© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition 2 - 1 CE 102 Statics Chapter 2 Statics

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Statics

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2 - 1

CE 102 Statics

Chapter 2

Statics of Particles



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2 - 2

Contents

Introduction

Resultant of Two Forces

Vectors

Addition of Vectors

Resultant of Several Concurrent Forces

Sample Problem 2.1

Sample Problem 2.2

Rectangular Components of a Force: Unit Vectors

Addition of Forces by Summing Components

Sample Problem 2.3

Equilibrium of a Particle

Free-Body Diagrams

Sample Problem 2.4

Sample Problem 2.5

Rectangular Components in Space

Sample Problem 2.6



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Introduction

• The objective for the current chapter is to investigate the effects of forces on particles:

- replacing multiple forces acting on a particle with a single equivalent or resultant force,

- relations between forces acting on a particle that is in a state of equilibrium.

• The focus on particles does not imply a restriction to miniscule bodies. Rather, the study is restricted to analyses in which the size and shape of the bodies is not significant so that all forces may be assumed to be applied at a single point.



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Resultant of Two Forces

• force: action of one body on another; characterized by its point of application, magnitude, line of action, and sense.

• Experimental evidence shows that the combined effect of two forces may be represented by a single resultant force.

• The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs.

• Force is a vector quantity.



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Vectors• Vector: parameters possessing magnitude and direction

which add according to the parallelogram law. Examples: displacements, velocities, accelerations.

• Vector classifications:- Fixed or bound vectors have well defined points of

application that cannot be changed without affecting an analysis.

- Free vectors may be freely moved in space without changing their effect on an analysis.

- Sliding vectors may be applied anywhere along their line of action without affecting an analysis.

• Equal vectors have the same magnitude and direction.

• Negative vector of a given vector has the same magnitude and the opposite direction.

• Scalar: parameters possessing magnitude but not direction. Examples: mass, volume, temperature



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Addition of Vectors

• Trapezoid rule for vector addition

• Triangle rule for vector addition

B

B

C

C

QPRBPQQPR

cos2222

• Law of cosines,

• Law of sines,

A

C

R

B

Q

A sinsinsin

• Vector addition is commutative,

PQQP

• Vector subtraction



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Addition of Vectors

• Addition of three or more vectors through repeated application of the triangle rule

• The polygon rule for the addition of three or more vectors.

• Vector addition is associative,

SQPSQPSQP

• Multiplication of a vector by a scalar



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Resultant of Several Concurrent Forces

• Concurrent forces: set of forces which all pass through the same point.

A set of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces.

• Vector force components: two or more force vectors which, together, have the same effect as a single force vector.



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Sample Problem 2.1

The two forces act on a bolt at A. Determine their resultant.

SOLUTION:

• Graphical solution - construct a parallelogram with sides in the same direction as P and Q and lengths in proportion. Graphically evaluate the resultant which is equivalent in direction and proportional in magnitude to the the diagonal.

• Trigonometric solution - use the triangle rule for vector addition in conjunction with the law of cosines and law of sines to find the resultant.



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Sample Problem 2.1

• Graphical solution - A parallelogram with sides equal to P and Q is drawn to scale. The magnitude and direction of the resultant or of the diagonal to the parallelogram are measured,

35N 98 R

• Graphical solution - A triangle is drawn with P and Q head-to-tail and to scale. The magnitude and direction of the resultant or of the third side of the triangle are measured,

35N 98 R



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Sample Problem 2.1• Trigonometric solution - Apply the triangle rule.

From the Law of Cosines,

155cosN60N402N60N40

cos222

222 BPQQPR

AA

R

QBA

R

B

Q

A

2004.15

N73.97

N60155sin

sinsin

sinsin

N73.97R

From the Law of Sines,

04.35



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Sample Problem 2.2

a) the tension in each of the ropes for = 45o,

b) the value of for which the tension in rope 2 is a minimum.

A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is 5000 N directed along the axis of the barge, determine

SOLUTION:

• Find a graphical solution by applying the Parallelogram Rule for vector addition. The parallelogram has sides in the directions of the two ropes and a diagonal in the direction of the barge axis and length proportional to 5000 N.

