Download CHAPTER 12 STATIC EQUILIBRIUM AND ELASTICITY • Conditions

By definition, an object is in equilibrium when it is either
at rest or moving with constant velocity, i.e., with no
CHAPTER 12
acceleration. The following are examples of objects in
static equilibrium ...
STATIC EQUILIBRIUM AND ELASTICITY
• Conditions for static equilibrium
! Center of gravity
! Examples of equilibrium
• Couples
Balanced bottle
Garden of the Gods
Colorado Springs
• Elasticity
! Stress and strain
! Young’s modulus
! Hooke’s Law and the elastic limit
! Shear modulus
Alexei
Therefore, two conditions are necessary for a body to be
in static equilibrium ...
• The net external force acting on the body is zero,
!
i.e., ∑ i Fi = 0,
Question 12.1: A see-saw consists of a board of length 4.0
so there is no translational motion
m pivoted at its center. A 28 kg child sits at one end of the
board. Where should a 40 kg child sit to balance the
• The net external torque about any point is zero,
!
i.e., ∑ i τi = 0,
so there is no rotational motion
Let’s do a problem as an illustration ...
seesaw, i.e., have the see-saw in static equilibrium?
First, identify all the forces ...
40kg
y
28kg
x
F
4m
d
FA
FB
The conditions for static and rotational equilibrium:
!
∑ i Fyi = 0 and ∑ i τi = 0.
Therefore, the pivot must supply an upward force so that
the net force on the board is zero, i.e.,
F − (28 kg)g + (40 kg)g = 0
∴F = (68 kg)g = 666.4 N.
Define ccw torques as positive and taking torques about
the pivot point we have:
(28 kg)g × (2 m) − (40 kg)g × d = 0
56 kg ⋅ m
∴d =
= 1.40 m.
40 kg
So, for equilibrium Cathy must stand 1.40 m from the
pivot point.
Question 12.2: At airports you see people running
around pulling their luggage on small trolleys.
Sometimes they arrange their cases as shown on the left;
sometimes as shown on the right. If FA and FB are the
forces necessary to hold the carts in the two cases, which
force is the smaller or are the equal?
FA
x3
FB
We assumed the weight force acts through the center of
mass ... strictly speaking it should be the center of
weight. What’s the connection between the two?
center of gravity
(center of weight)
mg
θ
Mg
x1
xCG
x2
O
Take torques about the contact point with the ground.
!
At equilibrium ∑ i τi = 0, i.e.,
F.x 3 − Mg.x1 − mg.x 2 = 0.
∴F =
(Mg.x1 + mg.x 2 )
x3
.
M, m, x1 and x 3 are the same on the left and on the
right, but x 2 (left) > x 2 (right) .
∴FA > FB,
so FB is the smaller force.
Is there any advantage in reducing
(or increasing) the angle θ?
"
W=
∑ i mi g i
O
xi "
w i = mi g i
Take some origin, O, and “break” the object into small
elements of mass mi. The total torque about O is:
τ! = ∑i mi gi x i = Wx CG ,
where x CG is the center of gravity (weight) defined as
∑ wx ∑ wx
x CG = i i i = i i i .
W
∑ i wi
(This is similar to the definition of the center of mass.)
The center of mass is at the same point as the center of
weight, i.e., x CM ≡ x CG , only if g is constant, then
τ! = (∑ i mi x i ) g = Mgx CM = WxCM .
O
!
W=
∑ i mi g i
(a)
If the origin is taken at the center of gravity, then the
object produces zero torque about that point.
The center of gravity is then ... the point about which the
gravitational force on the object produces zero torque
no matter what the orientation of the object. It could
also be called ... the center of weight!
If g is constant over the object then the center of gravity
and the center of mass occur at the same point.
Question 12.3: Why is it you cannot touch your toes
without falling over if you have your heels against a wall
as in (a) ... yet, ordinarily, you have no problem, as in
(b)?
If you don’t believe me ... try it!
1m
2m
CG
CG
(a)
4 kg
(b)
I
physics!
20 kg
You can only be in equilibrium if your center of gravity
(weight) is above the pivot point, i.e., your feet. In order
Question 12.4: A sign hangs in front of a store. The sign
to do that, you have to be able to shift your rear end
has a mass of 20 kg and it hangs at the end of a
“backwards”, as in (b). If the wall is in the way then you
horizontal rod of length 2 m and mass 4 kg, which is
can’t, as in (a), and your center of gravity produces a
hinged at the wall. The rod is supported by a wire
torque about your feet (the pivot point), which tends to
attached to a point on the wall 1 m above the rod. (a)
rotate you forward.
