Static Force Analysis

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EME 2056 Theory of Machines

EME 2056 Theory of Machines

Chapter 5:

Static Force Analysis

Lectured by: K.W. Liew

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Learning Outcome of Chapter 5Learning Outcome of Chapter 5

Analyse the static force of the linkage and slider-crank mechanism.

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ContentsContents

5.1. Introduction5.2. Forces5.3. Moment of a Force5.4. Free-Body Diagrams5.5. Static Equilibrium5.6. Two- and Three-Force Members5.7. Force Analysis5.8. Sliding Friction Force

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5.1 Introduction5.1 Introduction

The general function of any machine is to transmit motion and forces from an actuator to the components that perform the desired task.

Example:Escalator used in many commercial building –

Electrical power is fed into motors, which drive mechanisms that move and fold the stairs. The task is to safely and efficiently move people up and down at multilevel buildings.

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A critical task in the design of machines is to ensure that the strength of the links and joints is sufficient to withstand the forces imposed on them. Therefore, a full understanding of the forces in the various components of a machine is vital.

This chapter deal with the force analysis in mechanisms without accelerations, or where the accelerations can neglected. This condition is termed static equilibrium.

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Static equilibrium is applicable in many machines where movement is relatively slow.

Examples: clamps, latches, support linkages, and many hand-operated tools, such as pliers and cutters.

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5.2 Forces5.2 Forces

A force is a vector quantity that represents a pushing or pulling action on a part.

Being a vector, this force is defined by a magnitude and a direction of the pulling action.

Unit SI, the primary unit used is the Newton (N).

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Scalar Notation-

Since the x and y axes are designated positive

and negative directions.-

The magnitude and directional sense of the

rectangular components of a force can be expressed in terms of algebraic scalars.Eg: Sense of direction along positive x and y axes

yx FFF


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Scalar NotationEg: Sense of direction along positive x and negative y axes

yx FFF '''


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Scalar Notation-

Head of a vector arrow = sense of the

vector graphically (algebraic signs not used)-

Vectors are designated using boldface

notations-

Magnitudes (always a positive quantity)

are designated using italic symbols


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Cartesian Vector Notation-

Cartesian unit vectors i

and j

are used to designate

the x and y directions

-

Unit vectors i

and j

have dimensionless magnitude of unity ( = 1 )

-

Their sense are indicated by a positive or negative sign (pointing in the positive or negative x or y axis)

-

Magnitude is always a positive quantity, represented by scalars Fx and Fy


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Cartesian Vector Notation F

= Fx i

+ Fy j F’ = F’x i

+ F’y(-j)

F’ = F’x i

–

F’y j


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Coplanar Force Resultants To determine resultant of several coplanar

forces:-

Resolve force into x and y components

-

Addition of the respective components using scalar algebra


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Coplanar Force Resultants Example: Consider three coplanar

forces

Cartesian vector notationF1

= F1x i + F1y j

F2

= -

F2x i + F2y j

F3

= F3x i –

F3y j


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Coplanar Force Resultants Vector resultant is therefore

FR

= F1

+ F2

+ F3

= F1x i

+ F1y j -

F2x i

+ F2y j + F3x i

–

F3y j= (F1x -

F2x + F3x )i

+ (F1y + F2y –

F3y )j

= (FRx )i

+ (FRy )j


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Coplanar Force Resultants If scalar notation are used

FRx = (F1x -

F2x + F3x )FRy = (F1y + F2y –

F3y )

In all cases,FRx = ∑FxFRy = ∑Fy

*

Take note of sign conventions


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Coplanar Force Resultants -

Positive scalars = sense of direction

along the positive coordinate axes-

Negative scalars = sense of direction

along the negative coordinate axes-

Magnitude of FR

can be found by Pythagorean Theorem

RyRxR FFF 22


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Coplanar Force Resultants -

Direction angle θ

(orientation of the force)

can be found by trigonometry

Rx

Ry

FF1tan


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Example 5.1The link is subjected to two forces F1

and F2

. Determine the magnitude and orientation of the resultant force.


