Question

A 40.0 m long uniform bridge span weighs 5.00 * 104 N. A car weighing 1.50 * 104 N is parked 12.0 m from the right pier. What force does it exert on the left pier?

Answer 1

Are we looking for the force of bridge plus car or car only. The bridge weight (force) is divided equal between both piers, assume they are at the same height (horizontal bridge. To get the force the care applies to the piers note that the clockwise and counter-clockwise torques due to the car alone must balance. Torque = force times lever arm, and the torques are upward F from left pier times the length of the bridge (40m) which must equal the downward weight of the car times 12m. This is a lever with mechanical advantage of 40/12.