• The angle for minimum tension in rope 2 is determined by applying the Triangle Rule and observing the effect of variations in .

• Find a trigonometric solution by applying the Triangle Rule for vector addition. With the magnitude and direction of the resultant known and the directions of the other two sides parallel to the ropes given, apply the Law of Sines to find the rope tensions.



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Sample Problem 2.2

• Graphical solution - Parallelogram Rule with known resultant direction and magnitude, known directions for sides.

N2600N3700 21 TT

• Trigonometric solution - Triangle Rule with Law of Sines

105sin

N5000

30sin45sin21 TT

N2590N3660 21 TT



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Sample Problem 2.2

• The angle for minimum tension in rope 2 is determined by applying the Triangle Rule and observing the effect of variations in .

• The minimum tension in rope 2 occurs when T1 and T2 are perpendicular.

30sinN50002T N25002 T

30cosN50001T N43301 T

3090 60



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Rectangular Components of a Force: Unit Vectors

• Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components.

Fx and Fy are referred to as the scalar components of

jFiFF yx

F

• May resolve a force vector into perpendicular components so that the resulting parallelogram is a rectangle. are referred to as rectangular vector components and

yx FFF

yx FF

and

• Define perpendicular unit vectors which are parallel to the x and y axes.

ji

and



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Addition of Forces by Summing Components

SQPR

• Wish to find the resultant of 3 or more concurrent forces,

jSQPiSQP

jSiSjQiQjPiPjRiR

yyyxxx

yxyxyxyx

• Resolve each force into rectangular components

x

xxxxF

SQPR

• The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces.

y

yyyy

F

SQPR

x

yyx R

RRRR 122 tan

• To find the resultant magnitude and direction,



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Sample Problem 2.3

Four forces act on bolt A as shown. Determine the resultant of the force on the bolt.

SOLUTION:

• Resolve each force into rectangular components.

• Calculate the magnitude and direction of the resultant.

• Determine the components of the resultant by adding the corresponding force components.



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Sample Problem 2.3SOLUTION:

• Resolve each force into rectangular components.

9.256.96100

0.1100110

2.754.2780

0.759.129150

4

3

2

1

F

F

F

F

compycompxmagforce

22 3.141.199 R N6.199R

• Calculate the magnitude and direction.

N1.199

N3.14tan 1.4

• Determine the components of the resultant by adding the corresponding force components.

1.199xR 3.14yR



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Equilibrium of a Particle• When the resultant of all forces acting on a particle is zero, the particle is

in equilibrium.

• Particle acted upon by two forces:

- equal magnitude

- same line of action

- opposite sense

• Particle acted upon by three or more forces:

- graphical solution yields a closed polygon

- algebraic solution

00

0

yx FF

FR

• Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line.



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Free-Body Diagrams

Space Diagram: A sketch showing the physical conditions of the problem.

Free-Body Diagram: A sketch showing only the forces on the selected particle.



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Sample Problem 2.4

In a ship-unloading operation, a 3500-N automobile is supported by a cable. A rope is tied to the cable and pulled to center the automobile over its intended position. What is the tension in the rope?

SOLUTION:

• Construct a free-body diagram for the particle at the junction of the rope and cable.

• Apply the conditions for equilibrium by creating a closed polygon from the forces applied to the particle.

• Apply trigonometric relations to determine the unknown force magnitudes.



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Sample Problem 2.4

SOLUTION:

• Construct a free-body diagram for the particle at A.

• Apply the conditions for equilibrium.

• Solve for the unknown force magnitudes.

58sin

N3500

2sin120sinACAB TT

N3570ABT

N144ACT



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Sample Problem 2.5

It is desired to determine the drag force at a given speed on a prototype sailboat hull. A model is placed in a test channel and three cables are used to align its bow on the channel centerline. For a given speed, the tension is 40 N in cable AB and 60 N in cable AE.

Determine the drag force exerted on the hull and the tension in cable AC.

SOLUTION:

• Choosing the hull as the free body, draw a free-body diagram.

• Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero.

• Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions.



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Sample Problem 2.5

SOLUTION:

• Choosing the hull as the free body, draw a free-body diagram.

