Chapter 11
Equilibrium and Elasticity
11.1
Conditions for Equilibrium
An object is said to be in equilibrium if two conditions are satisfied:
• First condition for equilibrium. The sum of all external forces
acting on the body is zero:
!
⃗ = 0.
F
(11.1)
• Second condition for equilibrium. The sum of all torques (due to
external forces about any point) on the body is zero:
!
⃗τ = 0.
(11.2)
Although we shall only apply these equilibrium conditions to bodies at rest,
the same conditions applies to bodies in uniform translational motion (without rotation).
149
CHAPTER 11. EQUILIBRIUM AND ELASTICITY
150
Example. Which situation satisfies both the first and second conditions
CHAPTER 11. EQUILIBRIUM AND ELASTICITY
151
for equilibrium?
(i) a seagull gliding at a constant angle below the horizontal
and at constant speed;
(ii) an automobile crankshaft turning at an increasing angular speed in
the engine of a parked car;
(iii) a thrown baseball that does not rotate as it sails through the air.
11.2
Center of Gravity
Consider a collection of particles which make up a given object, then their
center of mass is then given by
"
mi⃗ri
m1⃗r1 + m2⃗r2 + m3⃗r3 + ...
⃗rcm =
= i
,
(11.3)
m1 + m2 + m3 + ...
M
where
M=
!
mi .
(11.4)
i
Now let us consider gravitational torque caused by gravitational forces with
axis of rotation in the origin
⃗τ =
!
⃗τi
i
=
!
i
=
⃗ri × (mi⃗gi )
! #⃗ri mi $
i
M
× (M⃗gi )
(11.5)
In general the gravitational acceleration changes with attitude, but in a
special case when gravitational acceleration is constant the torque is given
by
$
#"
ri
i mi⃗
× (M⃗g )
⃗τ =
M
= ⃗rcm × (M⃗g ) .
(11.6)
In fact we have already used this result (without proving it first) by assuming
that the torque due to gravitational force can always be assumed to be applied
to the point of the center of mass. It turns out that even in a more general
case when the gravitational acceleration decreases with altitude
⃗τ = ⃗rcg × (M⃗g) .
(11.7)
CHAPTER 11. EQUILIBRIUM AND ELASTICITY
152
where ⃗rcg is the center of gravity.
Finding the center of gravity (theoretically) is easy when the object is
symmetric, but what if the object has irregular shape? In this case one can
preform a simple experiment of suspending an object from a given point.
Since the object is in equilibrium the vertical line extending from these
points of suspension must pass through the center of gravity or otherwise
there would be a non-zero torque due to gravity. By preforming the same
experiment for two distinct points a pair of such lines can be constructed and
then the point where the lines intersect should be the center of gravity.
CHAPTER 11. EQUILIBRIUM AND ELASTICITY
153
Example 11.1. A uniform plank of length L = 6.0 m and mass M =
90 kg rests on sawhorses separated by D = 1.5 m and equidistant from the center of the plank. Cousin Throckmorton wants to stand on the right-hand end
of the plank. If the plank is to remain at rest, how massive can Throckmorton
be?
CHAPTER 11. EQUILIBRIUM AND ELASTICITY
154
This is an equilibrium problem and the (rotational) equilibrium condition
!
⃗τ = 0
(11.8)
can be applied to any point. For sufficiently large mass of Throckmorton the
plank would start rotating around point S.
Solution #1: Therefore, the equilibrium condition (11.8) states
⃗τP + ⃗τT = 0
or
D
Mg −
2
which can be solved for the mass
#
L D
−
2
2
m=M
$
(11.9)
mg = 0
D
.
L−D
(11.10)
(11.11)
Solution #2: Alternatively the equilibrium condition can be written as
⃗τcm = ⃗rcm × ((M + m) ⃗g) = 0
(11.12)
τcm = rcm,x (M + m) g = 0
(11.13)
where or in scalar from
where the origin is chosen to be at point S. The only way the above equation
can be satisfied is when
&
%
&
%
M − D2 + m L2 − D2
0 = rcm,x =
(11.14)
M +m
or when the numerator vanishes, i.e.
