Question

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

Options

• `(Wx)/d`

• `(Wd)/d`

• `(W(d - x))/x`

• `(W(d - x))/d`

Answer 1

`(W(d - x))/d`

Explanation:

N_A + N_B = W

Torque balances at the CM of the rod

N_Ax = N_B(d – x)

∴ N_Ax = (W – N_A)(d – x)

∴ N_Ax = Wd – Wx – N_Ad + N_Ax

∴ N_Ad = W(d – x)

∴ N_A = `(W(d - x))/d`