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

As he weight w balances the normal reactions
So , `w = N_(1) +N_(2)`
Now balancing torque about the COM .
i.e anti- clockwise moment = clockwise moment
` N_(1)x = = N_(2) (d-x)`
Putting the value of`N_(2)` from eq (i) , we get `N_(1) x = N_(2) (d-x)`
Puting the value of `N_(2)` from eq . (i) , we get
`N_(1)x = N_(2)(d-x)`
`rArr N_(1) x = wd - wx - N_(1)d +N_(1)x`
`rArr N_(d) w (d-x) `
` rArr N_(1) = (w(d-x))/d`