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

A. `(W(d- x))/(x)`
B. `(W( d - x))/(d)`
C. `(Wx)/(d)`
D. `(Wd)/(x)`

Answer

B

Explanation

Given situation is shown in figures .
`N_(1)` = Normal reaction on A
`N_(2)` = Normal reaction on B
W = Weight of the rod

In vertical equilibrium
`N_(1) + N_(2) = W " " ….. (i)`
Torque balance about centre of mass of the rod ,
`N_(1) x = N_(2) (d - x)`
Putting value of `N_2` from equation (i)
`N_(1)x = (W - N_1) (d - x)`
`implies N_1 x = Wd - Wx - N_1 d + N_1 x implies N_1 d = W (d - x)`
`therefore N_(1) = (W (d - x))/(d)`