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

Chapter 23 Statics

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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