Chapter 12, Number 2
The conditions for equilibrium come from Newton's Second Law and its rotational
analog. We need the summation of forces to be equal to zero and the summation of
torques to be equal to zero for a static system.
The force equation gives us two scalar equations: one for the x components and one for
the y components.
We use the standard coordinate representation with the positive y
direction up and the positive x direction to the right.
It follows that:
F
x.net
F
x
R
x
0
F
y.net
F
y
R
y
F
g
0
To use the torque equation we find the torque due to each force individually and then sum
them.
(Note: as the book indicates, we will consider counterclockwise rotations as
positive torques).
Using the definition
r
F
we can find the magnitude of each
torque:
r
F
r
F
sin
where
is the angle between
r
and
F
.
Then the toruqe due to
F
x
is
Fx
F
x
lsin
because it forces a clockwise rotation
while the torque due to
F
y
is given by
Fy
F
y
l
cos
since the angle between
F
y
and the radius vector is
2
.
Now because the radius of the force R is zero,
we have
Rx
Ry
0
.
Now the last torque to consider is that due to the graviational
force.
The angle between this force and a radius is
2
so
sin
cos
and
the torque becomes
Fg
F
g
l
2
cos
.
The final equation is:
net
Fx
Fy
Fg
F
x
lsin
F
y
l
cos
F
g
l
2
cos
0
So our conditions for equilibrium are:
F
x.net
F
x
R
x
0
F
y.net
F
y
R
y
F
g
0
net
Fx
Fy
Fg
F
x
lsin
F
y
l
cos
F
g
l
2
cos
0
Chapter 12, Number 3
We want to find the torque about P when the support at O is not needed.
(Think about
this like a scale.
We would like to move mass 2 until the system is balanced. )
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Note that in this drawing, all forces are in the
y
direction and the radii are all in the
x
direction.
Thus, the magnitude of each torque is given by the magnitude of the
force mulitplied by the radius.
We call counterclockwise rotations positive and note that
the normal at P exerts no torque.
It follows that:
n
O
l
2
d
m
1
g
l
2
d
m
b
g d
m
2
g x
0
If we set
n
O
0
we find:
m
1
g
l
2
d
m
b
g d
m
2
g x
0
.
Hence we arrive at the answer
x
m
1
m
b
d
m
1l
2
m
2
If we were instead given a distance x, we could caluclate the necessary torque at O and
use Newton's Second Law to determine the magnitues of
n
O
and
n
P
.
Chapter 12, Number 12
Since the system is in equilibrium the torque about any point must be zero.
We can set
the pivot to be the edge of the beam that touches the pole.
This choice is convenient as it
means the torque due to the beam is zero since it acts at the pivot point.
Let L represent
the length of the beam, which we should be able to eliminate from our final expressions.
(a)We apply the rotational analog of Newton's Second Law, being careful to consider the
angle between the vectors when taking the cross product.
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 Fall '06
 Winnokur
 Physics, Force, Mass, escape

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