Equilibrium
A body is truly in equilibrium when it has no tendency to turn or
move.
This means that:
When a body is in equilibrium:
The sum of the anticlockwise moments about any point is equal and opposite to the sum of the anticlockwise moments about that point.
and also:
The resultant force in any direction is zero.
It is easy to show
this using the light loaded beam of length L shown in Figure 1. Notice that F
1 will turn
the beam anticlockwise about the pivot (P) while F
2 will turn it clockwise.
Using
the principle of moments and taking moments about the pivot:
Clockwise moments = Anticlockwise moments F1d1 = F2d2
Balancing the vertical forces:
Sum of the vertical forces is zero. F1+ F2 – R = 0
Notice the
minus sign since R acts in the opposite direction from F
1 and
F
2.
Example problem
F1 = 20 N, d2 = 15 cm, d2 = 50 cm
Find the value of F2 and the reaction at the pivot such that the beam is in equilibrium.
(a) Take moments about the pivot:
Clockwise moments = 20x0.15 = F2x0.5 = Anticlockwise moments
Therefore: F2 = 6 N
(b) Taking the sum of the vertical forces to be zero: F1 + F2 - R = 0
R = 26 N
The same result would be obtained by taking moments about ANY point.
Further example
A more
complex example where there are three forces acting on the rod is shown below (Figure 2).
Using the principle of
moments and taking moments about the pivot:
Clockwise
moments = F1d1 + F2d2 = F3d3 = Anticlockwise
moments
Moments may be
taken about any point
(a) If we take moments about end A :
Anticlockwise
moments = LR/2 = [L/2 + d
1] F
1+ [L/2 + d
2] F
2 + [L/2 –
d
3] F
3 = Clockwise moments
(b) If we take moments about end B
:
Anticlockwise moments = [L/2 - d
1] F
1+ [L/2 - d
2] F
2 +
[L/2 + d
3] F
3 = LR/2 = Clockwise moments
Combining equations (a)
and (b) gives:
Clockwise moments = F
1d
1 + F
2d
2 =
F
3d
3 = Anticlockwise moments
This is the identical result that we obtained by
taking moments about the pivot.
Example problems
Let L = 100 cm, F1 = 20 N, F3 = 10 N, d1 = 10 cm, d2 = 70 cm, d3 = 45 cm
Find the values of F2 and R such that the beam is in equilibrium.
Take moments about the pivot:
(a) Clockwise moments = 20x0.1 + F2x0.7 = 10x0.45 = Anticlockwise moments
2 + F2x0.7 = 10x0.45 giving F2 = [4.5 – 2]/0.7 = 3.6 N
(b) But F1 + F2 + F3 - R = 0 and so R = 20 + 3.6 + 10 = 33.6 N