Question

The uniform bar shown below weighs 40 N and is subjected to the

forces shown. Find the magnitude, location, and direction of the force
needed to keep the bar in equilibrium

Answer 1

The magnitude of the force is 90 N, the location of the force is 0.7 m from the left end of the bar and the direction of the force is Upwards.

∑F = 0

60 N force acting downwards at the left end of the bar

80 N force acting downwards at 0.6 m from the left end of the bar

50 N force acting upwards at 0.4 m from the left end of the bar

60 N + 80 N - 50 N -F = 0

F= 90 N

The location of the force is

∑τ = 0

60 N * 1 m + 80 N * 0.6 m - 50 N * 0.4 m -F * x = 0

x = 0.7 m

The direction of the force needed to keep the bar in equilibrium must be upwards because the other three forces are all acting downwards.