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.. And on B is......

Options

A. `(wx)/d`
B. `(wd)/x`
C. `(w(d-x))/x`
D. `(w(d-x))/d`

Answer

D

Explanation

As the weight w balances the normal reactions.

So, `w=N_(1) + N_(2)`...................(i)
Now, balancing torque about the COM,
i.e. Anti-clockwise momentum = clockwise momentum
`rArr N_(1)x= N_(2)(d-x)`
Putting the vluie of `N_(2)` from Eq. (i), we get
`N_(1)x = (w-N_(1))(d-x)`
`rArr N_(1)x = wd -wx - N_(1)d +N_(1)x`
`rArr N_(1)d = w(d-x)`
`rArr N_(1) = (w(d-x))/d`