A metal bar `70 cm` long and `4.00 kg` in mass is supported on two knife edges placed `10 cm` from each end. A `6.00 kg` weight is suspended at `30 cm` from one end. Find the reactions at the knife edges. Assume the bar to be of uniform cross-section and homogeneous.

Explanation

Figure 7.26 shows the rod AB, the positions of the knife edges `K_(1) and K_(2)` , the centre of 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 = 30 cm, PG = 5 cm, AK_(1)= BK_(2) = 10 cm and K_(1)G = K_(2)G = 25 cm`. Also, W= weight of the rod = `4.00 kg and W_(1)`= suspended load = `6.00 kg, R_(1) and R_(2)` are the normal reactions of the support at the knife edges.
For translational equilibrium of the rod,
`R_(1)+R_(2)-W_(1)-W=0` (i)
Note `W_(1) and W` act vertically down and `R_(1) and R_(2)` act vertically up.
For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments about is G. The moments of `R_(2) and W_(1)` are anticlockwise (+ve), whereas the moment of `R_(1)` is clockwise (-ve).
For rotational equilibrium.
`-R_(1)(K_(1)G)+W_(1)(PG)+R_(2)(K_(2)G)=0` (ii)
It is given that `W = 4.00g N and W_(1) = 6.00g N`, where g = acceleration due to gravity. We take `g = 9.8 m//s^(2)`.
With numerical values inserted, from (i)
`R_(1)+R_(2)-4.00g-6.00g=0`
or `R_(1)+R_(2)=10.00gN` (iii)
`=98.00N`
From (ii), `– 0.25 R_(1) + 0.05 W_(1) + 0.25 R_(2) = 0`
or `R_(1) – R_(2) = 1.2g N = 11.76 N` (iv)
From (iii) and (iv), `R_(1) = 54.88 N`,
`R_(2) = 43.12 N`
Thus the reactions of the support are about 55 N at `K_(1) and 43 N` at `K_(2)`.