What is the tension in the wire? (b) What is the
magnitude and direction of the force the rod exerts on
the wall?
1m
!
F
!
T
Ty
Tx
2m
y
!
F
1m
Ty
!
T
2m
Tx
y
θ
x
Fy
4 kg
Fx
Force of the wall
on the rod
I
physics!
x
Fy
4 kg
Fx
20 kg
20 kg
The rod must supply a force on the wall ... how do we
know that? However, we have no idea in what direction
so choose it arbitrarily to begin with. Note, by Newton’s
3rd Law the force the rod exerts on the wall is equal and
opposite to the force the wall exerts on the rod.
(a) For static and rotational equilibrium:
!
∑ i Fxi = 0, ∑ i Fyi = 0 and ∑ i τi = 0.
Take torques about the hinge ... a great idea since we
!
don’t know anything about F. Then
Ty × (2 m) − (4 kg)g × (1 m) − (20 kg)g × (2 m) = 0
∴Ty = 215.6 N.
I
physics!
But
Ty
Tx
= tan θ =
(1 m) 1
= .
(2 m) 2
∴ Tx = 2 Ty , i.e., Tx = −431.2 N.
T = Tx2 + Ty2 = (−431.2 N)2 + (215.6 N)2
= 482.1 N
!
Note ... we found T even though we
!
know nothing about F! Of course, the
wire must be strong enough to
support a force of 482.1 N.
!
F
1m
!
T
Ty
Tx
y
2m
!
F
1m
!
T
Ty
Tx
2m
y
x
Fy
4 kg
Fx
I
physics!
20 kg
(b) For static and rotational equilibrium:
x
Fy
4 kg
Fx
I
physics!
20 kg
y
!
∑ i Fxi = 0, ∑ i Fyi = 0 and ∑ i τi = 0.
So ∑ i Fxi = Fx + Tx = Fx − 431.2 N = 0
∴Fx = 431.2 N,
and ∑ i Fyi = Fy + Ty − (20 kg)g − (4 kg)g = 0
∴Fy = 19.6
!
F = (431.2ˆi + 19.6ˆj) N
Fy
φ
Fx
N.
⎛ Fy ⎞
∴ φ = tan −1⎜ ⎟ = 2.6"
⎝ Fx ⎠
The force of the rod on the wall is therefore
!
F = −(431.2ˆi + 19.6ˆj) N.
x
!
Note: we could get F another way ... (a) by taking
torques about the right hand end of the rod, since the
!
system is in static equilibrium, ∑ i τi = 0 about any point,
i.e., (4 kg)g × (1 m) − Fy × (2 m) = 0.
∴Fy = 19.6 N.
(b) By taking torques about the top end of the wire,
i.e., Fx × (1 m) − (4 kg)g × (1 m) − (20 kg)g × (2 m) = 0.
∴Fx = 431.6 N.
P
ℓ
Question 12.5: A square plate is made by welding
together four smaller square plates, each of side ℓ. Each
of the smaller squares is made from a different material
so they have different weights, as shown in the figure.
(a) Find the position of the center of gravity, relative to
the point (0, 0). (b) If the plate is suspended from the
60 N
20 N
gravity (weight), with respect
ℓ
(0,0)
50 N
30 N
ℓ
ℓ
to (0, 0) is given by
∑ wx
x CG = i i i
W
⎛
ℓ
ℓ
3ℓ
3ℓ⎞
⎜ (60 N) × + (50 N) × + (20 N) × + (30 N) × ⎟
⎝
2
2
2
2⎠
=
(60 N) + (50 N) + (20 N) + (30 N)
point P, what is the angle between the vertical and the
= 0.8125ℓ,
left-hand side of the plate?
and y CG =
P
(a) By definition, the center of
∑ i wi y i
W
ℓ
60 N
20 N
ℓ
50 N
30 N
⎛
3ℓ
ℓ
3ℓ
ℓ⎞
⎜ (60 N) × + (50 N) × + (20 N) × + (30 N) × ⎟
⎝
2
2
2
2⎠
=
(60 N) + (50 N) + (20 N) + (30 N)
ℓ
ℓ
= 1.000ℓ.
(0, 0)
Therefore, the coordinates of the CG relative to (0, 0) are
(0.8125ℓ, ℓ).
(b) When hung from P, a vertical line through P passes
through the CG.
ℓ φ
frictionless, vertical wall. If the coefficient of static
∴tan φ =
0.8125ℓ
(0, 0)
Question 12.6: A uniform ladder rests against a
P
CG
0.8125ℓ
,
ℓ
i.e., φ = 39.1" .
friction between the bottom of the ladder and the floor is
0.3, what is the smallest angle at which the ladder will
remain stationary?
