A uniform heavy beam of length 10 m and mass 50 kg is supported horizontally by two pillars at its ends, A and B. A block of mass 30 kg rests on the beam at a distance of 2 m from pillar A, as shown in the figure. What is the normal reaction force exerted by pillar B on the beam? (Take g = 10 m s ⁻² )

Step-by-step solution

Let N_A and N_B be the normal reaction forces exerted by pillars A and B, respectively. The forces acting on the beam are: 1. Weight of the uniform beam, W_ beam = 50 10 = 500 N , acting at its center of mass, which is at a distance of 5 m from pillar A. 2. Weight of the block, W_ block = 30 10 = 300 N , acting at a distance of 2 m from pillar A. 3. Normal reaction N_B acting upwards at a distance of 10 m from pillar A. Since the beam is in rotational equilibrium, the net torque about any point must be zero. Taking