Calculus-Based Physics by Jeffrey W. Schnick - HTML preview

Chapter 23 Statics

then a free body diagram:

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

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