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Chapter 12
Static Equilibrium and Elasticity
2
Static Equilibrium
Equilibrium implies that the object moves with both constant
velocity and constant angular velocity relative to an observer in an
inertial reference frame.
Will deal now with the special case in which both of these
velocities are equal to zero
This is called static equilibrium.
Static equilibrium is a common situation in engineering.
The principles involved are of particular interest to civil engineers,
architects, and mechanical engineers.Introduction
3
Elasticity
We can discuss how objects deform under load conditions.
An elastic object returns to its original shape when the deforming
forces are removed.
Various elastic constants will be defined, each corresponding to a
different type of deformation.
Introduction
4
Rigid Object in Equilibrium
In the particle in equilibrium model a particle moves with
constant velocity because the net force acting on it is zero.
With real (extended) objects the situation is more complex .
The objects often cannot be modeled as particles.
For an extended object to be in equilibrium, a second condition of
equilibrium must be satisfied.
This second condition involves the rotational motion of the extended
object.
Section 12.1
5
Torque Reminder
Use the right hand rule to determine
the direction of the torque.
The tendency of the force to cause a
rotation about O depends on F and
the moment arm d.
The net torque on a rigid object
causes it to undergo an angular
acceleration.
r F
Section 12.1
6
Conditions for Equilibrium
The net external force on the object must equal zero.
If the object is modeled as a particle, then this is the only condition
that must be satisfied .
The net external torque on the object about any axis must be zero.
This is needed if the object cannot be modeled as a particle.
These conditions describe the rigid object in equilibrium analysis
model.
ext 0F
ext 0 τ
Section 12.1
(12.1)
(12.2)
7
Equilibrium NotesTranslational Equilibrium
The first condition of equilibrium is a statement of translational equilibrium.
It states that the translational acceleration of the object’s center of mass must be zero.
This applies when viewed from an inertial reference frame.
Rotational Equilibrium
The second condition of equilibrium is a statement of rotational equilibrium.
It states the angular acceleration of the object to be zero.
This must be true for any axis of rotation.
Section 12.1
8
Static vs. Dynamic Equilibrium
In this chapter, we will concentrate on static equilibrium.
The object will not be moving.
vCM = 0 and = 0
Zero net torque does not mean an absence of rotational motion.
Dynamic equilibrium is also possible.
The object would be rotating with a constant angular velocity.
The object would be moving with a constant vCM.
Section 12.1
9
Equilibrium EquationsWe will restrict the applications to situations in which all the forces lie
in the xy plane.
These are called coplanar forces since they lie in the same plane.
This restriction results in three scalar equations.
There are three resulting equations:
Fx = 0
Fy = 0
z = 0
The location of the axis for the torque equation is arbitrary.
Section 12.1
(12.3)
10
Quick Quiz 12.1
Consider the object subject to the two forces of equal magnitude in Figure 12.2. Choose the correct statement with regard to this situation. (a) The object is in force equilibrium but not torque equilibrium. (b) The object is in torque equilibrium but not force equilibrium. (c) The object is in both force equilibrium and torque equilibrium. (d) The object is in neither force equilibrium nor torque equilibrium
Answer: (a)
11
Quick Quiz 12.2
Consider the object subject to the three forces in Figure 12.3 (page 350). Choose the correct statement with regard to this situation. (a) The object is in force equilibrium but not torque equilibrium. (b) The object is in torque equilibrium but not force equilibrium. (c) The object is in both force equilibrium and torque equilibrium. (d) The object is in neither force equilibrium nor torque equilibrium
Answer: (b)
12
Center of Mass
An object can be divided into many small particles.
Each particle will have a specific mass and specific coordinates.
The x coordinate of the center of mass will be
Similar expressions can be found for the y and z coordinates.
CM
i ii
ii
m xx
m
Section 12.2
13
Center of Gravity
All the various gravitational forces acting on all the various mass elements are equivalent to a single gravitational force acting through a single point called the center of gravity (CG).
Each particle contributes a torque about an axis through the origin equal in magnitude to the particle’s weight multiplied by its moment arm.
1 1 2 2 3 3CG
1 2 3
m x m x m xx
m m m
Section 12.2
(12.4)
14
Center of Gravity, cont
The torque due to the gravitational force on an object of mass M is
the force Mg acting at the center of gravity of the object.
If g is uniform over the object, then the center of gravity of the
object coincides with its center of mass.
If the object is homogeneous and symmetrical, the center of
gravity coincides with its geometric center.
Section 12.2
15
Quick Quiz 12.3
A meterstick is hung from a string tied at the 25-cm mark. A 0.50-kg object is hung from the zero end of the meterstick, and the meterstick is balanced horizontally. What is the mass of the meterstick? (a) 0.25 kg (b) 0.50 kg (c) 0.75 kg (d) 1.0 kg (e) 2.0 kg (f) impossible to determine
Answer: (b)
16
17
Problem-Solving Strategy – Equilibrium Problems
Conceptualize Identify all the forces acting on the object.
Image the effect of each force on the rotation of the object if it were the only force acting on the object.
Categorize Confirm the object is a rigid object in equilibrium.
The object must have zero translational acceleration and zero angular acceleration.
Analyze Draw a diagram.
Show and label all external forces acting on the object.
Section 12.3
18
Problem-Solving Strategy – Equilibrium Problems, 2Analyze, cont Particle under a net force model
The object on which the forces act can be represented in a free body diagram as a dot because it does not matter where on the object the forces are applied.
Rigid object in equilibrium model
Cannot use a dot to represent the object because the location where the forces act is important in the calculations.
