Equilibrium

EQUILIBRIUM OF RIGID BODIES Page 1
CHAPTER 2
EQUILIBRIUM OF RIGID BODIES
CONTENT OF THE TOPIC:
I) Equilibrium of co-planar forces
 Analytical and graphical conditions of equilibrium
 Analytical conditions of equilibrium
 graphical conditions of equilibrium
 Different types of supports and their reactions
 Free body diagram
 Concept
 Examples based on to draw F.B.D.
 Different type of supports
 Lami’s Theorem
 Concept
 Procedure to solve the problems
 Problems based on Lami’s Theorem
II) Friction
 Concept of Friction
 Problems on horizontal plane and inclined plane
 Problems on ladder
III) Types of Problems
 Problems on equilibrium
 Problems on compound beam and frame with hinged joint
 Problems on pulleys
 Friction problem on inclined plane
 Problems on ladder

Equilibrium of Force:
Any system of forces that keeps the body at rest is said to be in equilibrium i.e. the state of the
body is not affected by the action of the force system in equilibrium. Equilibrium is applicable
to those systems of forces whose resultant action is zero.
Equilibrant:
1) A force that brings the system of forces in equilibrium is known as Equilibrant.
2) It is always equal, opposite and collinear with the resultant of the system.
3) When a system of forces is in equilibrium, its Equilibrant and resultant, both are zero.
Free Body:
Consider a body is resting against various supports and suppose all such supports are replaced by
their reactions exerted on the body, such body is known as free body.
Free Body Diagram:
A diagram of the body in which the body under consideration is freed from all the contact
surfaces and all the forces acting on it (including the reactions at contact surfaces) are drawn is
called a free body diagram.
Add figure
Application of Free Body Diagram (F.B.D.):
1) A Free Body Diagram (F.B.D.) is helpful in analyzing the equilibrium of a constrained
body.
2) A Free Body Diagram (F.B.D.) is helpful in finding reactions at the supports and internal
forces in the members of frames and trusses.
Moment Law of Forces:

Algebraic sum of the moments of all the forces taken about any point in the plane of
forces should be zero.
∑M = 0
Consider two forces P and Q acting on a body as shown in Fig. below. Let the angle
between the two forces be θ. The diagonal AC of the parallelogram ABCD represents the
resultant. Drop a perpendicular CE to AB.
Now, the resultant R of P and Q is given by,
R = AC
R2 = AE2 + CE2
R = √𝐴𝐸2 + 𝐶𝐸2
R = √(𝐴𝐵 + 𝐵𝐸)2 + 𝐶𝐸2
But
AB = P
BE = BC cos θ = Q cos θ
CE = BC sin θ = Q sin θ
R = √(𝑃 + 𝑄 𝑐𝑜𝑠𝜃)2 + (𝑄 𝑠𝑖𝑛𝜃)2
R = √
tan α =
𝐶𝐸
𝐴𝐸
tan α =
𝑄 sin 𝜃
𝑃+𝑄 cos 𝜃
α = tan-1(
𝑄 sin 𝜃
𝑃+𝑄 cos 𝜃
) ---------------------------------- (2)
Particular cases:
1) θ = 900 R= √𝑃2 + 𝑄2
2) θ = 00 R= P + Q
3) θ = 1800 R= P - Q
Lami’s Theorem:
“If three forces acting at a point are in equilibrium, each force will be proportional to the sine of
the angle between the other two forces.”

Fig. 1
Suppose three forces P, Q and R are acting at a point O and they are in equilibrium as shown in
Fig. 1 above.
Let, α = Angle between force P and Q
β = Angle between force Q and R
γ = Angle between force R and P
Then according to the Lami’s theorem
P α sine of the angle between Q and R α sin β
𝑃
sin(𝛽)
= constant
Similarly,
𝑃
sin(ϒ)
= constant
𝑃
sin(𝛼)
= constant
𝑃
sin(𝛽)
=
𝑃
sin(ϒ)
=
𝑃
sin(𝛼)
--------(1)
Proof of Lami’s Theorem:
The three forces are acting on a point, are in equilibrium and hence they can be represented by
the three sides of the triangle taken in the same order. Now draw the force triangle as shown in
following Figure (a).

