Conditions
Equilibrium In chapter 4 we explored the conditions for
mechanical equilibrium
We found that the necessary condition for
mechanical equilibrium is the absence of a net
external force
In that case Newtons First Law stipulates that an
object in mechanical equilibrium stays at rest or
moves with constant velocity
But most often we want to explore the conditions
that are needed for an object to stay at rest, in
static equilibrium
Physics for Scientists &
Engineers 1
Spring Semester 2012
Lecture 30
Static Equilibrium
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The requirement of no translational or rotational motion
means that the linear and angular velocities of this object
are zero at all times
Since the values of the linear and angular velocities do not
change in time
A famous example for such a collection of objects in static
equilibrium is shown below
Part of what makes
this sculpture by
Alexander Calder
so amazing is that
the eye does not
want to accept
that the
configuration it is
in is a stable
static equilibrium
Physics for Scientists&Engineers 1
The linear and angular accelerations are zero at all times
Fnet = ma = 0 and net = I = 0
Static Equilibrium Condition 1:
An object can stay in static equilibrium only if the net force acting
on it is zero.
Static Equilibrium Condition 2:
An object can stay in static equilibrium only if the net torque acting
on it is zero. Even if Newtons First Law is satisfied and an object
has no translational motion, it will still rotate if it experiences a net
torque.
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Equilibrium Conditions
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If an object is supported from a pin
located directly above its center of mass,
then the object stays balanced
It does not move or rotate
For static equilibrium with zero net torque, the net torque has to be
zero for any pivot point chosen
A clever choice of pivot point can be the key to a quick
solution
For example, we can select the point where an unknown force acts as
the pivot point
Then the unknown force would not contribute to the net torque
because it would have a zero moment arm
Physics for Scientists&Engineers 1
Physics for Scientists&Engineers 1
Equilibrium Conditions
Torque is always defined with respect to a pivot point
The pivot point is the intersection of the plane determined
by the force and moment arm vectors and the axis of
rotation
When we compute the net torque, the pivot point must be
the same for all forces
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Equilibrium Conditions
Equilibrium Conditions
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Why?
The only two forces acting on the object
are the force of gravity and the normal force
The two forces cancel each other and produce
no torque
If the object is supported in the same way
from a pin but its center of mass is not
below the support point, a net torque acts
Object is not in static equilibrium
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Experimentally
Experimentally Determining
the Center of Mass
Experimentally Determining
the Center of Mass
We can locate the center mass
for many objects supporting the
object on two fingers such that
the center of mass is located
somewhere between them
Slowly slide the fingers closer to
each other
The finger closer to the center of
mass exerts a higher normal
force and thus a high friction
force
When the fingers meet, you have
determined the center of mass
We can locate the center of mass of an
object experimentally by suspending
the object from different points
First, support the object in such a way
that it can rotate freely
Use a plumb bob to mark the vertical
line
Support the object from a second point
such that it can rotate freely and mark
the vertical line
The center of mass will lie at the
intersection of the lines
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Equilibrium Equations
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Seesaw
The equations for equilibrium include zero net
force
The seesaw is a playground toy, which consists of a
pivot and a bar, of mass M, that is placed over it so
that its ends can move up or down freely as shown
below
n
Fnet, x = Fi , x = F1, x + F2, x + ... + Fn, x = 0
i =1
n
Fnet, y = Fi , y = F1, y + F2, y + ... + Fn, y = 0
i =1
And zero net torque
net = counter-clockwise,i clockwise, j = 0
i
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j
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Seesaw
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Seesaw
If you place a an object of mass m1 on one end at a
distance r1 from the center axle, then obviously
that end goes down, simply because of the force
and torque that the object exerts on it
Question 1:
Where do you have to place an
object of mass m2 to get
the see saw to be balanced,
so that the bar is horizontal
and neither touches
the end ground?