Let the ladder, length ℓ, rest at an angle θ against the
wall. Identify all of the force acting on the ladder.
"
F1
Since the wall is
frictionless, the force of
ℓ
"
Fn
"
w
θ
"
fs
ℓ sin θ
the wall acting on the
"
ladder is F1, which is ⊥ to
!
F1
the wall, i.e., a normal
force. The force exerts
two forces on the ladder; a
ℓ cosθ ℓ cosθ
"
2
2
normal force Fn and a
"
"
static frictional force f s. The weight of the ladder is w,
which acts through the CG, i.e., the middle of the ladder.
For static and rotational equilibrium:
#
∑ i Fxi = 0, ∑ i Fyi = 0 and ∑ i τi = 0.
and
Taking torques about the foot of the ladder:
!
⎛ℓ
⎞
∑ i τi = w⎜ cosθ⎟ − F1(ℓsin θ) = 0 ... (3)
⎝2
⎠
∴ ∑ i Fxi = f s − F1 = 0 ... ... ... (1)
∑ i Fyi = Fn − w = 0 ... ... ... (2)
!
w
ℓ sin θ
θ
!
fs
ℓ cosθ
F1 = f s = µ sFn = µ s w
∴θ = tan −1 12µ
s
(
= 59.0# .
2
w
.
2F1
But from (1) and (2):
ℓ
!
Fn
∴tan θ =
ℓ cosθ
2
)
Again, identify all the forces acting on the ladder.
Summing the vertical and horizontal forces we get
!
F1
!
Fn
!
fs
∑ Fy = Fn − w + f s ... (1)
!
w
and
∑ Fx = −F1 ... ... ... (2).
Question 12.7: In the previous problem there was no
Now eq (1) can be zero (if f s = w − Fn ). However, eq
friction at the wall but there was static friction between
(2) can never be zero if the ladder is leaning against the
the ground and the foot of the ladder. If, instead, there
wall as it consists of only one term ( F1). Consequently,
was friction at the wall but no friction at the ground,
the ladder can never be in equilibrium, i.e., if there’s no
what difference would it make?
friction with the ground it will always slip!
We ignore the arrangement when the ladder is vertical!
h
L
For ‘balance’ the CG of the two
2
boxes must lie directly over the
edge of the table, i.e., the ‘pivot’
xCG
10 kg
Question 12.8: A box X, of mass 10 kg, is placed on a
box Y, of mass 5 kg, so that the center of box X lies
directly over the left hand edge of box Y. If the two
boxes are then placed over the edge of a table, what is the
maximum possible overhang, h, without the boxes falling,
if the sides of the boxes is 0.50 m?
point.
5 kg
L
Let the length of the side of the
2
boxes be L, and find the CG with
respect to the center line of box X:
∑ i mi gx i (10 kg)g × 0 + (5 kg)g × L 2 L
x CG =
=
= .
(15 kg)g
6
∑i mi g
Therefore, total overhang is,
L L 2L
h= + =
= 0.33 m.
6 2
3
OR ... take torques about O, above the edge of the table.
h
1.0 m
O
2.0 m
10 kg
5 kg
Question 12.9: The figure shows a cart of mass 100 kg
L
2
At equilibrium, ∑ i τi = 0.
L
∴(10 kg)g × ⎛ h − ⎞ − (5 kg)g × (L − h ) = 0,
⎝
2⎠
i.e., 10h − 5L − 5L + 5h = 0.
10L 2(0.5 m)
∴h =
=
= 0.33m.
15
3
loaded with six identical packing cases, each of mass
50 kg. How is the total weight of the cart and packing
cases distributed between the left hand and the right hand
wheels?
1.0 m
A couple
"
x1
2.0 m
50 kg 100 kg 150 kg
FL
FR
At equilibrium: ∑ Fy = 0, ∑ Fy = 0 and ∑ τ = 0.
∴ (FL + FR ) = −(50 kg)g + (100 kg)g
+ (150 kg)g + (100 kg)g = 0,
i.e., ∴ (FL + FR ) = 3924 N.
Taking torques about the axle of the right hand wheel
(ccw is positive):
(50 kg)g × (2 m) + (100 kg)g × (1 m)
+ (100 kg)g × (1 m) − FL × (2 m) = 0.
∴FL = (150 kg)g × (1 m) = 1471.5 N
so
FR = 3924 N − 1471.5 N = 2452.5 N.