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SolutionScalar Notation

N

NNF

FFN

NNF

FF

Ry

yRy

Rx

xRx

8.582

45cos40030sin600

:8.236

45sin40030cos600

:


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SolutionResultant Force

From vector addition,Direction angle θ

is

N

NNFR

6298.5828.236 22

9.678.2368.582tan 1

NN


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SolutionCartesian Vector NotationF1

= { 600cos30°i

+ 600sin30°j

} NF2

= { -400sin45°i

+ 400cos45°j

} N

Thus, FR

= F1

+ F2

= (600cos30°N -

400sin45°N)i

+ (600sin30°N + 400cos45°N)j

= {236.8i

+ 582.8j}N


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Moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis.

Case 1Consider horizontal force Fx

, which acts perpendicular to the handle of the wrench and is located a distance dy from the point O

5.3 Moment of a Force – Scalar Formulation


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Fx

tends to turn the pipe about the z axis

The larger the force or the distance dy , the greater the turning effect

Torque

–

tendency of rotation caused by Fx

or simple moment (Mo

) z



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Moment axis (z) is perpendicular to shaded plane (x-y)

Fx

and dy lies on the shaded plane (x-y)

Moment axis (z) intersects the plane at point O



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Case 2Apply force Fz

to the wrench

Pipe does not rotate about z axis

Tendency to rotate about x axis

The pipe may not actually rotate Fz

creates tendency for rotation so moment (Mo

) x

is produced



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Case 2

Moment axis (x) is perpendicular to shaded plane (y-z)

Fz

and dy lies on the shaded plane (y-z)



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Case 3Apply force Fy

to the wrench

No moment is produced about point O

Lack of tendency to rotate as line of action passes through O

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In General

Consider the force F

and the point O which lies in the

shaded plane

The moment MO

about point O, or about an axis passingthrough O and perpendicularto the plane, is a vector quantity

Moment MO

has its specified magnitude and direction

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Magnitude

For magnitude of MO

,MO

= Fdwhere d = moment arm

or perpendicular

distance

from the axis at point O to its line of action of the force

Units for moment is N.m

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Direction

The direction of MO

is specified by using “right hand rule”

The thumb points along the moment axis to give the direction and sense of the moment vector

Moment vector is upwards and perpendicular to the shaded plane

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Direction

MO

is shown by a vector arrow with a curl to distinguish it from force vectorExample (Fig b)

MO

is represented by the counterclockwise curl, which indicates the action of F. The arrowhead of the curl shows the sense of rotation caused by F.

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Direction

Moment always acts about an axis perpendicular to the plane containing F and d

Moment axis intersects the plane at point O

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Resultant Moment of a System of Coplanar Forces

Resultant moment, MRo = addition of the moments of all the forces algebraically since all moment forces are collinear

MRo = ∑Fdtaking counterclockwise to be positive

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Resultant Moment of a System of Coplanar Forces

A counterclockwise curl is written along the equation to indicate that a positive moment if directed along the + z axis and negative along the – z axis

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Example 5.2For each case, determine the moment of the force about point O

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Solution

Line of action is extended as a dashed line to establish moment arm d

Tendency to rotate is indicated and the orbit is shown as a colored curl

)(.5.37)75.0)(50()()(.200)2)(100()(

CWmNmNMbCWmNmNMa

o

o

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Solution

)(.4.42)45sin1)(60()(

)(.229)30cos24)(40()(

CCWmNmNMd

CWmNmmNMc

o

o

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Example 5.3Determine the moments of the 800N force acting on the frame about points A, B, C and D.