25.60

75.1m 4

m 7tan

56.20

375.0m 4

m 1.5tan

• Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero.

0 DAEACAB FTTTR



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Sample Problem 2.5

• Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions.

jT

iFT

R

iFF

jT

jTiT

jTiTT

ji

jiT

AC

DAC

DD

ACAC

ACACAC

AB

609363.084.19

3512.073.34

0

N 06

9363.03512.0

56.20cos56.20sin

N 84.19N 73.34

26.60cosN 4026.60sinN 40



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Sample Problem 2.5

jT

iFT

R

AC

DAC

609363.084.19

3512.073.34

0

This equation is satisfied only if each component of the resultant is equal to zero.

609363.084.1900

3512.073.3400

ACy

DACx

TF

FTF

N 66.19

N 9.42

D

AC

F

T



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• The vector is contained in the plane OBAC.

F

• Resolve into horizontal and vertical components.

yh FF sin

F

yy FF cos

• Resolve into rectangular components.

hF

sinsin

sin

cossin

cos

y

hz

y

hx

F

FF

F

FF



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• With the angles between and the axes,F

kji

F

kjiF

kFjFiFF

FFFFFF

zyx

zyx

zyx

zzyyxx

coscoscos

coscoscos

coscoscos

• is a unit vector along the line of action ofand are the direction cosines for

F

F

zyx cos and,cos,cos



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Direction of the force is defined by the location of two points,

222111 ,, and ,, zyxNzyxM

d

FdF

d

FdF

d

FdF

kdjdidd

FF

zzdyydxxd

kdjdid

NMd

zz

yy

xx

zyx

zyx

zyx

1

and joining vector

121212



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Sample Problem 2.6

The tension in the guy wire is 2500 N. Determine:

a) components Fx, Fy, Fz of the force acting on the bolt at A,

b) the angles x, y, zdefining the direction of the force

SOLUTION:

• Based on the relative locations of the points A and B, determine the unit vector pointing from A towards B.

• Apply the unit vector to determine the components of the force acting on A.

• Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.



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Sample Problem 2.6SOLUTION:

• Determine the unit vector pointing from A towards B.

m 3.94

m30m80m40

m30m80m40

222

AB

kjiAB

• Determine the components of the force.

kji

kji

FF

N 795N 2120N1060

318.0848.0424.0N 2500

kji

kji

318.0848.0424.0

3.94

30

3.94

80

3.94

40



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Sample Problem 2.6• Noting that the components of the unit vector are

the direction cosines for the vector, calculate the corresponding angles.

kji

kji zyx

318.0848.0424.0

coscoscos

5.71

0.32

1.115

z

y

x

33

Problem 2.7

The direction of the 75-lbforces may vary, but theangle between the forcesis always 50o. Determinethe value of for whichthe resultant of the forcesacting at A is directedhorizontally to the left.

240 lb

30o

50o75 lb

75 lb

A

34

Solving Problems on Your Own

1. Determine the resultant R of two or more forces.

2. Draw a parallelogram with the applied forces as two adjacent sides and the resultant as the included diagonal.

3. Set the resultant, or sum of the forces, directed horizontally.

The direction of the 75-lbforces may vary, but theangle between the forcesis always 50o. Determinethe value of for whichthe resultant of the forcesacting at A is directedhorizontally to the left.

240 lb

30o

50o75 lb

75 lb

A

Problem 2.7

35

Problem 2.7 Solution

Determine the resultant R of twoor more forces.

We first Replace the two 75-lbforces by their resultant R1, usingthe triangle rule.

50o

25o

25oR1

R1 = 2(75 lb) cos25o = 135.95 lb

R1 = 135.95 lb +25o

240 lb

30o

50o75 lb

75 lb

A

36

Draw a parallelogram with the applied forces as two adjacentsides and the resultant as the included diagonal. Set the resultant, or sum of the forces, directed horizontally.


30o +25o

R2

R1 = 135.95 lb240 lb

Consider the resultant R2

of R1 and the 240-lb forceand recall that R2 must behorizontal and directed tothe left.

