Chapter 11
Equilibrium and
Elasticity
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Goals for Chapter 11
• To study the conditions for equilibrium of a body
• To understand center of gravity and how it relates to a
body’s stability
• To solve problems for rigid bodies in equilibrium
Goals for Chapter 11
• To analyze situations involving tension, compression,
pressure, and shear
• Real materials are not truly rigid. They are elastic
and do deform to some extent.
• To investigate what happens when a body is stretched
so much that it deforms or breaks
Conditions for equilibrium
• First condition:
The sum of all external forces is equal to zero:
Fx = 0
• Second condition:
Fy = 0
Fz = 0
The sum of all torques about any (and ALL!)
given point(s) is equal to zero.
Conditions for equilibrium
Conditions for equilibrium
• First condition:
The sum of all external forces is equal to zero:
Fx = 0
• Second condition:
Fy = 0
Fz = 0
The sum of all torques about any (and ALL!)
given point(s) is equal to zero.
Center of gravity
• Treat body’s weight as
though it all acts at a
single point—
the center of gravity…
Center of gravity
• If ignore variation of
gravity with altitude…
Center of Gravity
(COG) is same as
center of mass.
Center of gravity
• We can treat a body’s
weight as though it all
acts at a single point—
the center of gravity…
mr

m r  m r  m r  


m  m  m  
m
i i
rcm
11
2 2
1
2
3 3
i
3
i
i
Center of gravity
Center of gravity
Center of gravity & Stability
Walking the plank
• Uniform plank (mass = 90 kg, length = 6.0 m) rests on
sawhorses 1.5 m apart, equidistant from the center.
• IF you stand on the plank at the right edge, what is the
maximum mass m so that the plank doesn’t move?
Walking the plank
• Ask WHAT WILL HAPPEN?
• It will rotate CLOCKWISE!
• If it rotates, where will it rotate around? ID Axis!
• It will rotate around S!
Walking the plank
• What is the maximum mass “m” for stability?
Origin at c
Center of
Gravity at s
Walking the plank
m r  m2 r2  m3r3  
rcm  1 1

m1  m2  m3  
mr
m
i i
i
i
i
Rcm = [M(@0) + m (@ ½ L)]
M+m
= m( ½ L)
=½D
(M+m)
Center of Gravity is the point where the gravitational
torques are balanced.
Walking the plank
m r  m2 r2  m3r3  
rcm  1 1