#
$
#
$
D
L D
M −
+m
−
=0
2
2
2
(11.15)
which leads to the same solution as (11.11).
Example. A rock is attached to the left end of a uniform meter stick
that has the same mass as the rock. In order for the combination of rock in
meter stick to balance atop the triangular object (see figure below), how far
from the left end of the stick should the triangular object be placed?
CHAPTER 11. EQUILIBRIUM AND ELASTICITY
155
(i) less than 0.25 m
(ii) 0.25 m
(iii) between 0.25 m and 0.5 m
(iv) more than 0.5 m.
11.3
Rigid-Body Equilibrium Problems in 2D
For problem in two dimensions we can choose a coordinate system so that
the vectorial equilibrium conditions (11.1) and (11.2) are replaced by three
scalar conditions
!
Fi,x = 0
(11.16)
i
!
Fi,y = 0
(11.17)
i
!
τi,z = 0.
(11.18)
i
Example 11.2. An auto magazine reports that a certain sports car has 53%
of its weight on the front wheel. The distance between the axles is D = 2.46 m.
How far in front of the rear axle is the car’s center of gravity? (Let’s denote
is by Lcg )
CHAPTER 11. EQUILIBRIUM AND ELASTICITY
156
The (translational) equilibrium condition (11.17) implies
nr + nf − w = 0
(11.19)
where
nf = 0.53w
nr = 0.47w
(11.20)
Moreover the (rotational) equilibrium condition (11.18) implies
−Lcg nr + (D − Lcg ) nf = 0
Dnf
nf + nr
= (2.46 m) 0.53 = 1.30 m.
Lcg =
Lcg
(11.21)
Example 11.3. Sir Lancelot, who weights 800 N, is assaulting a castle by
climbing a uniform ladder that is 5.0 m long and weights 180 N. The bottom
of the ladder rests on a ledge and leans across the moat in equilibrium against
a frictionless, vertical castle wall. The ladder makes an angle of 53.1◦ with
the horizontal. Lancelot pauses one-third of the way up the ladder.
(a) Find normal and friction forces on the base of the ladder.
(b) Find the minimum coefficient of static friction needed to prevent slipping at the base.
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CHAPTER 11. EQUILIBRIUM AND ELASTICITY
(c) Find the magnitude and direction of the contact force on the base of
the ladder.
a) The (translational) equilibrium conditions (11.16) and (11.17) imply
fs − n1 = 0
n2 − (180 N) − (800 N) = 0
(11.22)
or
n2 = 980 N
fs = n1
and the (rotational) equilibrium condition (11.18) (around the point B) implies
#
$
#
$
5.0 m
5.0 m
◦
◦
(5.0 m) sin (75 ) n1 +
cos (53.1 ) (180 N)+
cos (53.1◦ ) (800 N) = 0.
2
3
(11.23)
or
n1 = 268 N
fs = 268 N.
(11.24)
b) Thus to prevent slipping the minimal coefficient of friction must be
fs = µs n2
fs
268 N
µs =
=
= 0.27
n2
980 N
c) The magnitude and directions of the contact force is
'
'
2
2
fs + n2 = (268 N)2 + (980 N)2 = 1020 N
FB =
#
$
980 N
θ = arctan
= 75◦ .
268 N
(11.25)
(11.26)
CHAPTER 11. EQUILIBRIUM AND ELASTICITY
11.4
158
Stress, Stain, and Elastic Moduli
So far we considered (mostly) rigid bodies, which is an idealized model of
extended objects. In reality all objects can deform when external forces are
applied to them. Examples of of deformations includes
• stretching,
• squeezing,
• twisting
and we want to develop a theoretical framework to discuss such deformations.
