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Chapter 23
Statics
155
23
Statics
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
232
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 2
nd
Law for translational motion
∑
=
F
h
h
m
1
a
and Newton’s 2
nd
Law for Rotational motion
∑
=
τ
a
h
I
1
a
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
0
=
a
h
, Newton’s 2
nd
Law for translational motion boils down to
0
=
∑
F
h
(231)
and when
0
=
a
a
, Newton’s 2
nd
Law for Rotational motion becomes
0
=
∑
τ
h
(232)
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:
∑
=
0
x
F
∑
=
0
y
F
∑
=
0
z
F
∑
=
0
x
<
∑
=
0
y
<
∑
=
0
z
<
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View Full DocumentChapter 23
Statics
156
In many cases, all the forces lie in one and the same plane, and if there are any torques aside
from the torques resulting from the forces, those torques are about an axis perpendicular to that
plane.
If we define the plane in which the forces lie to be the xy plane, then for such cases, the
set of six scalar equations reduces to a set of 3 scalar equations (in that the other 3 are trivial 0=0
identities):
∑
=
0
x
F
(233)
∑
=
0
y
F
(234)
∑
=
0
z
<
τ
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 Fall '08
 RABE
 Physics, Force

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