A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is
Wdx
W(d−x)d
W(d−x)/x
Wxd
Solution:
Taking moment about point B NAd - W(d-x) = 0 NA = W(d-x)/d hence correct answer is option B.