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The force that m1 exerts on the bar is simply m1g,
acting downward
The same is true for force that m2 exerts on the
bar
The weight of the bar is Mg acting at
the center of mass of the bar
The final force is the
normal force N exerted
by the support of the
seesaw acting on the axle of the seesaw
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2
Seesaw
Standing on a board
Our equilibrium force equation is
A construction supervisor of mass M = 90.6 kg is standing on
a board of mass m = 27.0 kg. Two sawhorses at a distance l =
4.40 m apart support the board. If the man stands a
distance x1 = 1.188 m away from the left-hand sawhorse as
shown, what is the magnitude of the force that the board
exerts on that sawhorse?
Fnet, y = Fi , y = m1g m2 g Mg + N = 0
i
N = g( m1 + m2 + M )
Our equilibrium torque equation is
net = clockwise,i counter-clockwise, j
i
j
= m2 gr2 sin 90 m1gr1 sin 90 = 0
m2 r2 = m1r1
m
r2 = r1 1
m2
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Standing on a board
FR
pivot
mg
clockwise, j
= FL
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counter-clockwise,i
i
net = counter-clockwise,i clockwise, j = 0
i
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Stacking Blocks
= mg + Mg x2
2
j
mg + Mg x2 FL = 0
2
mg + Mg x2 FL = 0
2
Mg x2 + mg
2
FL =
90.6kg 9.81
Physics for Scientists&Engineers 1
j
FL =
Mg
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Standing on a board
Find the force on the board exerted by the left sawhorse
Draw a free-body diagram
The net torque is zero net = counter-clockwise,i clockwise, j = 0
i
j
Choose a pivot point
FL
Physics for Scientists&Engineers 1
Calculate the torques
M = 90.6 kg, m = 27.0 kg. l = x1 + x2 = 4.40 m and x1 = 1.188
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FR
FL
mg
pivot
Mg
m
m 4.4m
[ 4.4 1.188] m + 27.0kg 9.81 2
s2
s2
= 781N
4.4m
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Stacking Blocks
Start with one block
If the block has length and uniform mass density, then it
center of mass is located at (1/2)
We can use the definition of the center of gravity to
produce a surprising result
Lets start with a stack of seven identical blocks
Surprising result:
Last block is
completely
over the edge of
the table!
How far can we push out the leading edge of the block
without the pile of blocks falling over?
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We call the x-coordinate of the center of mass x1
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Stacking Blocks
Stacking Blocks
We call the x-coordinate of the center of mass of the
second block x2
The center of mass of the combined system of two blocks is
xm +x m
x12 = 1 1 2 2 m1 = m2 x12 = 1 ( x1 + x2 )
2
m1 + m2
Block 1 can stay at rest as long as at least half of it is
supported from below and can stick out an infinitesimal
amount less than (1/2) and remain at rest
Now add a second identical block
In the limiting case that the center of mass of the first
block is still supported from below by the second block
x1 = x2 + 1
2
Combining these two equations
gives us
x12 = 1 ( x1 + x2 ) =
2
x12 = 1 (( 1 + x2 ) + x2 )
2
2
x12 = x2 + 1
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Stacking Blocks
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Stacking Blocks
Shifting x12 to the very edge of the third block gives us
Now add a third block
x12 = x3 + 1
2
Combining this with our result for two blocks, we get
x12 = x3 + 1 = x3 + 1 x2 = x3 + 1
4
4
2
The center of mass of the three blocks is
x12 ( 2 m ) + x3 m
2m + m
x123 = 2 x12 + 1 x3
3
3
x123 =
The top two blocks will not topple if the combined center of
mass of x12 is supported from below
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x123 =
2
3
( x3 + 1 ) + 13 x3
2
x123 = x3 + 1
3
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Stacking Blocks
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Stacking Blocks
Now add a fourth block and require that the top three
blocks are supported from below
Combining our equations so far we get
x123 = x3 + 1 = x4 + 1 x3 = x4 + 1
3
2
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x123 = x4 +
1
2
We could keep going on to the fifth block, but lets think
If we have (n 1) blocks supported by the nth block, then
the coordinates of the (n 1)st and nth block are related as
xn 1 = xn +
2n 2
We can add up all the terms and find out how far away x1
can be from the edge of the table
n
x1 = xn +1 + ( 1 )
2
i =1
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1
i
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4
Stacking Blocks
We get the amazing result that x1 can move infinitely far
from the edge of the table
Here are our seven blocks
The last block is completely over the edge
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