EM025 CHAPTER 2 ENGINEERING MECHANICS

Example 3 :
The members of a truss are pin connected at O. Determine the
magnitude of F1 and F2 for equilibrium. Set θ = 60°.

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SOLUTION : →∑FX = 0
F₁ cos 60°+ F₂ sin 70°- 5 cos 30°- 7(4/5) = 0

F₁ (0.5) + F₂ (0.939) = 9.93 ………………i

↑∑FY =0
F₂ cos 70° -F₁ sin 60°+ 5 sin 30°- 7(3/5) = 0

F₂ (0.342) - F₁ (0.866)=1.70 ………….ii

► F₁ = 1.83kN 52
► F₂ = 9.6kN

2.3 MOMENT

a) Define concept of moment in engineering problem.
b) Apply concept of moment in engineering problem.
c) Apply couple moment concept by using FBD related to

engineering problem limited to 2D system only.

DEFINITION

Moment is the tendency of a force to cause rotation
about a point or around an axis. It is also called a
torque or twist moment that tendency of a force to
rotate a body about the axis. It is a vector, so its has
both magnitude and direction (right hand rule)

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APPLICATION OF MOMENT

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APPLICATION OF MOMENT

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MOMENT OF A FORCE IN 2D

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The moment, M, of a force about a point provides a measure of the
tendency for rotation. the direction of MO is either clockwise or
counter-clockwise, depending on the tendency for rotation.

Mo = F.d 58

The units of a Moment are N·m in the SI system

Example of situation:
• Force F tends to rotate the beam clockwise about A with

moment, MA = FdA
• Force F tends to rotate the beam counterclockwise

about B with moment, MB = FdB
• Hence support at A and B prevents the rotation

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Problem & Solution

Example 1

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

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Solution:

Mo = (100) (2)
= 200 N.m ()

Mo = (50) (0.75)
= 37.5 N.m ()

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Solution:

Mo = (40) (4 + 2 kos 30˚)
= 229.28 N.m ()

Mo = (60) (1 sin 45˚)
= 42.43 N.m ()

Mo = (7k) (3)
= 21k N.m ()

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Example 2

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

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Solution: = (800) (2.5)
MA = 2000 N.m ()

MB = (800) (1.5)
= 1200 N.m ()

MC = (800) (0)
= 0 N.m

MD = (800) (0.5)
= 400 N.m ()

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Example 3
Determine the resultant moment of the four forces

acting on the rod about point O.

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Solution:

 Mo = F.d
= (50)(2) + 60(0) + 40 (4+3 cos 30˚) – 20 (3 cos30˚)
= 334 N.m ()

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COUPLE MOMENT

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INTRODUCTION

• A special case of moments is a couple. A couple consists of:
- two parallel forces
- same magnitude
- opposite direction
- separated by perpendicular distance (d)

• It does not produce any translation, only rotation.
• The resultant force of a couple is zero.

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APPLICATION OF COUPLE MOMENT

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Scalar Formulation
• Magnitude of couple moment : M = Fd
• Direction and sense are determined by right hand rule
• In all cases, M acts perpendicular to plane containing the forces

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Equivalent Couples

• Every set of forces and moments has an equivalent force couple
system.

• This is a single force and pure moment (couple) acting at a single 71
point that is statically equivalent to the original set of forces and
moments.

• Two couples are equivalent if they produce the same moment

• Since moment produced by the couple is always perpendicular to
the plane containing the forces, forces of equal couples either lie
on the same plane or plane parallel to one another

Example 1 :
Determine the moment of the couple acting on the member.

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Solution (method 1)
• Resolve each force

Fx = 4/5(150kN) = 120kN Fy = 3/5(150kN) = 90kN

• Principle of Moment about point D, 73
MD = 120kN(0m) + 90kN(2m) - 90kN(5m) - 120kN(1m)
= 390kN

Solution (method 2)
• Principle of Moment about point A,

MA = 90kN(3m) + 120kN(1m)
= 390kN

*Note:

- Same result if take moment about point B

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Note:
• Couple in figure A can be replaced by two couple in figure B.
• M is a free vector and acts on any point on the member as in

figure C

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Example 2 :
Two couples act on the beam. Determine the magnitude of F so that
the resultant couple moment is 450 N.m, counterclockwise. Where
on the beam does the resultant couple moment act?

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Solution 77

 MR = ∑M
F cos 30ᵒ (0.5) + 200 (0.6) = 450

F = 762 N

 The resultant couple moment is a free vector. It can act at any
point on the beam.

