Auniform horizontal beam of weight 481 n and length 3.32 m has two weights hanging from it. one weight of 381 n is located 0.8798 m from the left end; the other weight of 281 n is located 0.8798 m from the right end. what must be the magnitude of the one additional force on the beam that will produce equilibrium?

1143 N at 1.59 m from the left end

Explanation:

For the system to produce equilibrium, the total force and moment must be 0. Since the total weight downward is

481 + 381 + 281 = 1143 N

Therefore the magnitude of the force acting upward to balance this system must be the same of 1143 N

That alone is not enough, we also need the position of the force for the total moment to be 0.

Let x be the length from the this upward force to the left side. And let the left point be the point of reference for moment arm:

481 * 3.32/2 + 381 * 0.8798 + 281*(3.32 - 0.8798) - 1143*x = 0

x = (481*1.66 + 381 * 0.8798 + 281*2.4402)/1143 = 1.59m

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