5.3 Equations of Equilibrium For equilibrium of a rigid body in 2D, ∑F x = 0; ∑F y = 0; ∑M O = 0 ∑F x and ∑F y represent the algebraic sums of the x and

5.3 Equations of Equilibrium5.3 Equations of Equilibrium

For equilibrium of a rigid body 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 body

∑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 body


Alternative Sets of Equilibrium Equations For coplanar equilibrium problems, ∑Fx = 0;

∑Fy = 0; ∑MO = 0 can be used Two alternative sets of three independent

equilibrium equations may also be used∑Fa = 0; ∑MA = 0; ∑MB = 0

When applying these equations, it is required that a line passing through points A and B is not perpendicular to the a axis


Alternative Sets of Equilibrium EquationsConsider FBD of an arbitrarily shaped

bodyAll the forces on FBD may be

replaced by an equivalent resultant force FR = ∑F acting at point A and a

resultant moment MRA = ∑MA If ∑MA = 0 is satisfied, MRA = 0


Alternative Sets of Equilibrium Equations If FR satisfies ∑Fa = 0, there is no

component along the a axis and its line of axis is perpendicular to the a axis

If ∑MB = 0 where B does not lies

on the line of action of FR, FR = 0

Since ∑F = 0 and ∑MA = 0, the

body is in equilibrium

Alternative Sets of Equilibrium Equations A second set of alternative equations is

∑MA = 0; ∑MB = 0; ∑MC = 0 Points A, B and C do not lie on the

same line Consider FBD, if If ∑MA = 0, MRA = 0 ∑MA = 0 is satisfied if line of action of FR

passes through point B ∑MC = 0 where C does not lie on line AB FR = 0 and the body is in equilibrium



Procedure for AnalysisFree-Body Diagram Establish the x, y, z coordinates axes in

any suitable orientation Draw an outlined shape of the body Show all the forces and couple

moments acting on the body Label all the loadings and specify their

directions relative to the x, y axes


Procedure for AnalysisFree-Body DiagramThe sense of a force or couple

moment having an unknown magnitude but known line of action can be assumed

Indicate the dimensions of the body necessary for computing the moments of forces


Procedure for AnalysisEquations of Equilibrium Apply the moment equation of

equilibrium ∑MO = 0 about a point O that lies on the intersection of the lines of action of the two unknown forces

The moments of these unknowns are zero about O and a direct solution the third unknown can be obtained


Procedure for AnalysisEquations of Equilibrium When applying the force equilibrium ∑Fx

= 0 and ∑Fy = 0, orient the x and y axes along the lines that will provide the simplest resolution of the forces into their x and y components

If the solution yields a negative result scalar, the sense is opposite to that was assumed on the FBD


Example 5.6Determine the horizontal and vertical components of reaction for the beam

loaded. Neglect the weight of the beam in the calculations.


SolutionFBD 600N force is represented by its x and y

components 200N force acts on the beam at B and is

independent of the force components Bx and By, which

represent the effect of the pin on the beam


SolutionEquations of Equilibrium

A direct solution of Ay can be obtained by applying ∑MB = 0 about point B

Forces 200N, Bx and By all create zero moment about B

NB

BN

M

x

x

B

424

045cos600

;0


Solution

NB

BNNNN

F

NA

mAmNmNmN

M

y

y

y

y

y

B

405

020010045sin600319

;0

319

0)7()2.0)(45cos600()5)(45sin600()2(100

;0


SolutionChecking,

NB

mBmN

mNmNmN

M

y

y

A

405

0)7()7)(200(

)5)(100()2.0)(45cos600()2)(45sin600(

;0


Example 5.7The cord supports a force of 500N and wraps

over the frictionless pulley. Determine the tension

in the cord at C and the horizontal and vertical components at pin A.


