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Question: A hiker, who weighs 1165 N, is strolling through the woods and crosses a small horizontal bridge. The bridge i?

A hiker, who weighs 1165 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs 3050 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?

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Answer:

You can use the requirement that the torques, about any point, must ballance.

If you take torques about the support nearest the hiker then the cclw torques of weight of bridge & hiker = clw torque of far support. This will give the force exerted by the far support;

N2L = (L/5)(1165) + (L/2)(3050)
N2 = 1758 N

Next take torques about the far support, to get force exerted by the near support;;

N1L = (4L/5)(1165) + (L/2)(3050)
N1 = 2457 N

As a check, the two upward support forces must equal the total weight;

2457 + 1758 = 4215 = 1165 + 3050