Physics 204A Class Notes
111
Section 12 – Equilibrium
Now we have a fairly complete answer to our question, “what do objects do and why do they do
it?” We will spend the rest of the semester applying our knowledge to a variety of new situations.
In
this chapter, we will focus our attention on objects that are at rest and remain at rest.
Such objects are
said to be in “static equilibrium.”
This is highly desired for buildings, bridges and many other
structures.
An object in static equilibrium has no acceleration and no angular acceleration.
According to the
Second Laws for Translation and Rotation,
!
r
F
=
m
r
a
" !
r
F
=
0
and
!
r
"
=
m
r
#
$ !
r
"
=
0
.
That is, the net force on the object and the net torque on the object must be zero.
The rest of this section
consists of example problems on increasing complexity.
Example 12.1:
A 5.00kg piñata is hung from the middle of a 4.00m long rope.
Find the tension
in the rope if it sags 20.0cm.
Given: m = 5.00kg,
l
= 4.00m, and s = 20.0cm
Find: F
t1
= ? and F
t2
= ?
The angle that the tension makes with the horizontal can be
found using the definition of the sine,
sin
"
=
s
l
2
=
2
s
l
#
"
=
arcsin
2
s
l
$
%
&
’
(
)
=
arcsin
2 0.2
(
)
4
$
%
&
’
(
)
#
"
=
5.74˚
Applying the Second Law to the point where the rope connects
to the piñata,
"
F
x
=
ma
x
#
F
t
2
cos
$
%
F
t
1
cos
$
=
0
#
F
t
2
=
F
t
1
=
F
t
Since the piñata is in the middle of the rope, both angles are equal and the tension in both sides
must be the same.
Looking at the ydirection,
!
F
y
=
ma
y
"
F
t
sin
#
+
F
t
sin
# $
mg
=
0
"
2F
t
sin
#
=
mg
"
F
t
=
mg
2sin
#
Putting in the numbers,
F
t
=
(5.00)(9.80)
2sin 5.74˚
!
F
t
=
245N
.
This problem did not require the use of torques.
Now, let’s look at a problem that involves some torques.
x
y
F
g
F
t1
F
t2
s
θ
θ
l
2
l
2
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Physics 204A Class Notes
112
Example 12.2: A 20.0kg board 4.00m long is supported at each end.
A 70.0kg person stands
1.00m from the left side.
Find the force exerted by each support.
Given: m = 20.0kg, M = 70.0kg, x
1
= 1.00m,
x
2
= 2.00m, and x
3
= 4.00m
Find: F
L
= ? and F
R
= ?
Note that choosing the origin where one of the unknown
forces acts simplifies the algebra of the solution because the
torque about the origin caused by this unknown force will be
zero.
So this unknown won’t appear in the equation
produced by the Second Law of Rotation.
Applying the Second Law for Rotation to the board about
the origin using the convention that counterclockwise torques are positive,
!"
o
=
I
# $
(0)F
L
%
x
1
Mg
%
x
2
mg
+
x
3
F
R
=
0
.
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 Fall '09
 Kagan
 Physics, Force, Physics 204A Class, 204A Class Notes

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