F
oy
O
F
ox
τ
o
L/2
F = mg
g
followed by the application of the equilibrium conditions to the free body diagram:
∑ F =
→
0
F = 0
ox
That was quick. Let’s see what setting the sum of the vertical forces equal to zero yields:
∑F = 0
↑
F − F = 0
oy
g
F = F
oy
g
F =
oy
g
m
Now for the torque equilibrium condition:
∑τ = 0
O<
− L
τ
F = 0
o
2 g
− L
τ
g
m
= 0
o
2
1
τ = m
o
gL
2
The wall exerts an upward force of magnitude mg, and a counterclockwise (as viewed from that
1
position for which the free end of the bar is to the right) torque of magnitude mgL on the bar.
2
1
61