We are going to need to apply the torque equilibrium condition to the beam so I am going to add
moment arms to the diagram. My plan is to sum the torques about point O so I will depict
moment arms with respect to an axis through point O.
r⊥ = x = 1.80 m
F N
F Py
r = x/2
⊥
Fg
O
F
Px
F
F
N
g
Now let’s apply the equilibrium conditions:
∑ F =
→
0
F
= 0
P x
There are three unknown force values depicted in the free body diagram and we have already
found one of them! Let’s apply another equilibrium condition:
∑ F = 0
↑
F − F + F = 0
P y
g
N
F − mg + F = 0 (23-6)
P y
N
There are two unknowns in this equation. We can’t solve it but it may prove useful later on.
Let’s apply the torque equilibrium condition.
∑τ = 0
O<
− x F + xF = 0
2 g
N
− x mg + xF = 0
2
N
1
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