A uniform plank of mass 10 kg and length 4 m rests horizontally on two supports placed exactly at its ends A and B. A block of mass 20 kg is placed on the plank at a distance of 1 m from end A. The magnitude of the normal reaction exerted by the support at end B is (take g = 10 m/s ^2 )

Step-by-step solution

Let N_A and N_B be the normal reactions at supports A and B respectively. The forces acting on the plank are: 1. Weight of the plank, W_p = 10 10 = 100 N , acting at its center of mass ( 2 m from end A). 2. Weight of the block, W_b = 20 10 = 200 N , acting at 1 m from end A. 3. Upward normal reactions N_A and N_B at the ends. For rotational equilibrium, the net torque about any point must be zero. Taking torque about end A to eliminate N_A : _A = 0 N_B 4 - W_p 2 - W_b 1 = 0 N_B 4 = (100 2) + (200 1) N_B 4 = 200 + 2