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SolutionScalar Analysis

Line of action of F passes through C)(.400)5.0)(800(

.0)0)(800()(.1200)5.1)(800()(.2000)5.2)(800(

CCWmNmNM

mkNmNMCWmNmNMCWmNmNM

D

C

B

A

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5.3 Moment of a Force - Vector Formulation


Moment of force F about point O can be expressed using cross product

MO = r X Fwhere r represents position vector from O to any pointlying on the line of action of F

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Magnitude

For magnitude of cross product,

MO = r F sinθwhere θ

is the angle measured between

tails of r and F

Treat r as a sliding vector. Since d =r sinθ, MO = r F sinθ

= F (r sinθ) = Fd

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Direction

Direction and sense of MO are determined by right-hand rule - Extend r to the dashed position - Curl fingers from r towards F- Direction of MO is the same as the direction of the thumb

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Direction*Note:

- “curl” of the fingers indicates the sense of rotation- Maintain proper order of rand F since cross product is not commutative

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Cartesian Vector Formulation

For force expressed in Cartesianform,

where rx , ry , rz represent the x, y, zcomponents of the position vectorand Fx , Fy , Fz represent that of the force vector

zyx

zyxO

FFFrrrkji

FXrM

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With the determinant expended, MO = (ryFz – rzFy )i – (rxFz - rzFx )j + (rxFy – yFx )k

MO is always perpendicular to the plane containing r and F

Computation of moment by cross product is better than scalar for 3D problems

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Resultant moment of forces about point O can be determined by vector addition

MRo = ∑(r x F)

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5.4 Free-Body Diagrams5.4 Free-Body Diagrams

FBD is the best method to represent all the known and unknown forces in a system.

FBD is a sketch of the outlined shape of the body, which represents it being isolated or free from its surroundings.

Necessary to show all the forces and couple moments that the surroundings exert on the body so that these effects can be accounted for when equations of equilibrium are applied.

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Support ReactionsEx: Pin

The pin passes through a hold in the beam and two leaves that are fixed to the ground

Prevents translation of the beam in any direction Φ

The pin exerts a force F on the beam in this direction

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Weight and Center of Gravity

When a body is subjected to gravity, each particle has a specified weight

For entire body, consider gravitational forces as a system of parallel forces acting on all particles within the boundary

The system can be represented by a single resultant force, known as weight W of the body

Location of the force application is known as the center of gravity

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Weight and Center of Gravity

Center of gravity occurs at the geometric center or centroid for uniform body of homogenous material

For non-homogenous bodies and usual shapes, the center of gravity will be given

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Idealized Models

When an engineer perform a force analysis of any object, he or she considers a corresponding analytical or idealized model that gives results that approximate as closely as possible the actual situation.

To do this, careful selection of supports, material behavior and object’s dimensions are needed for trusty results.

In complex cases may require developing several different models for analysis.

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Procedure for Drawing a FBD1. Draw Outlined Shape

Imagine body to be isolated or cut free from its constraints

Draw outline shape

2. Show All Forces and Couple Moments

Identify all external forces and couple moments that act on the body

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Procedure for Drawing a FBD

Usually due to - applied loadings- reactions occurring at the supports or at points of contact with other body- weight of the body

To account for all the effects, trace over the boundary, noting each force and couple moment acting on it

3. Identify Each Loading and Give Dimensions

Indicate dimensions for calculation of forces

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Procedure for Drawing a FBD

Known forces and couple moments should be properly labeled with their magnitudes and directions

Letters used to represent the magnitudes and direction angles of unknown forces and couple moments

Establish x, y and coordinate system to identify unknowns

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Example 5.4Draw the free-body diagram of the foot lever. The operator applies a vertical force to the pedal so that the spring is stretched 40mm and the force in the short link at B is 100N.

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Solution

Lever loosely bolted to frame at A

Rod at B pinned at its ends and acts as a short link

For idealized model of the lever,

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Solution

Free-Body Diagram

Pin support at A exerts components Ax and Ay on the lever, each force with a known line of action but unknown magnitude

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Solution

Link at B exerts a force 100N acting in the direction of the link

Spring exerts a horizontal force on the leverFs = ks = 5N/mm(40mm) = 200N

Operator’s shoe exert vertical force F on the pedal

Compute the moments using the dimensions on the FBD

Compute the sense by the equilibrium equations

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Example 5.5Draw the free-body diagram of the unloaded platform that is suspended off the edge of the oil rig. The platform has a mass of 200kg.

