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 onefifth of the way along the bridge. What is the magnitude of the force that a concrete support exerts on the bridge at each end?

physics 
Jennifer,
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