EQUILIBRIUM OF RIGID BODIES

1. A rigid body is one where all parts of the body are fixed relative to each other (for
example, a pencil). Fluids and gases are non-rigid.
2. The translational motion of the centre of mass of a rigid body is governed by:
where is the external force.
Similarly, for rotational motion, where is the net external torque.
ext
ext
dP F F F net
dt
dL
dt
τ τ τ
= =
= =
Σ
Σ
􀁇 􀁇 􀁇 􀁇
􀁇
􀁇 􀁇 􀁇
3. A rigid body is in mechanical equilibrium if the linear momentum and angular
momentum have a constant value. i.e., 0 0.
refers to 0 a
P
L dP dL
dt dt
P
= =
=
both
and Static equilibrium
􀁇
􀁇 􀁇 􀁇
􀁇
nd 0.
4. As an example of static equilibrium, consider a beam resting on supports:
L =
􀁇
1 2 We want to find the forces and with which the supports push on the rod
in the upwards direction. First, balance forces in the vertical direction:
y
F F
y
F
( )( ) ( )( ) ( )( ) ( )( )
1 2
1 2
1 2
0
Now demand that the total torque vanishes:
0 / 4 / 2 0
From these two conditions you can solve for and ,
y
F F Mg mg
F F L Mg L mg L
F F
τ
= + − − =
= + − − =
Σ
Σ
( ) ( )
1 2
3 2 2
, and .
4 4
5. Angular momentum and torque depend on where you choose the origin of your
coordinates. However, I shall now prove that for a body in equi
M mg M mg
F F
+ +
= =
1 2 1 1 2 2
librium, the choice of
origin does not matter. Let's start with the origin O and calculate the tor
que about O,
Now, if we
O N N N τ =τ +τ + ⋅ ⋅ ⋅ +τ =r×F+r ×F + ⋅ ⋅ ⋅ +r ×F
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
take a second point P, then all distances will be measured from P and each
L
L / 4 M
m
1 F 2 F
i r􀁇
CG r􀁇
O x
y
z
CG
i m g􀁇
M g􀁇
( ) ( ) ( ) 1 1 2 2
1 1 2 2
vector will be shifted by an amount . Hence the torque about P is,
P
P P P N P N
N N
r
r r F r r F r r F
r F r F r F
τ = − × + − × + ⋅ ⋅ ⋅ ⋅ + − ×
= ⎡ × '2B × +⋅⋅⋅+ ×
􀁇
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
( )
( )
1 2
1 2
but 0, for a body in translational equilibrium
P P P N
O P N
O P ext
ext
r F r F r F
r F F F
r F
F
τ
τ
⎣ ⎦⎤ − ⎣⎡ × + × + ⋅ ⋅ ⋅ + × ⎦⎤
= −⎡⎣ × + + ⋅ ⋅ ⋅ + ⎤⎦
= −⎡⎣ × ⎤⎦
= ∴
Σ
Σ
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 􀁇
􀁇
.
6. Let us use the equilibrium conditions to do something of definite practical importance.
Consider the balance below which is in equilibrium when two known weights are hung
as shown.
P O τ􀁇 =τ􀁇
1 2 We want to know m in terms of m and m .
( ) ( ) 1 2
1
1 2
2
Taking the torques about the knife edge in the two cases, we have:
and
or .
Remarkably, we do not need the values
mgx m g L x m gx mg L x
m m m mm
m m
= − = −
⇒ = =
of or .
7. The centre of gravity is the average location of the weight of an
object. This is not quite the same as the centre of mass of a body (see lecture 12) but
suppos
x L
Centre of Gravity.
e the gravitational acceleration has the same value at all points of a body.
Then: 1) The weight is equal to , and 2)the centre of gravity coincides with
the centre of mass. Remember tha
g
M g
􀁇
􀁇
t weight is force, so the CG is really the centre of
gravitational force acting on the body. The net force on the whole body = sum of forces
over all individual particles, . If
i ΣF=Σmg g 􀁇 􀁇 􀁇
( ) ( )
has the same value at all points of the body, then
. So the net torque about
the origin is
Hence, . So the torque
due to gravit
i
i i ii
cm cm
F g m Mg
O r m g m r g
Mr g r Mg
τ
τ
= =
= × = ×
= × = ×
Σ Σ
Σ Σ Σ
Σ
􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇
y about the centre of mass of a body
(i.e. at 0 ) is zero !! cm r􀁇 =
L x
mg 1 m g
L x
2 m g mg
L
W
x
θ φ
+
CM
A
B O
W
C
D
1 R
2 R
1 1 μ R
2 2 R μ
θ
8. In the demonstration I showed, you saw how to find the CG of an irregular object
by simply suspending it on a pivot,
9. Let's solve a problem of static equilibrium: A non-uniform bar of weight W is suspended
at rest in a horizontal position by two light cords. Find the distance from the left-hand
end to the
x
( )
1 2
2 1
2 1 2
center of gravity.
Call the tensions and . Put the forces in
both directions equal to zero,
a) sin sin 0 (horizontal)
b) cos cos 0 (vertical)
sin
Solution : T T
T T
T T W T W
φ θ
φ θ
θ φ
− =
+ − = ⇒ =
+
( ) ( )
( )
2
2
The torque about any point must vanish. Let us choose that point to be one end of the bar,
cos 0 cos cos
sin
10.Here is another problem of the same k
Wx T L x T L L
W
φ φ
φ
θ φ
− + = ⇒ = =
+
1 2 2 2 1 1
ind: find the least angle at which the rod can lean
to the horizontal without slipping.
Considering the translational equilibrium of
the rod, and . This
Solution :
R R R R W
θ
=μ +μ =
2 ( )
1 2
2 22
2 22
2 2 2
gives,
. Now consider rotational equilibrium
1
about the point A:
or, cos cos sin .
2
This gives cos sin from which tan
2
R W
R OB W OD R OA
R AB W AB R AB
R W R
μ μ
μ
θ
θ μ θ
θ μ θ θ
=
+
× = × + ×
× = × + ×
⎛⎜ − ⎞⎟=
⎝ ⎠ ( )
2
2
2 2 1 2
1 2
2
2
2 with .
1
Using this value of , we get tan 1- .
2
10. In the lecture you heard about:
a)Stable equilibrium: object returns to its original positi
R W R W
R
R
μ μ μ
μ μ
θ
μ
−
= =
+
=
Types of Equilibrium.
on if displaced slightly.
b)Unstable equilibrium: object moves farther away from its original position if
displaced slightly.
c)Neutral equilibrium: object stays in its new position if displaced slightly.