m1  m2  m3  
mr
m
i i
i
i
i
Rcm = [M(@0) + m (@ ½ L)]
M+m
So… m =
MD
(L – D)
= m( ½ L)
=½D
(M+m)
= 90 kg (1.5/6-1.5) = 30 kg
Solving rigid-body equilibrium problems
• Make a sketch; create a coordinate system and draw all
normal, weight, and other forces.
• Specify a positive direction for rotation, to establish the
sign for all torques in the problem. Be consistent.
• Choose a “convenient” point as your reference axis for
all torques. Remember that forces acting at that point
will NOT produce a torque!
• Create equations for Fx = 0 Fy = 0 and t  0
Solving rigid-body equilibrium problems
• Car with 53% of weight on front wheels and 47% on
rear. Distance between axles is 2.46 m. How far in
front of the rear axle is the center of gravity?
Solving rigid-body equilibrium problems
• Suppose you measure from the REAR wheels….
• Forces at the axis of rotation create NO torque!
t = 0 =>
[0.47w x 0 meters] – wLcg
+[0.53w x 2.46m] = 0
So..
Lcg = 1.30 m
Solving rigid-body equilibrium problems
• Suppose you measure from the FRONT wheels….
• Forces at the axis of rotation create NO torque!
t = 0 =>
- [0.47w x 2.46 meters] + w L’cg
+[0.53w x 0 meters] = 0
L’cg
So..
L’cg = 1.16 m
L’ + L = 1.16 + 1.30 = 2.46 m
Ladder Problems!
• Lots of variables:
• Length of ladder L
• Mass of ladder (and its center of gravity)
• Angle of ladder against wall
• Normal force of ground
• Static Friction from ground
• Normal force of wall
• Static Friction of wall
• Weight of person on ladder
• Location of person on ladder
Ladder Problems!
• Lots of variables:
• Length of ladder L
• Mass of ladder (and its center of gravity)
• Angle of ladder against wall
• Normal force of ground
Wall
(normal
force and
friction!)
• Static Friction from ground
• Normal force of wall
cog of
ladder
• Static Friction of wall
• Weight of person on ladder
• Location of person on ladder
Ground
(normal
force and
friction!)
Ladder Problems!
• Equilibrium involves Fx = 0
Fy = 0
tz = 0
• 3 Equations!
• Solve for up to 3 unique
unknowns!
• Use key words/assumptions to
narrow choices
• “uniform ladder”
• “just about to slip”
• “frictionless”
Ground
(normal
force and
friction!)
Wall
(normal
force and
friction!)
Ladder Problems!
fs (wall)
N (wall)
N
(ground)
Ground
(normal
force and
friction!)
Weight
(ladder)
Angle q
fs
(ground)
Wall
(normal
force and
friction!)
Ladder Problems!
Torques about
point 1
Ground forces
create no torque
about point 1!
N
(ground)
Angle q
1
fs
(ground)
Ladder Problems!
Torques about
point 1
Line of
Action
Weight drawn from
center of gravity
Torque = Wladder x d
Clockwise!
= - Wladder x (1/2 L sinq)
Weight
(ladder)
+
Angle q
1
d
Lever
Arm
Ladder Problems!
Torques about
point 1
Line of
Action
fs (wall)
N (wall)
Angle q
Lever
Arm =
Lsinq
Line of
Action
Angle q
1
Lever
Arm =
Lcosq
Ladder Problems!
• Fx = 0 = +fs (ground) – N (wall)
• Fy = 0 = N (ground) + fs (wall) – Wladder
• t1 = 0
= +[N (wall) Lsinq] CCW
+ [fs (wall) Lcosq] CCW
- [Wladder Lsinq] CW
Will the ladder slip?
• 800 N man climbing ladder 5.0 m long that weighs 180 N.
Find normal and frictional forces that must be present at
the base of the ladder.
• What is the minimum coefficient of static friction at the
base to prevent slipping?
• What is the magnitude and direction of the contact force
on the base of the ladder?
Will the ladder slip?
• 800 N man climbing ladder 5.0 m long that weighs 180 N.
Find normal and frictional forces that must be present at
the base of the ladder.
• What is the minimum coefficient of static friction at the
base to prevent slipping?
Will the ladder slip?
Fx = 0 =>
fs – n1 = 0
Will the ladder slip?
Fx = 0 =>
fs – n1 = 0
Fy = 0 =>
n2 – wperson – wladder = 0
Will the ladder slip?
Fx = 0 =>
fs – n1 = 0
Fy = 0 =>
n2 – wperson – wladder = 0
t (about ANY point) = 0 =>
+n1(4m) – wp(1.5m) –wl(1.0m) = 0
Will the ladder slip?
Fx = 0 =>
fs – n1 = 0
Fy = 0 =>
n2 – wperson – wladder = 0
t (about ANY point) = 0 =>
+n1(4m) – wp(1.5m) –wl(1.0m) = 0
So
n1 = 268 N
Will the ladder slip?
Fx = 0 =>
fs – n1 = 0
Fy = 0 =>
n2 – wperson – wladder = 0
t (about ANY point) = 0 =>
+n1(4m) – wp(1.5m) –wl(1.0m) = 0
So
n1 = 268 N & ms(min) = fs/n2 = .27
Will the ladder slip? – Example 11.3
• 800 N man climbing ladder 5.0 m long that weighs 180 N.
Find normal and frictional forces that must be present at
the base of the ladder.
• What is the minimum coefficient of static friction at the
base to prevent slipping?
• What is the magnitude and direction of the contact force
on the base of the ladder?
Example 11.4: Equilibrium and pumping iron
Given weight w and angle between tension and horizontal,
Find T and the two components of E
Strain, stress, and elastic moduli
• Stretching, squeezing, and twisting a real body causes it to deform.
• Stress is force per unit area
• Measured in N/m2 or “Pascals” or lbs/sq. in. or “PSI”
•1 PSI ~ 7000 Pa
•Typical tire pressure ~ 32 PSI ~ 200 kPA
• Same units as PRESSURE in fluids/gases
• Stress is an internal force on an object that produces (or results
from) a Strain
Strain, stress, and elastic moduli
• Stretching, squeezing, and twisting a real body causes it to deform.
• Stress is force per unit area
• Strain is the relative change in size or shape of an object because of
externally applied forces.
• Strain is the fractional deformation due to the stress.
• Strain is dimensionless – just a fraction.
• Linear strain = Dl/l0 (tensile strain)
Strain, stress, and elastic moduli
• Stretching, squeezing, and twisting a real body causes it to deform.
• Stress is force per unit area
• Strain is the fractional deformation due to the stress.
• Elastic modulus is stress divided by strain.
• The direct (linear) proportionality of stress and strain is called
Hooke’s law:
STRESS / STRAIN = Elastic Modulus
Strain, stress, and elastic moduli
• The direct (linear) proportionality of stress and strain is called
Hooke’s law.
• Similar to spring/rubber bands stretching:
• Apply external force F, see spring/band deform a distance
Dx from the original length x.
•Apply that force F over the entire cross sectional area of the
rubber band (or thickness of the spring wire)
•F/A creates a STRESS
Strain, stress, and elastic moduli
• The direct (linear) proportionality of stress and strain is called
Hooke’s law.
• Similar to spring/rubber bands stretching:
•Apply that force F over the entire area of the rubber band
(or thickness of the spring wire) F/A creates a STRESS
•Deformation of band in length Dx/x is the resulting Strain
• F = kDx
•F/A = Stress = kDx/A = YDx/x = Y X Strain
Strain, stress, and elastic moduli
• The direct (linear) proportionality of stress and strain is called
Hooke’s law.
• Similar to spring/rubber bands stretching:
• Tensile Stress/Strain = Y = Young’s Modulus
•F/A = Stress = kDx/A = YDx/x = Y X Strain
•Y = kx/A = Force/Area (Pascals)
Tensile and compressive stress and strain
• Tensile stress = F /A & tensile strain = Dl/l0
(stretching under tension)
• Young’s modulus is tensile stress divided by tensile strain, and is
given by Y = (F/A)(l0/Dl)
Tensile and compressive stress and strain
• Tensile stress = F /A & tensile strain = Dl/l0
Young’s modulus is tensile stress divided by tensile strain, and is
given by Y = (F/A)(l0/Dl)
• Compressive stress & compressive strain defined similarly.
Some values of elastic moduli
Tensile stress and strain
• Body can experience both tensile & compressive stress
at the same time.
Bulk stress and strain
• Pressure in a fluid is force
per unit area: p = F/A.
• Bulk stress is pressure
change Dp & bulk strain is
fractional volume change
DV/V0.
• Bulk modulus is bulk stress
divided by bulk strain and is
given by B = –Dp/(DV/V0).
Sheer stress and strain
• Sheer stress is F||/A and
sheer strain is x/h.
• Sheer modulus is sheer
stress divided by sheer
strain, and is given by
S = (F||/A)(h/x).
Elasticity and plasticity
• Hooke’s law applies up to point a.
• Table 11.3 shows some approximate breaking stresses.