2.4 EQUILIBRIUM OF RIGID BODY IN TWO
DIMENSION (2D)

a) Describe the concept of rigid body in engineering practice.

b) Identify the differences between particle and rigid body.

c) Draw the diagram that reflects existing forces by applying FBD
concepts.

d) Solve the force system using concept of rigid body equilibrium

INTRODUCTION

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CONCEPT

• Rigid body is an idealized extended solid whose size and shape
are definitely fixed and remain unaltered when forces are
applied.

• A body which is currently stationary will remain stationary if
the resultant force and resultant moment are zero for all the
forces and couples applied on it.

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DIFFERENCE BETWEEN
PARTICLE AND RIGID BODY

EQUILIBRIUM OF PARTICLE EQUILIBRIUM OF RIGID BODY

• Made up from one point. • Can be in any shape.
• Has no dimensions, like a • Dimension is very

point in geometry. important.
• Equations : • Equations :

∑Fx = 0 ∑Fx = 0
∑Fy = 0 ∑Fy = 0
∑MO = 0

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PROCEDURE FOR ANALYSIS

1. Free-Body Diagram
• Draw an outlined shape of the body.
• Establish the x and y coordinates.
• Show all the forces acting on the body.
• Label all the loadings and specify their directions.
• Indicate the dimensions of the body.

2. Equations of Equilibrium 82
• Moment equation ∑MO= 0.
• Force equation ∑Fx = 0 and ∑Fy = 0.
• iIsf tohpepsoosliutetitoontyhiaetldwsaasnaesgsuatmiveedroensutlhtescFaBlaDr., the sense

THE PROCESS OF SOLVING RIGID BODY
EQUILIBRIUM PROBLEMS

For analyzing an actual physical system:
• Create an idealized model and it separate from surroundings.
• Draw a free-body diagram (FBD) showing all the external (active

and reactive) forces.
• Apply the equations of equilibrium to solve for any unknowns.

Idealized Model

Physical System Free-body Diagram 83

SUPPORT REACTIONS

• If the support prevents the translation of a body in
a given direction, then a force is developed on the
body in that direction.

• If rotation is prevented, a couple moment is exerted
on the body.

• Consider the three ways a horizontal member, beam is
supported at the end:
- roller/cylinder/rocker/smooth surface
- pin/hinge
- fixed support
- cable

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Roller/Cylinder/Rocker/ Smooth contact surface

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1.Roller/Cylinder/Rocker/ Smooth contact surface

• Prevent t he beam from translating in the vertical
direction.

• Can only exerts a force on the beam in the vertical
direction.

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Pin/ Hinge

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2.Pin/ Hinge

• The pin/hinge 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/hinge exerts a force F on the beam in this direction.

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Fixed Support

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3.Fixed Support

• This support prevents both translation and rotation of
the beam.

• A couple and moment must be developed on the beam at its
point of connection

• Force is usually represented in x and y components.

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Cable

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4.Cable

• The reaction is a tension force which acts away from the
members in the direction of the cable.

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Example 1 :

Determine the horizontal and vertical components of reaction for
the beam loaded. A is a rocket and B is a pin. Neglect the weight
of the beam in the calculations.

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Solution:
FBD

• 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,represent the effect of the pin on the beam.
• Ay, represent the effect of the rocker on the beam.

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Solution: 95

→ Ʃ FX = 0
0 = 600 COS 45˚ – BX
BX = 424.26 N

↑Ʃ FY = 0
0 = AY + BY – 600 SIN 45˚ – 100 – 200

AY + BY = 724.26 N  I
 MB = 0

0 = AY(7)–100(2) – 600 SIN 45˚(5)+600 COS 45˚(0.2)
AY = 319.49 N
INSERT AY= 319.49 N INTO EQUATION I
319.49 + BY = 724.26
BY = 404.77 N

Example 2 :

The 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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Solution : 97

 M A = 0;
− 90N.m − 60N (1m) + NB (0.75m) = 0
NB = 200N

+ →  Fx = 0;
Ax − 200sin 30 N = 0
Ax = 100N

+   Fy = 0;
Ay − 60N − 200 cos30 N = 0
Ay = 233N

Example 3 :

The 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.

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Solution :

 M A = 0;

M A − 52 12  N (0.3m) − (30 sin 60 N )(0.7m) = 0
 13 

M A = 32.6N.m

+ →  Fx = 0;

Ax − 52 5 N + 30 cos60 N = 0
 13 

Ax = 5.00N

+   Fy = 0;

Ay − 5212  N − 30 sin 60 N = 0
 13 
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Ay = 74.0N

Problem 4 :

The uniform truck ramp has a weight of 1600N 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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