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.. And on B is......
A. `(w x)/(d)`
B. `(w d)/(x)`
C. `(w(d-x))/(x)`
D. `(w(d-x))/(d)`

Answer 1

As the weight w balances the normal reaction.

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 value 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)`