Posted by Physics on .
A hiker, who weighs 1120 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs 3775 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?
Let F1 and F2 be the forces on the two ends of the bridge.
The sum of the forces in the y direction is 0:
F1 + F2 - 1120 - 3775 = 0
The sum of the torques about each support is zero:
1120*(L/5) + 3775*(L/2) - F2*L = 0
Dividing by L:
1120/5 + 3775/2 - F2 = 0
Solve for F2, then solve for F1