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 a distance `x` from `A`.

Options

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

Answer

D

Explanation

`N_(A)+N_(B)=W" "..........(i)`
As torque along centre of mass P is zero.
`:.tau=-N_(A)x+N_(B)(d-x)=0`
`N_(A)x=N_(B)(d-x)`
Put value of `N_(B)` in equation (i)
`(N_(A)x)/((d-x))+N_(A)=W`
`M_(A)((d)/((d-x)))=W`
`N_(A)=(W(d-x))/(d)`