# A rod of weight W is supported by two parallel knife edges A and B

Problem: 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__________  (IIT JEE 1997)

Solution: The forces on the rod are its weight $W$ and normal reactions $N_A$ and $N_B$.

In equilibrium condition, net force on the rod is zero and the net torque about its centre of mass C is also zero i.e., \begin{align} &N_A+N_B-W=0,\nonumber\\ &N_B (d-x)-N_A x=0.\nonumber \end{align} Solve to get $N_A={(d-x)W}/{d}$ and $N_B={xW}/{d}$.