O•
"
F1
ℓ
"
x2
"
F2
Two forces form a couple if they are equal but opposite
and their line of action is separated (by a distance ℓ):
"
"
i.e., F1 = F2 = F.
Then the torque about any point O is:
"
"
" "
"
τ = ( x1 × F1 ) + ( x 2 × F2 )
"
i.e., τ = x1F − x 2F = (x1 − x 2 )F = ℓF
So the torque produced by a couple is the same about
all points in space, i.e., no matter where O is located ...
it depends only on their perpendicular separation.
Stress and strain (Young’s modulus)
Area A
L
F
F
However, the strain is reversible over only a limited
range of stress ...
tensile
stress
ΔL + L
Elastic limit
Fracture
Plastic behavior
Objects deform when subjected to a force. In the case of
the length of a metal bar or wire ...
The stress ⇒ F A , and the strain ⇒ ΔL L .
F
stress
A
Young’s modulus ⇒ Y =
=
ΔL
strain
L
( )
( )
Units: Force/area ⇒ N ⋅ m−2 (scalar)
• Thomas Young (1773-1829)
The example above is tensile stress. There is also
compressive stress. Young’s modulus is often the same in
both cases, but there are exceptions (e.g., bone and
concrete).
Hooke’s Law
(linear)
Plastic
deformation
strain
... up to the elastic limit. Beyond that the material
shows plastic behavior until it reaches the point of
fracture (tensile strength).
The initial linear behavior is known as Hooke’s Law ...
after Robert Hooke (1635-1703).
DISCUSSION PROBLEM [12.1]:
Ysteel = 200 × 109 N ⋅ m−2 and YAl = 70 × 109 N ⋅ m−2 .
An aluminum wire and a steel wire
L
with the same lengths (L) and
diameters (d), are joined to form a
wire of length 2L. One end of the
wire is fixed to the ceiling and the
other is attached to a weight (w).
L
Neglecting the weight of the wires,
which one of the following
w
statements is true?
A: The strains in the two wires are the same.
B: The stresses in the two wires are the same.
C: The stress in the aluminum wire is greater than the
stress in the steel wire.
D: The stress in the aluminum wire is less than the
stress in the steel wire.
Question 12.10: A steel wire of length 1.50 m and
diameter 1.00 mm is joined to an aluminum wire of
identical dimensions to make a composite wire of length
3.00 m. (a) What is the length of the composite wire if it
is used to support a mass of 5 kg? (b) What is the
maximum load the composite wire could withstand?
Assume the weights of the wires are negligible.
Aluminum:
YAl = 70 × 109 N ⋅ m−2 .
Tensile strength = 90 × 106 N ⋅ m−2 .
Steel:
Ysteel = 200 × 109 N ⋅ m−2 .
Tensile strength = 520 × 106 N ⋅ m−2 .
( )
( )
F
stress
A .
(a) The Young’s modulus is Y =
=
ΔL
strain
L
d = 1.00mm
1.50m
aluminum
1.50m
steel
The stress in each wire is:
(5 kg)g
F =
A π × (0.5 × 10 −3 m)2
= 6.25 × 107 N ⋅ m−2 .
( )( )
But ΔL = L Y F A .
∴ΔL = (6.25 × 107 N ⋅ m−2 ) L Y .
5kg
• Aluminum:
ΔL1 = (6.25 × 107 N ⋅ m−2 )
(1.50 m)
(70 × 109 N ⋅ m−2 )
= 1.34 × 103 m ⇒ 1.34 mm.
• Steel:
ΔL2 = (6.25 × 107 N ⋅ m−2 )
(1.50 m)
(200 × 109 N ⋅ m−2 )
= 0.47 × 103 m ⇒ 0.47 mm.
Therefore, the increase in length is 0.00181 m, so the new
length is 3.00181 m.
(b) The tensile strength of aluminum is less than that of
steel. Consequently, as the load increases, it will be the
aluminum wire that fails.
Mg
6
−2
A = 90 × 10 N ⋅ m .
∴M = (90 ×106 N ⋅ m−2 ) A g
The maximum stress is F A =
= (90 × 106 N ⋅ m−2 )
π(0.5 ×10 −3 m)2
9.81 m/s2
= 7.21 kg.
Shear modulus
Area
F
L
θ
F
x
Δx
F
θ
The shear strain is
The stress is
F
A
Δx
= tanθ
L
(note which area!)
Shear modulus ⇒ Ms =
=
shear stress
shear strain
(F A ) = (F A ) .
(Δx L) tanθ
Units: Force/area ⇒ N ⋅ m−2 (scalar).
=A