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Solution

Idealized model considered in 2D because by observation, loading and the dimensions are all symmetrical about a vertical plane passing through the center

Connection at A assumed to be a pin and the cable supports the platform at B

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Solution

Direction of the cable and average dimensions of the platform are listed and center of gravity has been determined

Free-Body Diagram

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Solution

Platform’s weight = 200(9.81) = 1962N

Force components Ax and Ay along with the cable force T represent the reactions that both pins and cables exert on the platform

Half of the cables magnitudes is developed at A and half developed at B

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Example 5.6The free-body diagram of each object is drawn. Carefully study each solution and identify what each loading represents.

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Solution

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Solution

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Example 5.7Draw the free-body diagrams of each part of the smooth piston and link mechanism used to crush recycled cans.


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Solution

Member AB is a two force member

FBD of the parts


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Solution

Since the pins at B and D connect only two parts together, the forces are equal but opposite on the separate FBD of their connected members

Four components of the force act on the piston: Dx and Dy represent the effects of the pin and Nw is the resultant force of the floor and P is the resultant compressive force caused by can C


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Example 5.8Draw the free-body diagrams of the bucket and the vertical boom of the back hoe. The bucket and its content has a weight W. Neglect the weight of the members.


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Solution

Idealized model of the assembly

Members AB, BC, BE and HI are two force members


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Solution

FBD of the bucket and boom

Pin C subjected to 2 forces, force of the link BC and force of the boom

Pin at B subjected to three forces, force by the hydraulic cylinder and the forces caused by the link

These forces are related by equation of force equilibrium


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5.5 Static Equilibrium5.5 Static Equilibrium

Static Equilibrium – all links that are at rest or moving at constant velocity.

Conditions for an object to be in static equilibrium:

1. The resultant, of all external forces acting on the object is equivalent to zero and does not cause it to translate.

2. The moment due to any external force is cancelled by the moments of the other forces acting on the object and do not cause it to rotate about any point.

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5.5 Static Equilibrium5.5 Static Equilibrium

For equilibrium of an object in 2D, ∑Fx

= 0; ∑Fy

= 0; ∑MO

= 0

∑Fx

and ∑Fy

represent the algebraic sums of the x and y components of all the forces acting on the object.

∑MO

represents the algebraic sum of the couple moments and moments of the force components about an axis perpendicular to x-y

plane and

passing through arbitrary point O, which may lie on or off the object.

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5.6 Two- and Three-Force Members


Simplify some equilibrium problems by recognizing members that are subjected top only 2 or 3 forces

Two-Force Members

When a member is subject to no couple moments and forces are applied at only two points on a member, the member is called a two-force member

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Two-Force MembersExample

Forces at A and B are summed to obtain their respective resultants FAand FB

These two forces will maintain translational and force equilibrium provided FA is of equal magnitude and opposite direction to FB

Line of action of both forces is known and passes through A and B

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Two-Force Members

Hence, only the force magnitude must be determined or stated

Other examples of the two- force members held in equilibrium are shown in the figures to the right

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Three-Force Members

If a member is subjected to only three forces, it is necessary that the forces be either concurrent or parallel for the member to be in equilibrium

To show the concurrency requirement, consider a body with any two of the three forces acting on it, to have line of actions that intersect at point O

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Three-Force Members

To satisfy moment equilibrium about O, the third force must also pass through O, which then makes the force concurrent

If two of the three forces parallel, the point of currency O, is considered at “infinity”

Third force must parallel to the other two forces to insect at this “point”

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Bucket link AB on the back hoe is a typical example of a two-force member since it is pin connected at its end provided its weight is neglected, no other force acts on this member

The hydraulic cylinder BC is pin connected at its ends, being a two-force member.

