physics question?

Question

A hiker, who weighs 1180 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs 3840 N, and rests on two concrete supports, one at each end. He stops one-fifth of the way along the bridge. What is the magnitude of the force that a concrete support exerts on the bridge at each end?

Answer 1

The bridge is in both translational and rotational equilibrium That means that the net vertical force is zero, and the net torque about every point on the bridge is zero.

Force is easy; the total weight is 3840 N + 1180 N, so the sum of the two support forces must be 5020 N.

F1 + F2 = 5020 N.

Now for the torques. Let us choose the leftmost support as the center of rotation (you can select any point you wish). There are three torques: from the weight of the bridge itself (measured from the center), from the weight of the hiker, and from the support at the other end. We'll let clockwise torques be positive and counterclockwise torques be negative.

Let L be the length of the bridge.

Torque = 0 = (3840 N)(1/2 L) + (1180 N)(1/5 L) - F2 (L)

Don't know L? No problem, it cancels from every term. This gives you F2. Then F1 = 5020 N - F2.