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2) Calculate FA and Fg. Assume the beam is uniform and has a mass of 280 kg. (10 pts) 4300 N 3100 N 2200 N FA FB 2.0 m 4.0 m

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Answer #2

The following diagram shows the forces acting on the beam, is the weight of the beam. The distance between the forces are respectively.

Picture 1

In the given problem assume that the beam is uniform. The beam is in equilibrium condition so the net torque and net force must be zero.

Consider the forces in vertical direction.

…… (1)

Here, is force applied at B point, and is force applied at point.

Consider the torque is in equilibrium then net force.

Here, is the torque, is force applied at B point, is the distance between two forces.

Substitute for, for, for, for, for and 0 for

Substitute for in equation (1) and simplify,

Therefore the forces and

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Answer #1

DATE: (4300N)[2m) + (274490/5m) + (31000)m) + (22001) (9m) Follom) = 0 FB = 6072 N NE from eq @ was FA + 6072 = 12344 IFA =

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