Chapter 12
Equilibrium and Elasticity
Equilibrium and Elasticity
I. Equilibrium
- Definition
- Requirements
- Static equilibrium
II. Center of gravity
III. Elasticity
- Tension and compression
- Shearing
- Hydraulic stress
I. Equilibrium
- Definition: An object is in equilibrium if:
- The linear momentum of its center of mass is constant.
- Its angular momentum about its center of mass is
constant
Example: block resting on a table, hockey puck sliding
across a frictionless surface with constant velocity, the
rotating blades of a ceiling fan, the wheel of a bike traveling
across a straight path at constant speed.

- Static equilibrium: P  0,
Objects that are not moving

L  0 either in TRANSLATION or
ROTATION
Example: block resting on a table.
Stable static equilibrium:
If a body returns to a state of static equilibrium after having
been displaced from it by a force  marble at the bottom of a
spherical bowl.
Unstable static equilibrium:
A small force can displace the body and end the equilibrium
.
(1) Torque about supporting
edge by Fg=0 because line of
action of Fg through rotation axis
 domino in equilibrium.
(1) Slight force ends equilibrium
 line of action of Fg moves to
one side of supporting edge 
torque due to Fg increases
domino rotation.
(3) Not as unstable as (1)  to topple it one needs to rotate it
through beyond balance position in (1).
.
- Requirements of equilibrium:



dP
P  cte  Fnet 
0
dt



dL
L  0,
 net 
0
dt
Balance of forces  translational
equilibrium
Balance of torques  rotational
equilibrium
- Equilibrium:
- Vector sum of all external torques that act on the body,
measured about any possible point must be zero.
- Vector sum of all external forces that act on body must be
zero.
Balance of forces  Fnet,x = Fnet,y = Fnet,z =0
Balance of torques  τnet,x = τnet,y = τnet,z =0
II. Center of gravity
Gravitational force on extended body  vector sum of the
gravitational forces acting on the individual body’s elements
(atoms) .
cog = Body’s point where the gravitational force “effectively” acts.
- This course initial assumption:
The center of gravity is at
the center of mass.
If g is the same for all elements of a body, then the
body’s Center Of Gravity (COG) is coincident with the
body’s Center Of Mass (COM).
Assumption valid for every day objects  “g” varies only slightly
along Earth’s surface and decreases in magnitude slightly with
altitude.
Each force Fgi produces a
torque τi on the element of
mass about the origin O,
with moment arm xi.
  r F   i  xi Fgi   net    i   xi Fgi
i
  xcog Fg  xcog  Fgi   net
i
i
xcog  Fgi   xi Fgi xcog  mi g i   xi mi g i  xcog  mi  xi mi
i
 xcog
i
1

 xi mi  xcom
M i
i
i
i
i
A baseball player holds a 36-oz bat (weight = 10.0 N) with one
hand at the point O . The bat is in equilibrium. The weight of the
bat acts along a line 60.0 cm to the right of O. Determine the
force and the torque exerted by the player on the bat around an
axis through O.
A uniform beam of mass mb and length ℓ supports blocks
with masses m1 and m2 at two positions. The beam rests
on two knife edges. For what value of x will the beam be
balanced at P such that the normal force at O is zero?
A circular pizza of radius R has a circular piece of radius R/2
removed from one side as shown in Figure. The center of
gravity has moved from C to C’ along the x axis. Show that
the distance from C to C’ is R/6. Assume the thickness and
density of the pizza are uniform throughout.
Pat builds a track for his model car out of wood, as in
Figure. The track is 5.00 cm wide, 1.00 m high and 3.00 m
long and is solid. The runway is cut such that it forms a
parabola with the equation
2
( x  3)
y
9
Locate the horizontal coordinate of the center of gravity of
this track.
Find the mass m of the counterweight needed to balance
the 1 500-kg truck on the incline shown in Figure. Assume
all pulleys are frictionless and massless.
A 20.0-kg floodlight in a park is supported at the end of a
horizontal beam of negligible mass that is hinged to a pole, as
shown in Figure. A cable at an angle of 30.0° with the beam
helps to support the light. Find (a) the tension in the cable and
(b) the horizontal and vertical forces exerted on the beam by
the pole.
States of Matter
• Matter is characterised as being solid,
liquid or gas
• Solids can be thought of as crystalline
or amorphous
• For a single substance it is normally the
case that
– solid state occurs at lower temperatures
than liquid state and
– liquid state occurs at lower temperatures
than gaseous state
Solids and Fluids
• Solids are what we have assumed all
objects are up to this point (rigid,
compact, unchanging, simple shapes)
• We will now look at bulk properties of
particular solids and also at fluid
• Fluids include both liquids and gases
– fluids assume the shape of their container
Deformation of Solids
• All states of matter (S,L,G) can be deformed
– it is possible to change the shape and volume of
solids and
– the volumes of liquids and gases
• Any external force will deform matter
– for solids the deformation is usually small in
relation to its overall size when ‘everyday’
forces are applied
Stress and Strain
• When force is removed, the object will usually
return to its original shape and size
– Matter is elastic
• We characterise the elastic properties of solids in
terms of stress (amount of force applied) and
strain (extent of deformation) that occur.
• The amount of stress required to produce a
particular amount of strain is a constant for a
particular material
– this constant is called the elastic modulus
stress
Elastic Modulus 
strain
Young’s Modulus (length elasticity)
• F creates a tensile stress of F/A.
– Units of tensile stress are Pascal
– 1 Pa = 1 N/m2
• Change in length created
by stress is
DL/Lo (no unit)
This quantity is the
tensile strain
F
A
Lo
Young’s modulus
• Young’s modulus is the ratio of tensile
stress to tensile strain.
FLo
F/A
Y

L / Lo AL
• Y has units of Pa
• Young’s modulus applies to rod or wire
under tension (stretching) or compression
What is happening in detail?
• Bonds between atoms are compressed or
put in tension
Elastic behaviour
• Notice that the Stress is proportional to
the Strain
– This is similar to the relation we had
between spring force and its extension
|F| = kx
– We can identify k with YA/Lo
F = kx so F/A = kx/A=YDL/Lo
Limits of elastic behaviour
Shear Modulus
• Shear modulus characterises a body’s
deformation under a sideways (tangential)
force
A
h
x
F/A
S
x / h
F
Bulk Modulus
• Response of body to uniform squeezing
• Bulk modulus is ratio of the change in
the normal force per unit area to the
relative volume change
F / A
B
V / V
P

V / V
F
Values for Y, S and B
Y (N/m2)
S (N/m2)
B (N/m2)
Al
7.0 x 1010
2.5 x 1010
7.0 x 1010
Water
-
-*
0.21 x 1010
Tungsten
35 x 1010
14 x 1010
20 x 1010
Glass
7 x 1010
3 x 1010
5.2 x 1010
– Note that liquids do not have Y defined. S for
liquids is called viscosity.
– In liquids and gases S and B are strongly
dependent on temperature (more later)
Example - Young’s modulus
• How much force must be applied to a 1 m long
steel rod that has end area 1 cm2 to fit it inside a
0.999 m long case? (Ysteel=20x1010N/m2)
1 cm2
Density
• Density of a pure solid, liquid, or gas is its
mass per unit volume
• Density r  m/V (units are kg/m3)
Pressure
• Pressure is the force per unit area
P  F/A
Pressure at any point can be measured
by placing a plate of known area in
(e.g.) a liquid and measuring the force
(compression) on a spring attached to it.