ = − + =

1 128 12 (4) 0P T e Tμπ− + =

For self-locking brake: 0P =

1 112 4

3

ln 3 1.0986

1.09860.3497

μ π

μ π

μ π

μπ

=

== =

= =

s

s

s

s

T T e

e

0.350sμ =

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