A hiker, who weighs 915 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is

Question

A hiker, who weighs 915 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs 4055 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?
(a) near end

(b) far end

Answer 1

before the hiker boarded the bridge was balanced with 4055/2 on each end.

When the hiker boarded, his weight moves the center of mass to the enar end. Walking 1/5 the length, the mass (weight) is distributed as follows:

near end: 4055/2 + 915*4/5
far end: 4055/2 +915/5

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Answer 2

the two supports must combine to support the weight of the hiker and the bridge also, if the bridge is not rotating, then we know the torques sum to zero for convenience, sum torques around the support nearest to the hiker; the torque due to the hiker is L/4 x 602N the torque due to the bridge is L/2 x 4180N the torque due to the right hand support is RL where L is the length of the bridge and R is the force exerted by the right support the torque due to the right support is in the opposite sense to the torques due to the weights, so we have L/4x602 + L/2 x 4180 = R x L or R = 602/4 + 4180/2 = 2240.5N since the total weight supported is 602+4180N = 4782N, the force exerted by the left support is left support force = 4782N - 2240.5N = 2541.5N