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162 PHYSICS

vary from one point of the body to the other. If 4.00 kg and W = suspended load = 6.00 kg;
1
the body is so extended that g varies from part R and R are the normal reactions of the
1 2
to part of the body, then the centre of gravity support at the knife edges.
and centre of mass will not coincide. Basically, For translational equilibrium of the rod,
the two are different concepts. The centre of R +R –W –W = 0 (i)
1 2 1
mass has nothing to do with gravity. It depends Note W and W act vertically down and R
1 1
only on the distribution of mass of the body. and R act vertically up.
2
In Sec. 7.2 we found out the position of the For considering rotational equilibrium, we
centre of mass of several regular, homogeneous take moments of the forces. A convenient point
objects. Obviously the method used there gives to take moments about is G. The moments of
us also the centre of gravity of these bodies, if R and W are anticlockwise (+ve), whereas the
2 1
they are small enough. moment of R is clockwise (-ve).
1
Figure 7.25 illustrates another way of For rotational equilibrium,
determining the CG of an irregular shaped body –R (K G) + W (PG) + R (K G) = 0 (ii)
1 1 1 2 2
like a cardboard. If you suspend the body from It is given that W = 4.00g N and W = 6.00g
1
some point like A, the vertical line through A N, where g = acceleration due to gravity. We
2
passes through the CG. We mark the vertical take g = 9.8 m/s .
AA . We then suspend the body through other With numerical values inserted, from (i)
1
points like B and C. The intersection of the R + R – 4.00g – 6.00g = 0
1
2
verticals gives the CG. Explain why the method or R + R = 10.00g N (iii)
1
2
works. Since the body is small enough, the = 98.00 N
method allows us to determine also its centre From (ii), – 0.25 R + 0.05 W + 0.25 R = 0
1
1
of mass. or R – R = 1.2g N = 11.76 N 2 (iv)
1 2
From (iii) and (iv), R = 54.88 N,
1
u Example 7.8 A metal bar 70 cm long R = 43.12 N
2
and 4.00 kg in mass supported on two Thus the reactions of the support are about
knife-edges placed 10 cm from each end. 55 N at K and 43 N at K . t
A 6.00 kg load is suspended at 30 cm from 1 2
one end. Find the reactions at the knife- u Example 7.9 A 3m long ladder weighing
edges. (Assume the bar to be of uniform 20 kg leans on a frictionless wall. Its feet
cross section and homogeneous.) rest on the floor 1 m from the wall as shown
in Fig.7.27. Find the reaction forces of the
Answer wall and the floor.
Answer

Fig. 7.26

Figure 7.26 shows the rod AB, the positions
of the knife edges K and K , the centre of
1 2
gravity of the rod at G and the suspended load
at P.
Note the weight of the rod W acts at its
centre of gravity G. The rod is uniform in cross
section and homogeneous; hence G is at the
centre of the rod; AB = 70 cm. AG = 35 cm, AP Fig. 7.27
= 30 cm, PG = 5 cm, AK = BK = 10 cm and K G The ladder AB is 3 m long, its foot A is at
1 2 1
= K G = 25 cm. Also, W= weight of the rod = distance AC = 1 m from the wall. From
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