1. Determine the moment of a force of 25 N applied to a spanner at an effective length of 180 mm from the centre of a nut.
[4.5 N m] 2. A moment of 7.5 N m is required to turn a wheel. If a force of 37.5 N applied to the
rim of the wheel can just turn the wheel, calculate the effective distance from the rim to the hub of the wheel. [200 mm] 3. Calculate the force required to produce a moment of 27 N m on a shaft, when the effective distance from the centre of the shaft to the point of application of the
force is 180 mm. [150 N]
5.2
Equilibrium and the principle of
moments
If more than one force is acting on an object and the forces do not act at a single point, then the turning effect of the forces, that is, the moment of the forces, must be considered.
Figure 5.3 shows a beam with its support (known as its pivot or fulcrum) at P, acting vertically upwards, and forces F1 and F2 acting vertically downwards at distancesaandb, respectively, from the fulcrum. F1 F2 Rp a b P Figure 5.3
A beam is said to be inequilibriumwhen there is no tendency for it to move. There are two conditions for equilibrium:
(i) The sum of the forces acting vertically down- wards must be equal to the sum of the forces acting vertically upwards, i.e. for Figure 5.3, Rp=F1+F2
(ii) The total moment of the forces acting on a beam must be zero; for the total moment to be zero:
the sum of the clockwise moments about any point must be equal to the sum of the anticlock- wise, or counter-clockwise, moments about that point
This statement is known as the principle of moments.
Hence, taking moments aboutP in Figure 5.3, F2×b=the clockwise moment, and F1×a=the anticlockwise, or
counter-clockwise, moment Thus for equilibrium: F1a=F2b
Problem 3. A system of forces is as shown in Figure 5.4 F P d 5 N 7 N 140 mm 200 mm Figure 5.4
(a) If the system is in equilibrium find the distanced.
(b) If the point of application of the 5 N force is moved to pointP, distance 200 mm from the support, and the 5 N force is replaced by an unknown force F, find the value ofF for the system to be in equilibrium.
(a) From above, the clockwise momentM1 is due to a force of 7 N acting at a distance d from the support; the support is called thefulcrum, i.e.
M1=7 N×d
The anticlockwise momentM2is due to a force of 5 N acting at a distance of 140 mm from the fulcrum, i.e.
M2=5 N×140 mm
Applying the principle of moments, for the system to be in equilibrium about the fulcrum: clockwise moment=anticlockwise moment
i.e. 7 N×d=5×140 N mm
Hence, distance,d= 5×140 N mm 7 N =100 mm
(b) When the 5 N force is replaced by force F at a distance of 200 mm from the fulcrum, the new value of the anticlockwise moment is F ×200. For the system to be in equilibrium: clockwise moment = anticlockwise moment i.e.
(7×100)N mm=F×200 mm Hence,new value of force,
F = 700 N mm
200 mm =3.5 N
Problem 4. A beam is supported on its fulcrum at the pointA, which is at mid-span, and forces act as shown in Figure 5.5. Calculate (a) forceF for the beam to be in equilibrium, (b) the new position of the 23 N force whenF is decreased to 21 N, if equilibrium is to be maintained. 12 N F 23 N 20 mm 80 mm 100 mm d A Figure 5.5
(a) The clockwise moment,M1, is due to the 23 N force acting at a distance of 100 mm from the fulcrum, i.e.
M1=23×100=2300 N mm
There are two forces giving the anticlockwise moment M2. One is the force F acting at a distance of 20 mm from the fulcrum and the other a force of 12 N acting at a distance of 80 mm. Thus
M2=(F ×20)+(12×80)N mm Applying the principle of moments about the fulcrum:
clockwise moment=anticlockwise moments i.e. 2300=(F ×20)+(12×80) Hence F×20=2300−960
i.e. force,F= 1340
20 =67 N
(b) The clockwise moment is now due to a force of 23 N acting at a distance of, say,dfrom the fulcrum. Since the value ofF is decreased to 21 N, the anticlockwise moment is(21×20)+ (12×80)N mm.
Applying the principle of moments, 23×d=(21×20)+(12×80) i.e. distance,d= 420+960 23 = 1380 23 =60 mm
Problem 5. For the centrally supported uniform beam shown in Figure 5.6, determine the values of forcesF1 andF2 when the beam is in equilibrium.
F1 F2 R = 5 kN 3 m 7 m Figure 5.6 At equilibrium: (i) R=F1+F2 i.e. 5=F1+F2 (1) and (ii) F1×3=F2×7 (2)
From equation (1),F2 =5−F1. Substituting forF2 in equation (2) gives: F1×3=(5−F1)×7 i.e. 3F1=35−7F1 10F1=35 from which, F1=3.5 kN SinceF2=5−F1, F2=1.5 kN
Thus at equilibrium, forceF1=3.5 kN and force
F2 =1.5 kN
Now try the following exercise
Exercise 26 Further problems on equilib-