Law of sines:

sin(+25o)240 lb

=sin(30o)

135.95 lb

sin(+25o) = (240 lb) sin(30o)

135.95 lb= 0.88270

+ 25o = 61.97o

= 37.0o

37

Problem 2.8

A container of weight W = 1165 N is supportedby three cables as shown.Determine the tension ineach cable.600 mm

320 mm

450 mm360 mm

500 mm

DO

C

B

A x

y

z

38

A container of weight W = 1165 N is supportedby three cables as shown.Determine the tension ineach cable.


1. Draw a free-body diagram of the particle. This diagram shows the particle and all the forces acting on it.

2. Resolve each of the forces into rectangular components. Follow the method outlined in the text.

F = F = (dx i + dy j + dz k) Fd

600 mm

320 mm

450 mm360 mm

500 mm

DO

C

B

A x

y

z

Problem 2.8

39

3. Set the resultant, or sum, of the forces exerted on the particle equal to zero. You will obtain a vectorial equation consisting of terms containing the unit vectors i, j, and k. Three scalar equations result, which can be solved for the unknowns.


A container of weight W = 1165 N is supportedby three cables as shown.Determine the tension ineach cable.

600 mm

320 mm

450 mm360 mm

500 mm

DO

C

B

A x

y

z

Problem 2.8

40


F = 0

TAB + TAC + TAD + W = 0

AB = (450 mm)i + (600 mm)j AB = 750 mm

AC = (600 mm)j _ (320mm)k AC = 680 mm

AD = (_500 mm)i + (600 mm)j + (360 mm)k AD = 860 mm

600 mm

320 mm

450 mm360 mm

500 mm

DC

B

A x

y

zTAD TAB

TAC

W = _ (1165 N) j

O

Draw a free-body diagram of the particle.

41

( )450750

i +600750

j TAB

= (0.6 i + 0.8 j) TAB

TAB = TAB AB = TAB ABAB

=

= =600 mm

320 mm

450 mm360 mm

500 mm

DC

B

A x

y

zTAD TAB

TAC

W = _ (1165 N) j

O

TAD ADAD

= TAD = TAD AD = ( )TAD

500860

i +600860

j +360860

k

=( ) TAD

25 43

i + 30 43

j + 18 43

k

=

Resolve each of the forces into rectangular components.


TAC ACAC ( )680 j _

320680

k TAC= TAC = TAC AC = ( )1517

j _= 817

k600TAC

42

Substitution into F = 0,factor i, j, k and set theircoefficients to zero:

0.6 TAB _ 2543

TAD = 0 TAB = 0.9690 TAD (1)

0.8 TAB + 1517

TAC3043

TAD+ _ 1165 N = 0 (2)

817

_ TAC +1843

TAD = 0

+

TAC = 0.8895 TAD (3)

600 mm

320 mm

450 mm360 mm

500 mm

DC

B

A x

y

zTAD TAB

TAC

W = _ (1165 N) j

O

Set the resultant, or sum, ofthe forces exerted on theparticle equal to zero.


43

Substitution for TAB andTAC from (1) and (3)into (2):

( 0.8 x 0.9690 + x 0.8895 + )TAD _ 1165 N = 01517

3043

2.2578 TAD _ 1165 N = 0

From (1): TAB = 0.9690 (516 N)

From (3): TAC = 0.8895 (516 N)

TAD = 516 N

TAB = 500 N

TAC = 459 N

600 mm

320 mm

450 mm360 mm

500 mm

DC

B

A x

y

zTAD TAB

TAC

W = _ (1165 N) j

O


44

Problem 2.9

Cable AB is 65 ft long, andthe tension in that cable is3900 lb. Determine (a) thex, y, and z components ofthe force exerted by thecable on the anchor B, (b)the angles x, y, and z

defining the direction ofthat force.

O

x

y

z

56 ft

D

B

C

A

50o 20o

45


O

x

y

z

56 ft

D

B

C

A

50o 20o



1. Determine the rectangular components of a force defined byits magnitude and direction. If the direction of the force F isdefined by the angles y and , projections of F through theseangles or their components will yield the components of F.

