Transcript Monday, Apr. 1, 2002

1443-501 Spring 2002
Lecture #17
Dr. Jaehoon Yu
1. Conditions for Equilibrium
2. Center of Gravity
3. Elastic Properties of Solids
• Young’s Modulus
• Shear Modulus
• Bulk Modulus
Today’s Homework Assignment is the Homework #8!!!
2nd term exam on Wednesday, Apr. 10. Will cover chapters 10 -13.
Conditions for Equilibrium
What do you think does the term “An object is at its equilibrium” mean?
The object is either at rest (Static Equilibrium) or its center of mass
is moving with a constant velocity (Dynamic Equilibrium).
When do you think an object is at its equilibrium?
Translational Equilibrium: Equilibrium in linear motion
Is this it?
The above condition is sufficient for a point-like particle to be at its static
equilibrium. However for object with size this is not sufficient. One more
condition is needed. What is it?
Let’s consider two forces equal magnitude but opposite direction acting
on a rigid object as shown in the figure. What do you think will happen?
F
d
d
F 0
CM
-F
Apr. 1, 2002

The object will rotate about the CM. The net torque
 0
acting on the object about any axis must be 0.
For an object to be at its static equilibrium, the object should not
have linear or angular speed. v
0  0
1443-501 Spring 2002
Dr. J. Yu, Lecture #17
CM
2
More on Conditions for Equilibrium
To simplify the problems, we will only deal with forces acting on x-y plane, giving torque
only along z-axis. What do you think the conditions for equilibrium be in this case?
The six possible equations from the two vector equations turns to three equations.
F 0 F
F
x
0
y
0
  0 
0
z
What happens if there are many forces exerting on the object?
If an object is at its translational static equilibrium, and if the
net torque acting on the object is 0 about one axis, the net
torque must be 0 about any arbitrary axis.
r’
r5 O O’
Net Force exerting on the object  F  F 1  F 2  F 3      0

Net torque about O
O
 r 1  F 1  r 2  F 2  r 3  F 3       r i  F i 0
Position of force Fi about O’
Net torque about O’
Apr. 1, 2002

O'
 
'
i
r  r i  r'


 r '1F 1  r '2  F 2      r 1  r '  F 1  r 2  r '  F 2       r i  F i  r ' F i
1443-501
  rSpring
 F 2002
 r '0 
Dr.
J. Yu, Lecture #17
O'
i
i
O
0
3
Center of Gravity Revisited
When is the center of gravity of a rigid body the same as the center of mass?
Under the uniform gravitational field throughout the body of the object.
Let’s consider an arbitrary shaped object
The center of mass of this object is
CM
CoG
m x  m x
M
m
my
my




m
M

xCM 
i i
i i
i
yCM
i
i
i
i
i
Let’s now examine the case with gravitational acceleration on
each point is gi
Since the CoG is the point as if all the gravitational force is exerted
on, the torque due to this force becomes
m1 g1  m2 g 2    xCoG  m1 g1 x1  m2 g 2 x2    
If g is uniform throughout the body
Apr. 1, 2002
Generalized expression for
different g throughout the body
m1  m2    gxCoG  m1 x1  m2 x2    g
m x
1443-501 Spring 2002
m
Dr. J. Yu, Lecture #17
xCoG 
i i
i
 xCM
4
Example 12.1
A uniform 40.0 N board supports a father and daughter weighing 800 N and 350 N,
respectively. If the support (or fulcrum) is under the center of gravity of the board and the
father is 1.00 m from CoG, what is the magnitude of normal force n exerted on the board
by the support?
1m
F
MFg
x
n
MBg
D
MFg
Since there is no linear motion, this system
is in its translational equilibrium
F
F
Therefore the magnitude of the normal force
x
0
y
 MBg  MF g  MDg  n  0
n  40.0  800  350  1190N
Determine where the child should sit to balance the system.
The net torque about the fulcrum   M g  0  M g 1.00  M g  x  0
B
F
D
by the three forces are
Therefore to balance the system x  M F g 1.00m  800 1.00m  2.29m
the daughter must sit
MDg
350
Apr. 1, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #17
5
Example 12.1 Continued
Determine the position of the child to balance the
system for different position of axis of rotation.
Rotational axis
1m
F
MFg
x
n
x/2
MBg
D
MFg
The net torque about the axis of
rotation by all the forces are
  M B g  x / 2  M F g  1.00  x / 2  n  x / 2  M D g  x / 2  0
Since the normal force is
n  MBg  MF g  MDg
The net torque can   M B g  x / 2  M F g  1.00  x / 2 
be rewritten
 M B g  M F g  M D g   x / 2  M D g  x / 2
 M F g 1.00  M D g  x  0
Therefore
Apr. 1, 2002
x
MFg
800
1.00m 
1.00m  2.29m
MDg
350
1443-501 Spring 2002
Dr. J. Yu, Lecture #17
What do we learn?
No matter where the
rotation axis is, net effect of
the torque is identical.
6
Example 12.2
A person holds a 50.0N sphere in his hand. The forearm is horizontal. The biceps
muscle is attached 3.00 cm from the joint, and the sphere is 35.0cm from the joint. Find
the upward force exerted by the biceps on the forearm and the downward force exerted
by the upper arm on the forearm and acting at the joint. Neglect the weight of forearm.
FB
Since the system is in equilibrium, from
the translational equilibrium condition
F  0
O
 F  F  F  mg  0
l
mg
F
From the rotational equilibrium condition   F  0  F  d  mg  l  0
d
x
y
B
U
U
U
Thus, the force exerted by
the biceps muscle is
B
FB  d  mg  l
mg  l 50.0  35.0
FB 