Figure (a).
Applying the sine rule, we get
𝑃
sin(180−𝛽)
=
𝑄
sin(180−ϒ)
=
𝑅
sin(180−𝛼)
This can also be written
𝑃
sin(𝛽)
=
𝑄
sin(ϒ)
=
𝑅
sin(𝛼)
--------(2)
This is the same equation as in equation (1) above
Types of beam supports:
1) Simple support:
It is a theoretical case in which the ends of the beam are simply supported or rested over
the supports. The reactions are always vertical as shown in Fig.1 below
Fig.1 Simple Support

It opposes downward movement but allows rotation and horizontal displacement or
movement.
2) Pin or hinged Support:
In such case, the ends of the beam are hinged or pinned to the support as shown in Fig.2
below.
Fig.2 (A) Hinged Support Fig.2 (B) Hinged Support
The reaction may be either vertical or inclined depending upon the type of loading. If the
loads are vertical the reaction is vertical as shown in Fig. 2 (A) and when the applied
loads are inclined the reaction is inclined as shown in Fig. 2 (B).
The main advantage of hinged support is that the beam remains stable i.e. there is only
rotational motion round the hinge but no translational motion of the beam i.e. hinged
support opposes displacement of beam in any direction but allows rotation.
3) Roller Support:
In such cases, the end of the beam is supported on roller as shown in Fig. 3 below.
Fig. 3 Roller Support
The reaction is always perpendicular to the surface on which rollers rest or act as shown
in Fig. 3. The main advantage of the roller support is that, the support can move easily in

the direction of expansion or contraction of the beam due to change in temperature in
different seasons.
4) Fixed Support:
It is also called as Built-in-supports. It is rigid type of support. The end of the beam is
rigidly fixed in the wall as shown in Fig. 4 below.
Fig. 4 Fixed Support
It produces reactions Ra in any direction and a moment Ma as shown in Fig. 4 above.
Problems:
1) A system of connected flexible cables as shown in Fig. below. It is supporting two
vertical forces 200 N and 250 N at points B and D. Determine the forces in various
segments of the cable.
2) Find
a. Tension in portion AB, BC and CD string
b. Magnitude of P1 and P2
3) Find the reactions at the support for a bent supported and loaded as shown in Fig. below.
(January 2002 12 Mks).

Problems on Cylinder:
1) Two cylinders P and Q of 1m diameter and weighing 1000 N each are supported as shown in
Fig. below. Neglect friction at contact points. Find the reactions at A, B, C and D. (January
2001 12 Mks).
Problems on Pulleys:
1) Blocks A and B are connected by links and supported as shown in Fig. below. If block A
weighs 300 N and block B weighs 150 N. Find the maximum and minimum values of P for
which the blocks are just in equilibrium. µ = 0.25
Assignment No. 2
Equilibrium
Q1. Define and explain the term ‘Equilibrium’. What do you understand by ‘Equilibrant’?
Q2. State and explain ‘Principles of Equilibrium’ or ‘Equilibrium law’.
Q3. What are the different conditions of equilibrium?
Q4. What are the different conditions of equilibrium for:
A) Non-concurrent Force System
B) Concurrent Force System
Q5. What do you understand by ‘Free Body’ and ‘Free Body Diagram’? What is the application
of Free Body Diagram?
Q6. State ‘Lami’s Theorem’.
Q7. Draw the Free Body Diagram for following System:

Q9) Two smooth cylinders, each of weight 1000N and Diameter 30 cm connected by a
string of 40 cm & rest upon a smooth horizontal surface, supporting a third cylinder of
weight 2000N & diameter of 30 cm which is above the two cylinder. Find out the reactions,
pressure at contact surface and tension in the string.

1. Determine the reactions at 1 and 2. Assume all the surfaces to be smooth.
2. Determine the reactions at A and B. Assume all the surfaces to be smooth.
3. Two spheres each of weight 1000 N and of radius 25 cm rest in a horizontal channel of
width 90 cm as shown in following Fig. Find the reactions at point of contact. Assume all
the surfaces to be smooth.

4. Two identical rollers, each of weight 1000 N, are supported by a inclined plane and a
vertical wall as shown in Fig. below. Find the reactions at point of contacts 1, 2 and 3.
Assume all the surfaces to be smooth.
5. Two cylinders A and B, of weight 1000 N and 500 N respectively, are supported by a
inclined plane and a vertical wall as shown in Fig. below. The radius cylinders A and B
are 250 mm and 157 mm respectively. Find the reactions at point of contacts. Assume all

are 250 mm and 157 mm respectively.

Equilibrium

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