Problem 2.9

46




2. Determine the direction cosines of the line of action of a force.The direction cosines of the line of action of a force F aredetermined by dividing the components of the force by F.

cos x=Fx

F cos y=Fy

F cos z=Fz

F

O

x

y

z

56 ft

D

B

C

A

50o 20o

Problem 2.9

47


z

y

Determine the direction cosines ofthe line of action of a force.

O

x

B

A

Fy

Fz

F

Fx

20o

y

65 ftFrom triangle AOB:

cos y = 56 ft65 ft

= 0.86154

y = 30.51o

(a) Fx = _ F sin y cos 20o

= _ (3900 lb) sin 30.51o cos 20o

Fx = _1861 lb

Fy = + F cos y = (3900 lb)(0.86154) Fy = + 3360 lb

Fz = + (3900 lb) sin 30.51o sin 20o Fz = + 677 lb

56 ft

48


O

x

B

A

Fy

Fz

F

Fx

20o

y

65 ft

y

Determine the direction cosinesof the line of action of a force.

(b) cos x = Fx

F=

_1861 lb3900 lb

cos x = _ 0.4771

x = 118.5o

From above (a): y = 30.5o

cos z = Fz

F= + 677 lb

3900 lb= + 0.1736 z = 80.0o

56 ft

49

Problem 2.10

A

9 ft

5 ft

8.5 ft

12 ft 7.5 ft

Two cables are tiedtogether at C andloaded as shown.determine the tension(a) in cable AC,(b) in cable BC.

B

C

396 lb

50


Two cables are tiedtogether at C andloaded as shown.determine the tension(a) in cable AC,(b) in cable BC.

A

9 ft

5 ft

8.5 ft

12 ft 7.5 ft

B

C

396 lb



Problem 2.10

51

x

y

TAC

TBC

4

7.512

3.58.512.5

396 lb

FREE BODY C:

TBC = 0 Fx = 0 : 1212.5

TAC7.58.5+

TBC = 1.088 TAC

Fy = 0 : TAC 48.5+

3.512.5 TBC 396 lb = 0


Draw a free-body diagramof the particle.

Set the resultant, or sum, ofthe forces exerted on theparticle equal to zero.

52

x

y

TAC

TBC

4

7.512

3.58.512.5

396 lb

(a) Substitute for TBC:

TAC 48.5+

3.512.5 (1.088 TAC )

_ 396 lb = 0

(0.280 + 0.512) TAC _ 396 lb = 0

(b) TBC = 1.088 (500 lb)

TAC = 500 lb

TBC = 544 lb


53

Problem 2.11

Collars A and B are connectedby a 25-in.-long wire and canslide freely on frictionless rods.If a 60-lb force Q is applied tocollar B as shown, Determine(a) the tension in the wire whenx = 9 in., (b) the correspondingmagnitude of the force P requiredto maintain the equilibrium ofthe system.

yx

x

20 in

A

B

O

z

z

P

Q

54

yx

x

20 in

A

B

O

z

z

P

Q




Problem 2.11

55


yx

x

20 in

A

B

O

z

z

P

Q



Problem 2.11

56

Problem 2.11 Solutionyx

x

20 in

A

B

O

z

z

P

Q

AB = ABAB =

_ x i _ (20 in) j + z k25 in

Draw a free-body diagram of the particle.

Ny j

Nz k

TAB AB

P i

A

Free Body: Collar A

F = 0: P i + Ny j + Nz k + TAB AB = 0

Substitute for AB and set coefficients

of i equal to zero:

P _ TAB x

25= 0 (1)

57

Problem 2.11 SolutionNy j

Nx i

_ TAB AB

Q = (60 lb) k

Free Body: Collar B

F = 0: (60 lb) k + Nx i + Ny j _ TAB AB = 0

Substitute for AB and set coefficients

of k equal to zero:

60 _ TAB z

25= 0 (2)

(a) Since x = 9 in.: (9 in)2 + (20 in) 2 + z 2 = (25 in) 2 z = 12 in

From eq. (2): 60 _ TAB (12)

25= 0 TAB = 125.0 lb

(b) From eq. (1): P =(125.0 lb)(9 in)

25 inP = 45.0 lb

B

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition 2 - 1 CE 102 Statics Chapter 2 Statics