Question

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Answer 1

Using Force balance on beam in vertical direction:

Fy_net = 0

F1 + F2 - W - 1500 - 2050 = 0

here, W = weight of beam = m*g

m = mass of beam = 500 kg

g = surface gravity = 9.8 m/s^2

then,

F1 + F2 = 500*9.8 + 3550

F1 + F2 = 8450 N

Now apply torque balance about the right end of beam:

Net torque = 0

So, Counterclockwise torque = Clockwise torque

(Since torque = rxF = r*F*sin 90 deg = r*F)

F1*(3 + 11 + 6) = 1500*(11 + 6) + W*(11 + 6 + 3)/2 + 2050*6

F1 = [1500*17 + 500*9.8*20/2 + 2050*6]/20

F1 = 4340 N

Now

F1 + F2 = 8450 N

F2 = 8450 - 4340

F2 = 4110 N

Let me know if you've any query.