Static Equilibrium and Torque Homework Question

AI Thread Summary
A hiker weighing 985 N crosses a uniform bridge that weighs 3610 N, stopping one-fifth of the way along. The problem involves calculating the forces exerted by the concrete supports at both ends of the bridge. The calculations yield a force of 2590 N at the near end and 2010 N at the far end. The approach used involved setting the opposite support as the pivot point to calculate torque and checking the vertical forces for equilibrium. The discrepancy in the far end calculation suggests a need for reevaluation of the force distribution or rounding methods.
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Homework Statement



A hiker, who weighs 985 N is strolling through the woods and crosses a small horizontal bridge. the bridge is uniform, weighs 3610 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:

a) at the near end and
b) at the far end

Answer to a) 2590, b) 2010

Homework Equations



Static Equilibrium: ƩT = 0
T = Frsinθ


The Attempt at a Solution



Took opposite support as pivot point such that Fn x r is the CCW torque where r is the length of bridge
Expressed distance from the pivot point of each force in terms of r
Wrote net torque equation, crossed out r from both sides
Solved for Fn

a) Fn = 985 x 4/5 + 3610 x 1/2
Fn = 2593 N
b) Fn = 985 x 1/5 + 3610 x 1/2
Fn = 2002 N

I am not sure whether or not my approach to this problem is correct.
 
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Yes, looks good.
You can also sum the vertical forces - total must be zero - as a check.
2593 rounds to 2590, but I don't see how to get 2010 from 2002.
 
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