7. Equilibrium of a Rigid Body

Engineering Mechanics

Statics

Dr. Rami Zakaria

Mechanical Systems Engineering _ 2016

The difference between the forces on a particle

and the forces on a rigid-body is that the forces

may cause rotation of the body (because of the

moments created by the forces).

Forces on a particle

We say that a rigid body is in equilibrium when

the net force and the net moment about any

arbitrary point O is equal to zero.

F = 0 (no translation)

and MO = 0 (no rotation)

Forces on a rigid body

Note: We will discuss how to find the centre of gravity in another chapter (the location where the

weight force is applied).

Conditions for Rigid-Body Equilibrium

3. Apply the equations of equilibrium (E-O-E) to

solve for any unknowns.

1. Create an idealized model (above right).

2. Draw a free-body diagram (FBD) showing all the

external (active and reactive) forces.

Note.1: In this lecture we will only discuss 2D force systems.

Note2: Internal forces always cancel each others and they don’t create external effect. So in

a free body diagram we only draw the external forces.

Steps of Solving Rigid Body Equilibrium Problems

As a general rule, if a support prevents translation of a body in a given

direction, then a force is developed on the body in the opposite direction.

Similarly, if a support prevents rotation, then a couple moment is exerted

on the body in the opposite direction.

A few example sets of diagrams are shown above. See the other support reactions in your

textbook (Table 5-1).

Prevents translation in

one direction

Prevents translation

in two directions

Prevent translation in two directions

and also prevents rotation

Support Reaction

2. Show all the external forces and couple moments. These usually include:

a) applied loads/forces b) support reactions, and c) the weight of the body.

Idealized model Free-body diagram (FBD)

1. Draw an outlined shape. (imagine the body is isolated)

3. Label loads and dimensions on the FBD: All known forces and couple

moments should be labeled with their magnitudes and directions. For the

unknown forces and couple moments, use letters like Ax, Ay, MA, etc..

Indicate any necessary dimensions.

How to Draw a Free Body Diagram?

W is the weight (effect of gravity on the roll).

NA, NB are the surface reactions on the roll.

NA is the reaction of the surface.

B (BX, BY) is the pin force

Q. If a support only prevents translation of a body, then the support exerts a ______ on

the body.

A) Couple moment

B) Force

C) Both force and couple moment.

D) None of the above

W is the weight

Ax, Ay are the pin reaction

FBC is the hydrologic cylinder reation

Q. Internal forces are _________ shown on the free body diagram of a whole body.

A) Always

B) Often

C) Rarely (Sometimes)

D) Never

A body is subjected to a system of forces that lie in the x-y plane.

When in equilibrium, the net force and net moment acting on the body

are zero.

This 2-D condition can be represented by the three scalar equations:

Fx = 0 Fy = 0 MO = 0 where point O is any arbitrary point.

Please note that these equations are the ones most commonly

used for solving 2-D equilibrium problems. There are other

sets of equations that are rarely used (we will not discuss them).

Steps of Solving Free Body Equilibrium problems

1. If not given, establish a suitable x - y coordinate system.

2. Draw a free body diagram (FBD) of the object.

3. Apply the three equations of equilibrium (E-of-E) to solve for the unknowns.

EQUATIONS OF EQUILIBRIUM

IMPORTANT NOTES

1. If there are more unknowns than the number of independent equations, then

we have a statically indeterminate situation. We cannot solve these

problems using just statics.

2. The order in which we apply equations may affect the simplicity of the

solution. For example, if we have two unknown vertical forces and one

unknown horizontal force,

then solving FX = 0 first allows us to find the horizontal unknown quickly.

3. If the answer for an unknown comes out as negative number, then the sense

(direction) of the unknown force is opposite to that assumed when starting

the problem.

Q. Which equation of equilibrium allows you to

calculate FB right away?

A) FX = 0 B) FY = 0

C) MA = 0 D) Any one of the above.

AX A B

FB AY

100 lb

EXAMPLE

Plan:

1. Establish the x- y axes.

2. Draw a complete FBD of the boom.

3. Apply the E-of-E to solve for the unknowns.

Given: The 4kN load at B of

the beam is supported

by pins at A and C .

Find: The support reactions

at A and C.

Solution:

Using E-o-f E, we get:

+ FX = AX + 11.31 cos 45 = 0; AX = – 8.00 kN

+ FY = AY + 11.31 sin 45 – 4 = 0; AY = – 4.00 kN

MA = (FC sin 45) (1.5) – (4) (3) = 0

Fc = 11.31 kN

FBD of the beam:

AX

AY

A

1.5 m

C B

4 kN

FC

45°

1.5 m

Note that the negative signs means that the reactions have the

opposite direction to that shown on FBD.

GROUP PROBLEM SOLVING

Plan:

a) Establish the x – y axes.

b) Draw a complete FBD of the jib crane beam.

c) Apply the E-of-E to solve for the unknowns.

Given: The jib crane is supported by

a pin at C and rod AB. The

load has a mass of 2000 kg

with its center of mass

located at G.

Assume x = 5 m.

Find: Support reactions at B

and C.

Solution:

A rigid body is in equilibrium when the net force and the net moment

about any arbitrary point O is equal to zero.

F = 0 (no translation) and MO = 0 (no rotation)

Fx = 0 and Fy = 0

Summary:

- We also learned that before we solve any rigid body problem we

need to draw the ‘Free Body Diagram’ (FBD).

- We should put all external forces on the FBD.

Let’s look at some examples…

Note: The pin (B) prevents the movement in two directions, while the rocker (A)

prevents the movement in one direction only.

FBD

NB sin30

NB cos30

=