reactions at the knife-edges. (Assume the bar to be of uniform cross section and homogeneous.) Answer Fig. . Figure .
shows the rod AB, the positions of the knife edges K and K , 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 = cm. AG = cm, AP = cm, PG = cm, AK = BK = cm and K G = K G = cm.
Also, W = weight of the rod = . kg and W = suspended load = . kg; R and R are the normal reactions of the support at the knife edges. For translational equilibrium of the rod, R +R –W –W = (i) Note W and W act vertically down and R and R 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 and W are anticlockwise (+ve), whereas the moment of R is clockwise (-ve). For rotational equilibrium, –R (K G) + W (PG) + R (K G) = (ii) It is given that W = .
g N and W = . g N, where g = acceleration due to gravity. We take g = . m/s .
With numerical values inserted, from (i) R + R – . g – . g = or R + R = . g N (iii) = .
N From (ii), – . R + . W + . R = or R – R = .
g N = . N (iv) From (iii) and (iv), R = . N, R