SolutionFBD of the cord and pulley Principle of action: equal but opposite

reaction observed in the FBD Cord exerts an unknown load

distribution p along part of the pulley’s surface

Pulley exerts an equal but opposite effect on the cord


SolutionFBD of the cord and pulley Easier to combine the FBD of the pulley

and contracting portion of the cord so that the distributed load becomes internal to the system and is eliminated from the analysis



Tension remains constant as cord passes over the pulley (true for any angle at which the cord is directed and for any radius of the pulley

NT

mTmN

M A

500

0)2.0()2.0(500

;0


Solution

NA

NNA

F

NAx

NA

F

y

y

y

x

x

933

030cos500500

;0

250

030sin500

;0


Example 5.8The link is pin-connected at a and rest a smooth support at B. Compute the

horizontal and vertical components of reactions at pin

A


SolutionFBDReaction NB is perpendicular to the

link at BHorizontal and vertical

components of reaction are represented at A



NAx

NA

F

NN

mNmNmN

M

x

x

B

B

A

100

030sin200

;0

200

0)75.0()1(60.90

;0


Solution

NA

NNA

F

y

y

y

233

030cos20060

;0


Example 5.9The box wrench is used to tighten the bolt at A. If the wrench does not turn when the load is applied to the handle, determine the torque or moment applied to the bolt and the force of the wrench on the bolt.


SolutionFBDBolt acts as a “fixed support” it exerts

force components Ax and Ay and a torque MA on the wrench at A



NAx

NNA

F

mNM

mNmNM

M

x

x

A

A

A

00.5

060cos3013

552

;0

.6.32

0)7.0)(60sin30()3.0(13

1252

;0


Solution

NmM

mNmNM

MCCW

NA

NNA

F

A

A

y

y

y

y

6.32

07.060sin303.013

1252

;0

0.74

060sin3013

1252

;0


Solution Point A was chosen for summing the

moments as the lines of action of the unknown forces Ax and Ay pass through this point and these forces are not included in the moment summation

MA must be included

Couple moment MA is a free vector and represents the twisting resistance of the bolt on the wrench


Solution By Newton’s third law, the wrench

exerts an equal but opposite moment or torque on the bolt

For resultant force on the wrench,

For directional sense,

FA acts in the opposite direction on the bolt

1.8600.5

0.74tan

1.740.7400.5

1

22

N

N

NFA


Solution Checking,

0.8.51.6.32.2.19

0)7.0(0.74.6.32)4.0(1312

52

;0

mNmNmN

mNmNmN

MC

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.

5.3 Equations of Equilibrium

5.3 Equations of Equilibrium


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


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


Solution

NAx

NA

F

NF

mN

mNmF

M

x

x

BC

BC

A

7900

07900

;0

7900

0)0625.0(30sin2240

)2(30cos2240)5.0(

;0


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


Example 5.11The uniform smooth rod is subjected to a force

and couple moment. If the rod is supported at A by

a smooth wall and at B and C either at the top or bottom by rollers, determine the reactions at these supports. Neglect the weight of the rod.


SolutionFBD All the support reactions act normal to the

surface of contact since the contracting surfaces are smooth

Reactions at B and C are acting in the positive y’ direction

Assume only the rollers located on the bottom of the rod are used for support



030cos30cos300

;0

030sin30sin

;0

0)8)(30cos300()6(.4000)2(

;0

''

''

''

yy

y

xyy

x

yy

A

BCN

F

ABC

F

mNmCmNmB

M


Solution Note that the line of action of the force

component passes through point A and this force is not included in the moment equation

Since By’ is negative scalar, the sense of By’ is opposite to shown in the FBD

kNNC

kNNB

y

y

35.14.1346

10.1000

'

'


Solution Top roller at B serves as the support

rather than the bottom one

NA

ANN

x

x

173

030sin0.100030sin4.1346


Example 5.12The 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.


Solution Idealized model of the ramp Center of gravity located at the midpoint since

the ramp is approximately uniform

FBD of the Ramp

View Free Body Diagram



By the principle of transmissibility, locate T at C

md

md

NT

N

TmT

M A

0154.120sin

2

10sin

5985

0)30cos5.1(1600

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

;0


SolutionSince there are two cables

supporting the ramp,

T’ = T/2 = 2992.5N

5.4 Two- and Three-Force Members

5.4 Two- and Three-Force Members

Simplify some equilibrium problems by recognizing embers 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

Documents

5.3 Equations of Equilibrium For equilibrium of a rigid body in 2D, ∑F x = 0; ∑F y = 0; ∑M O = 0 ∑F x and ∑F y represent the algebraic sums of the x and