For each type of deformation there are two relevant physical quantities:
• stress which characterizes the cause of the deformation
• strain which characterizes the effect of the deformation
Very often the two quantities are directly proportional to each other and then
coefficient of proportionality is known as elastic modulus, defined as
Stress
= Elastic modulus
Strain
(11.27)
which is also known as Hooke’s law. In fact we have already seen an example
of this law when elastic force of an ideal spring was considered, i.e.
F
= k.
x
(11.28)
Tensile/compressive stress. For stretching/squeezing deformations
the tensile/compressive stresses are defined as
σ=
F⊥
A
(11.29)
and the tensile/compressive strains are defined as
ε=
l − l0
∆l
=
.
l
l
The units of the stress is
[Stress] =
[Force]
[Distance]2
(11.30)
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CHAPTER 11. EQUILIBRIUM AND ELASTICITY
and in SI system a unit of stress is measures in Pascals,
1 Pa ≡ 1 N/m2 .
but the strain is dimensionless.
In these two case the elastic modulus is known as Young’s modulus
Y =
F⊥ /A
F⊥ l0
σ
=
=
.
ε
∆l/l0
A∆l
(11.31)
This definition insures that Young’s modulus is positive if the perpendicular
component of force F⊥ is chosen to point outwards:
Tensile: F⊥ > 0 and ∆l > 0
Compressive: F⊥ < 0 and ∆l < 0
⇒
⇒
Y >0
Y >0
(11.32)
(Note that since the strain was dimensionless, the Young’s modulus has the
same units as stress.)
For many materials the tensile and compressive Young’s modulus are the
same and the Hooke’s law works very well. Here is a table of the modulus
for some of the materials
CHAPTER 11. EQUILIBRIUM AND ELASTICITY
160
However for composite materials (such as concrete or stone) the compressive stress can be quite large, while the tensile stress easily breaks the object
and thus the Hooke’s law does not hold.
Example 11.5. A steel 2.0 m long has a cross-sectional area of 0.30 cm2 .
It is hung by one end from a support, and a 550 kg milling machine is hung
from its other end. Determine the stress on the rod and the resulting strain
and elongation.
The stress is given by (11.29):
%
&
(550 kg) 9.8 m/s2
F⊥
=
= 1.8 × 108 Pa,
(11.33)
σ=
A
0.30 cm2
from the Hooke’s law (11.31):
ε=
σ
(1.8 × 108 Pa)
=
= 9.0 × 10−4
Y
(20 × 1010 Pa)
(11.34)
and from (11.30):
%
&
∆l = σ · l = 9.0 × 10−4 (2.0 m) = 1.8 mm.
(11.35)
Bulk stress. There situations when the same tensile/compressive stress
is applied to all spatial directions simultaneously. An example could be a
stress stress experienced by an object placed in water which can causes a
volume deformation of an object. Such an isotropic (i.e. the same in all
directions) stress is called (sometimes bulk stress) or pressure
p=
F⊥
A
(11.36)
and an isotropic strain is called bulk or volume strain
εbulk =
∆V
.
V0
(11.37)
where
∆V = Vf − Vi .
(11.38)
Note that in contrast to tensile/compressive stress the pressure does not have
a direction and thus is a scalar quantity. Units of pressure are either Pascals
defined by
1 Pa ≡ 1 N/m2 .
(11.39)
or Atmospheres
1 atm ≈ 105 Pa.
(11.40)
CHAPTER 11. EQUILIBRIUM AND ELASTICITY
161
or Pound-per-square-inch
1 psi ≈ 6900 Pa.
(11.41)
Confusing Point #1: The pressure is usually positive which is in conflict
with our choice for perpendicular direction to point outwards of an object.
This can be fixed by either stating that
p=−
F⊥
A
(11.42)
which is not how it was defined in the book. For most systems the pressure
is positive and the confusion can be avoided, but there are very important
systems (such as comic inflation or dark energy) where pressure is negative,
and thus the choice of the positive direction of F⊥ as well as the sign in the
definition of pressure is important.