 583N
d
3.00
Force exerted by the upper arm is FU  FB  mg  583  50.0  533N
Apr. 1, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #17
7
Example 12.3
A uniform horizontal beam with a length of 8.00m and a weight of 200N is attached to a wall by
a pin connection. Its far end is supported by a cable that makes an angle of 53.0o with the
horizontal. If 600N person stands 2.00m from the wall, find the tension in the cable, as well as
the magnitude and direction of the force exerted by the wall on the beam.
R

53.0o
53.0o
FBD
Rsin
Rcos
8m
From the rotational equilibrium
Using the
translational
equilibrium
Apr. 1, 2002
First the translational equilibrium,
using components
200N
600N
2m
T
F
F
Tsin53
Tcos53
 R cos  T cos 53.0  0
y
R sin   T sin 53.0  600 N  200 N  0
 T sin 53.0  8.00  600 N  2.00  200 N  4.00m  0

T  313N
R cos  T cos 53.0
And the magnitude of R is
R sin   T sin 53.0  600 N  200 N
 800  313  sin 53.0 


  tan 

71
.
7


 313 cos 53.0

1
x
T cos 53.0 313  cos 53.0
R

 582 N
cos
cos 71.1
1443-501 Spring 2002
Dr. J. Yu, Lecture #17
8
Example 12.4
A uniform ladder of length l and weight mg=50 N rests against a smooth, vertical wall. If
the coefficient of static friction between the ladder and the ground is ms=0.40, find the
minimum angle min at which the ladder does not slip.
First the translational equilibrium,
using components
P
l
FBD

mg
n
O
F
F
f
Thus, the normal force is
x
f  P  0
y
 mg  n  0
n  mg  50 N
The maximum static friction force
just before slipping is, therefore,
f smax  m s n  0.4  50 N  20 N  P
From the rotational equilibrium

 min
Apr. 1, 2002
l
  mg cos min  Pl sin  min  0
2
 mg 
1  50 N 

 tan 1 
  tan 
  51
 2P 
 40 N 
O
1443-501 Spring 2002
Dr. J. Yu, Lecture #17
9
Elastic Properties of Solids
We have been assuming that the objects do not change their
shapes when external forces are exerting on it. It this realistic?
No. In reality, the objects get deformed as external forces act on it,
though the internal forces resist the deformation as it takes place.
Deformation of solids can be understood in terms of Stress and Strain
Stress: A quantity proportional to the force causing deformation.
Strain: Measure of degree of deformation
It is empirically known that for small stresses, strain is proportional to stress
The constants of proportionality are called Elastic Modulus Elastic Modulus 
Three types of
Elastic Modulus
Apr. 1, 2002
1.
2.
3.
stress
strain
Young’s modulus: Measure of the resistance in length
Shear modulus: Measure of the resistance in plane
Bulk modulus: Measure of the resistance in volume
1443-501 Spring 2002
Dr. J. Yu, Lecture #17
10
Young’s Modulus
Let’s consider a long bar with cross sectional area A and initial length Li.
Li
Fex
After the stretch
F
Tensile Stress  ex
A
Young’s Modulus is defined as
Fex
Fex=Fin
A:cross sectional area
Tensile stress
Lf=Li+DL
Tensile strain
Tensile Strain 
Fex
Tensile Stress
A
Y

DL
Tensile Strain
Li
DL
Li
Used to characterize a rod
or wire stressed under
tension or compression
What is the unit of Young’s Modulus?
Force per unit area
1. For fixed external force, the change in length is
Experimental
proportional to the original length
Observations
2. The necessary force to produce a given strain is
proportional to the cross sectional area
Elastic limit: Maximum stress that can be applied to the substance
before it becomes permanently
1443-501deformed
Spring 2002
Apr. 1, 2002
Dr. J. Yu, Lecture #17
11
Shear Modulus
Another type of deformation occurs when an object is under a force tangential
to one of its surfaces while the opposite face is held fixed by another force.
A
Dx
h
After the stress
Fixed face
Shear stress
Shear strain
F=fs
Shear Stress 
Tangential Force
F

Surface Area the force applies A
Shear Strain 
Dx
h
F
Shear Stress
A
S


Shear Modulus is defined as
Dx
Shear Strain
h
Apr. 1, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #17
12
Bulk Modulus
F
Bulk Modulus characterizes the response of a substance to uniform
squeezing or reduction of pressure.
V
After the pressure change
F
V’
F
F
Normal Force
F
Volume stress
Pressure 

Surface Area the force applies A
=pressure
If the pressure on an object changes by DP=DF/A, the object will
undergo a volume change DV.
Bulk Modulus is
defined as
Because the change of volume is
reverse to change of pressure.
Apr. 1, 2002
DF
Volume Stress
DP
A
B


D
V
DV
Volume Strain
Vi
Vi
Compressibility is the reciprocal of Bulk Modulus
1443-501 Spring 2002
Dr. J. Yu, Lecture #17
13
Example 12.7
A solid brass sphere is initially under normal atmospheric pressure of 1.0x105N/m2. The
sphere is lowered into the ocean to a depth at which the pressures is 2.0x107N/m2. The
volume of the sphere in air is 0.5m3. By how much its volume change once the sphere is
submerged?
Since bulk modulus is
DP
B
DV
Vi
The amount of volume change is
DV  
DPVi
B
From table 12.1, bulk modulus of brass is 6.1x1010 N/m2
The pressure change DP is
DP  Pf  Pi  2.0 107 1.0 105  2.0 107
Therefore the resulting
2.0 107  0.5
4
3
D
V

V

V




1
.
6

10
m
f
i
volume change DV is
6.11010
The volume has decreased.
Apr. 1, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #17
14