Engineering Engineering FundamentalsFundamentalsEngineering Engineering

FundamentalsFundamentals

Session 9Session 9

Equilibrium• A body is in Equilibrium if it moves with

constant velocity. A body at rest is a special case of constant velocity i.e. v = 0 = constant.

• For a body to be in Equilibrium the resultant force (meaning the vector addition of all the forces) acting on the body must be zero.

• Resulting force = vector addition of force vectors

• A Force can be defined as 'that which tends to cause a particle to accelerate‘.

Equilibrium of Concurrent Forces

Force

F1

F2 Resultant R

Equilibrant E Equilibrant E

E

F1

F2

Fx

Fy R Equilibrant E are equal and opposite to Resultant R

E = -R

Particle Vs Rigid Body

• A particle has dimension = 0• A Rigid body is a non-particle body

and it does not deform (change shape). Concurrent

forces: all forces acting a the same point

Coplanar forces: all

forces lie on the same

plane

Conditions for Equilibrium

0Fn=i

1=ii

Explanation:

Sum of forces = 0,

Or

F1 + F2 + … + Fn = 0

F1 + F2 + F3= 0

Example

Conditions for Equilibrium

• Breaking down into x and y components

;0F and ;0Fn=i

1=ii y,

n=i

1=ii x,

;0FFF x,3x,2x,1 Example: For three forces acting on a particle

;0FFF y,3y,2y,1

Free Body Diagram

• Free body diagram isolates a rigid body to describe the system of forces acting on it.

R

mgR

R

R

Free Body Diagram

W

W

Wr Reaction

Wa Action

Free Body

Boundary

Definitions• System of Particles or Bodies

Two or more bodies or particles connected together are referred to as a system of bodies or particles.

• External Force External forces are all the forces acting on a body defined as a free body or free system of bodies, including the actions due to other bodies and the reactions due to supports.

Transmissibility of Force

FF F

Load and Reaction

• Loads are forces that are applied to bodies or systems of bodies.

• Reactions at points supporting bodies are a consequence of the loads applied to a body and the equilibrium of a body.

Tensile and Compressive Forces

Action Reaction

Compressive Force

Action Reaction

Tensile Force

• Pushing force on the body -- compressive force

• Pulling force on a body -- a tensile force

Procedure for drawing a free body diagram

• Step 1: Draw or sketch the body to be isolated

• Step 2: Indicate all the forces that act on the particle.

• Step 3: Label the forces with their proper magnitudes and directions

Example 1

Supportedpulley

PullingForce

Mass

Spring

PullingForce

Mass

Spring

mg

Fp

F1

F1

Fs

Fs

Example 2

Weight = 10 N

Ceiling Support

Ring

Weight = 10 N

Ceiling Support

Ring

Gravity Load =10 N

Reaction Force = 10 N

Action Load = 10 N

Free Body boundary

Example 3

Tug No.1

Tug No.2

Barge

60

25

Ring.

tow rope

tow rope

Solution• Resultant R of the two forces in two

ropes:

kN 06.2460cos3025cos10

xR

kN 75.2160sin3025sin10

yR

Solution

kN 43.3275.2106.24 22 R

1.4206.24

75.21tan 1

R = 32.43 kN

42.1

E = 32.43 kN

42.1

Equilibrant E = - R

Solution

Tug No.1

Tug No.2

Barge

60

25

Ring.

tow rope

tow rope

Resultant R is the sum of the actions of the tow ropes on the barge

Equilibrant E is the reaction of the barge to the ropes

E = - R

Moment and Couple

• Moment of Force• Moment M of the force F

about the point O is defined as:

M = F dwhere d is the perpendicular distance from O to F

• Moment is directional

dF

oM

dF

oM

Moment and Couple

d

F

r

M = F.d = F.r.cos

A

B

Moment = Force x Perpendicular Distance

Resultant of a system of forces

O

R

F 3

F 2

F 1

l 3

l 2

l 1

l

(a)

R

R x

R yF 3

F 2

F 1

F 1x

F 3y

F 3x

F 1yF 2x

F 2y

(b)

An arbitrary body subjected to a number of forces F1, F2 & F3.

Resultant R = F1 + F2 + F3

ComponentsRx = F1x + F2x + F3xRy = F1y + F2y + F3y

Resultant MomentO

R

F 3

F 2

F 1

l 3

l 2

l 1

l

(a)

R

R x

R yF 3

F 2

F 1

F 1x

F 3y

F 3x

F 1yF 2x

F 2y

(b)

Resultant moment Mo= Sum of Moments

Mo = F1 l1 + F2 l2 + F3 l3 = R l

Couple

• For a Couple R =F = 0 But Mo 0 Mo = F(d+l) - Fl = Fd Moment of couple is

the same about every point in its plane

d

F

O

F

l

d

F

O

F

l

Mo = F d

Example 4• Calculate the total (resultant)

moment on the body.

15 N

15 N 300 mm

30 N

30 N

170 mm

100 mm 50 mm

A

Example 4 (Solution)• Taking moments about the corner A

• Note that the forces form two couples or pure moments 3.6 Nm and 3.0 Nm(resultant force =0, moment is the same about any point).

1.01505.0303.01517.030 M

Nm 6.62.015120.030

Exercise

1. What is the moment of the 10 N force about point A (MA)?

A) 10 N·m B) 30 N·m C) 13 N·m

D) (10/3) N·m E) 7 N·m

• Ad = 3 m

F = 10 N

APPLICATIONS

What is the net effect of the two forces on the wheel?

APPLICATIONS

What is the effect of the 30 N force on the lug nut?

MOMENT IN 2-D

The moment of a force about a point provides a measure of the tendency for rotation (sometimes called a torque).

Moment

F=100

L=20

M=_____________

Moment

F=55N

F=32N

L=50cm

L=300mm

+ M=27.5N

M=-9.6N+

Given: A 400 N force is applied to the frame and = 20°.

Find: The moment of the force at A.

Plan:

1) Resolve the force along x and y axes.

2) Determine MA using scalar analysis.

EXAMPLE 1

EXAMPLE 1

Solution

+ Fx = -400 cos 20° N

+ Fy = -400 sin 20° N

+ MA = {(400 cos 20°)(2)

+ (400 sin 20°)(3)} N·m

= 1160 N·m

GROUP PROBLEM SOLVINGGiven: A 40 N force is

applied to the wrench.

Find: The moment of the force at O.

Plan: 1) Resolve the force along x and y axes.

2) Determine MO using scalar analysis.

Solution: + Fy = - 40 cos 20° N + Fx = - 40 sin 20° N

+ MO = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm = -7107 N·mm = - 7.11 N·m