Confusing Point #2: The bulk stress is defined not as pressure but as a
negative change in pressure as an object is moved from one substance (e.g.
air) to another (e.g. water), i.e.
σ bulk = ∆p.
(11.43)
∆p = pf − pi .
(11.44)
where
This point is somewhat confusing, but might be related to historical conventions.
Confusing Point #3: The corresponding elastic modulus known as bulk
modulus is defined with a minus sign to ensure its positivity
B=
σ bulk
−∆p · V0
=
.
εbulk
∆V
(11.45)
∆V
1
=−
.
B
∆p · V0
(11.46)
and the bulk compressibility
k≡
Of course two out of three confusion (i.e. #1 and #3) could have been
avoided if the pressure was defined with a minus sign as as in (11.42).
Example 11.6. A hydraulic press contains V0 = 0.25 m3 (250 L) of
oil. Find the decrease in the volume ∆V of the oil when it is subjected to a
pressure increase ∆p = 1.6 × 107 Pa (about 160 atm). The bulk modulus of
the oil is B = 5.0 × 109 Pa (about 5.0 × 104 atm) and its compressibility is
k = 1/B = 20 × 10−6 atm−1 .
CHAPTER 11. EQUILIBRIUM AND ELASTICITY
162
From the definition of bulk modulus we have
∆V = −
∆p · V0
1.6 × 107 Pa
=−
0.25 m3 = −8.0 × 10−4 m3
B
5.0 × 109 Pa
(11.47)
or alternatively from the definition of bulk compressibility
%
&%
&
∆V = −∆pV0 k = − (160 atm) 0.25 m3 20 × 10−6 atm−1 = −8.0×10−4 m3 .
(11.48)
Shear stress. In addition to tensile/compressive deformations due to
perpendicular component of the applied force, a parallel component of the
applied force can cause a deformation.
For such deformations we defined shear stress as
σ shear =
F||
A
(11.49)
εshear =
x
h
(11.50)
shear strain as
and shear modulus as
S=
F|| /A
F|| h
σshear
=
=
.
εshear
x/h
Ax
(11.51)
If we imagine a shear deformation of rectangular parallelepiped of dimensions
x, y, z then
Fx y
.
(11.52)
S=
xz∆x
CHAPTER 11. EQUILIBRIUM AND ELASTICITY
163
Example 11.7. Suppose a brass base plate of an outdoor sculpture experiences shear force in an earthquake. The vertically oriented square plate
is x = 0.5 cm thick. Whats is the force exerted on each of its edges is the
resulting displacement is ∆x = 0.16 mm?
From definition of shear modulus we have
F|| =
Sxz∆x
y
(11.53)
where
y=z
(11.54)
and thus
%
&
F|| = Sx∆x = 3.5 × 1010 Pa (0.0050 m) (0.00016 m) = 2.8×104 N. (11.55)
Example. A copper rod of cross-sectional area A = 0.500 cm2 and length
1.00 m is elongated by 2.00 × 10−2 mm, and a steel rod of the same crosssectional area but 0.100 m in length is elongated by 2.00 × 10−3 mm.
a) Which rod has greater tensile strain?
• the copper rod
• the steel rod
• the stain is the same
b) Which rod is under greater tensile stress?
• the copper rod
• the steel rod
• the stain is the same
The ratio of tensile strains is
∆lcopper lsteel
2.00 × 10−2 mm 0.100 m
εcopper
=
=
=1
εsteel
∆lsteel lcopper
2.00 × 10−3 mm 1.000 m
(11.56)
and of tensile stresses is
σcopper
Ycopper εcopper
11
=
= .
σsteel
Ysteel εsteel
20
(11.57)
CHAPTER 11. EQUILIBRIUM AND ELASTICITY
11.5
164
Elasticity and Plasticity
For larger deformations the linear dependence of stress as functions of stain
(11.27) does not hold. A typical dependence for metals under tension is
illustrated in
and for rubber under tension is illustrated by