Equilibrium
A body is truly in equilibrium when it has no tendency to turn or
move. This means no translation and no rotation.
When a body is in equilibrium:
The sum of the anticlockwise moments about any point is equal and opposite to the sum of the clockwise 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 F2d2 = F1d1
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). The rod is pivoted about its centre.
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
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