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The boom ABD is subjected to the weight of the suspended motor at D, the forces of the hydraulic cylinder at B, and the force of the pin at A. If the boom’s weight is neglected, it is a three-force member

The dump bed of the truck operates by extending the hydraulic cylinder AB. If the weight of AB is neglected, it is a two-force member since it is pin-connected at its end points

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5.7 Force Analysis5.7 Force Analysis

Example 5.9The link is pin-connected at a and rest a smooth support at B. Compute the horizontal and vertical components of reactions at pin A

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SolutionFBD

Reaction NB is perpendicular to the link at B

Horizontal and vertical components of reaction are represented at A

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SolutionEquations of Equilibrium

NAxNA

FNN

mNmNmNM

x

x

B

B

A

100030sin200

;0200

0)75.0()1(60.90;0

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Solution

NA

NNA

F

y

y

y

233

030cos20060

;0

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Example 5.10Placement of concrete from the truck is accomplished using the chute. Determine the force that the hydraulic cylinder and the truck frame exert on the chute to hold it in position. The chute and the wet concrete contained along its length have a uniform weight of 560N/m.


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Solution

Idealized model of the chute

Assume chute is pin connected to the frame at A and the hydraulic cylinder BC acts as a short link

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SolutionFBD

Since chute has a length of 4m, total supported weight is (560N/m)(4m) = 2240N, which is assumed to act at its midpoint, G

The hydraulic cylinder exerts a horizontal force FBC on the chute

Equations of Equilibrium

A direct solution of FBC is obtained by the summation about the pin at A

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Solution

NAxNA

FNF

mN

mNmF

M

x

x

BC

BC

A

790007900

;07900

0)0625.0(30sin2240

)2(30cos2240)5.0(

;0

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Solution

Checking,

0)0625.0(30sin2240)1(30cos2240

)30cos1(2240)5.0(7900

;0

2240

02240

;0

mNmN

mNmN

M

NA

NA

F

B

y

y

y

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Example 5.11The uniform truck ramp has a weight of 1600N ( ≈

160kg ) and is pinned to the body

of the truck at each end and held in position by two side cables. Determine the tension in the cables.

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Solution

Idealized model of the ramp

Center of gravity located at the midpoint since the ramp is approximately uniform

FBD of the Ramp

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SolutionEquations of Equilibrium

By the principle of transmissibility, locate T at C

md

md

NTN

TmT

M A

0154.120sin

210sin

59850)30cos5.1(1600

)30cos2(20sin)30sin2(20cos

;0

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SolutionSince there are two cables supporting the ramp,

T’ = T/2 = 2992.5N

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Example 5.12The lever ABC is pin-supported at A and connected to a short link BD. If the weight of the members are negligible, determine the force of the pin on the lever at A.

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Solution FBD

Short link BD is a two-force member, so the resultant forces at pins D and B must be equal, opposite and collinear

Magnitude of the force is unknown but line of action known as it passes through B and D

Lever ABC is a three-force member

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Solution FBD

For moment equilibrium, three non-parallel forces acting on it must be concurrent at O

Force F on the lever at B is equal but opposite to the force F acting at B on the link

Distance CO must be 0.5m since lines of action of F and the 400N force are known

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Solution Equations of

Equilibrium

Solving,

kNFkNF

FF

FNFF

F

A

A

y

A

x

32.107.1

045sin3.60sin

;0040045cos3.60cos

;0

3.604.07.0tan 1

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Example 5.13The hand exerts a force of 35N on the grip of the spring compressor. Determine the force in the spring needed to maintain equilibrium of the mechanism.

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Solution

FBD for parts DC and ABG

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SolutionLever ABG

Pin E

NFNFF

FFF

FFF

NFmmNmmFM

x

EFED

EFEAy

EA

EAB

140014060cos2;0

060sin60sin;0

1400)100(35)25(;0


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SolutionArm DC

NFmmmmF

M

s

s

C

62.600)75(30cos140)150(

;0


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5.8 Sliding Friction Force5.8 Sliding Friction Force

When friction force cannot be neglected in a machine analysis, an additional force, friction force, Ff , is observed.

Friction always acts to impede motion. So, a friction force acts on a sliding link (perpendicular to the normal force, N) and in a direction opposite to the motion.

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5.8 Sliding Friction Force5.8 Sliding Friction Force

For a stationary object, friction works to prevent motion until the maximum attainable friction has been reached.

This maximum value is a function of a coefficient of friction, .

The friction force value that acts on sliding components is calculated as

Ff = N

Documents

Static Force Analysis