A horizontal uniform bar of mass 220 kg and length L = 3.0 m is placed on two supports, labeled A and B, located as shown in the diagram. A block of mass 30 kg is placed on the right end of the bar.
Find the normal forces F_A and F_B exerted on the bar by the supports.
I'm sure we'll use torque, T = rFsinθ. We'll probably use center of mass, Σmx / M. Because the bar is uniform, though, we know the center of mass will be the mid-point of the bar. Torque will be determined by the gravity acting on the center of mass, which has equation F = mg. Since the object is static the sum of all torques will be 0.
The Attempt at a Solution
I treat the point B as if it were a pivot point, in which case the distance to the center of mass is 0.5m to the left. Then the torque applied is T1 = 0.5 * 220g = 110g. The block has distance 1m to the right of the pivot and so exerts a torque T2 = -1*30g in the opposite direction. The normal force due to support A will then satisfy 110g-30g+N=0 and this implies N = -80g.
However, this answer turns out to be wrong. Any idea what my mistake is?
Last edited by a moderator: