It bears repeating: Make sure that any force that enters the torque equilibrium equation
is multiplied by a moment arm, and that any pure torque (such as τ
o in the solution of
example 23-2 on page 151) that enters the torque equilibrium equation is NOT multiplied
by a moment arm.
For any rigid body, at any instant in time, Newton’s 2nd Law for translational motion
= 1
a
∑F
m
and Newton’s 2nd Law for Rotational motion
= 1
a
∑τ
I
both apply. In this chapter we focus on rigid bodies that are in equilibrium. This topic, the study
of objects in equilibrium, is referred to as statics. Being in equilibrium means that the
acceleration and the angular acceleration of the rigid body in question are both zero. When
a = 0 , Newton’s 2nd Law for translational motion boils down to
∑F = 0 (23-1)
and when a = 0 , Newton’s 2nd Law for Rotational motion becomes
∑τ = 0 (23-2)
These two vector equations are called the equilibrium equations. They are also known as the
equilibrium conditions. In that each of the vectors has three components, the two vector
equations actually represent a set of six scalar equations:
∑ F = 0
x
∑F = 0
y
∑ F = 0
z
∑τ = 0
x<
∑τ = 0
y<
∑τ = 0
z<
1
55