Establish a convenient coordinate system.
Find the components of the forces along the two axes.
Apply the first condition for equilibrium (F = 0).
Be careful of signs.
Section 12.3
19
Problem-Solving Strategy – Equilibrium Problems, 3Analyze, final
Choose a convenient axis for calculating the net torque on the rigid object.
Remember the choice of the axis is arbitrary.
Choose an axis that simplifies the calculations as much as possible.
A force that acts along a line passing through the origin produces a zero
torque.
Apply the second condition for equilibrium.
The two conditions of equilibrium will give a system of equations.
Solve the equations simultaneously.
Section 12.3
20
Problem-Solving Strategy – Equilibrium Problems, 4
Finalize
Make sure your results are consistent with your diagram.
If the solution gives a negative for a force, it is in the opposite
direction to what you drew in the diagram.
Check your results to confirm Fx = 0, Fy = 0, z = 0
Section 12.3
21
Horizontal Beam Example
Conceptualize The beam is uniform.
So the center of gravity is at the
geometric center of the beam.
The person is standing on the beam.
What are the tension in the cable and the
force exerted by the wall on the beam?
Categorize The system is at rest, categorize as a rigid
object in equilibrium.Section 12.3
22
Horizontal Beam Example, 2
Analyze
Draw a force diagram.
Use the pivot in the problem (at
the wall) as the pivot.
This will generally be
easiest.
Note there are three unknowns
(T, R, ).
Section 12.3
23
Horizontal Beam Example, 3
Analyze, cont.
The forces can be resolved into components.
Apply the two conditions of equilibrium to obtain three equations.
Solve for the unknowns.
Finalize
The positive value for indicates the direction of R was correct in
the diagram.
Section 12.3
24
Ladder Example
Conceptualize
The ladder is uniform.
So the weight of the ladder acts through its geometric center (its center of gravity).
There is static friction between the ladder and the ground.
Categorize
Model the object as a rigid object in equilibrium. Since we do not want the ladder to
slip
Section 12.3
25
Ladder Example, 2
Analyze
Draw a diagram showing all the
forces acting on the ladder.
The frictional force is ƒs = s n.
Let O be the axis of rotation.
Apply the equations for the two
conditions of equilibrium.
Solve the equations.
Section 12.3
26
Example 12.1
A seesaw consisting of a uniform board of mass M and length ℓ, supports at rest a father and daughter with masses mf and md, respectively, as shown in Figure 12.7. The support (called the fulcrum) is under the center of gravity of the board, the father is a distance d from the center, and the daughter is a distance, ℓ/2 from the center.
27
Example 12.1
(A) Determine the magnitude of the upward force exerted by the support on the board.
Solution:
(B) Determine where the father should sit to balance the system at rest.
Solution:( )( ) ( ) 0
2
2
f d
d
f
m g d m g
md
m
n
gMmmMggmgmn
Mggmgmn
dfdf
df
)(
0
28
Example 12.2
A uniform horizontal beam with a length of ℓ = 8.00 m and a weight of Wb = 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of = 53.0 with the beam (Fig. 12.8a). A person of weight Wp = 600 N stands a distance d = 2.00 m 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.
29
Example 12.2
30
Example 12.2
Solution:
(1) cos cos 0
(2) sin sin 0
x
y p b
F R T
F R T W W
( sin )( ) 02
2 (600 N)(2.00 m) (200 N)(4.00 m)313N
sin (8.00 m)sin 53.0
sinsintan
cos cos
z p b
p b
p b
T W d W
W d WT
W W TR
R T
31
Example 12.2
1
1
sin tan
cos
600 N 200 N (313 N)sin53.0 tan 71.1
(313 N)cos53.0
cos (313 N)cos53.0581 N
cos cos71.1
p bW W T
T
TR
32
Example 12.3
A uniform ladder of length ℓ, rests against a smooth, vertical wall (Fig.12.9a). The mass of the ladder is m, and the coefficient of static friction between the ladder and the ground is s = 0.40. Find the minimum angle min at which the ladder does not slip.
33
Example 12.3
Solution:
1
1 1 1min
max
sin cos 02
sintan tan
cos 2 2
1 1tan tan tan 51
2 2 2(0.40)
O
s
P mg
mg mg
P P
mg
P
max , max
(1) 0
(2) 0
(3)
(4)
(5)
x s
y
s
s s s
F f p
F n mg
P f
n mg
p f n mg
34
Example 12.4
(A) Estimate the magnitude of the force a person must apply to a wheelchair’s main wheel to roll up over a sidewalk curb (Fig. 12.10a). This main wheel that comes in contact with the curb has a radius r, and the height of the curb is h.
F
35
Example 12.4
36
Example 12.4
Solution:2 2 2
2
2
2
(1) ( ) 2
(2) (2 ) 0
2 (2 ) 0
2(3)
2
2
2 2
0.1 m(700 N)
2(0.3 m) 0.1 m
3 10 N
A
d r r h rh h
mgd F r h
mg rh h F r h
mg rh hF
r h
h r h hF mg mg
r h r h
F
37
Example 12.4
(B) Determine the magnitude and direction of .
Solution:
R
1 1
2
(3) cos 0
(4) sin 0
sintan
cos700 N
tan tan 70300 N
700 N8 10 N
sin sin70
x
y
F F R
F R mg
R mg
R Fmg
F
mgR
38
Elasticity
So far we have assumed that objects remain rigid when external
forces act on them.
Except springs
Actually, all objects are deformable to some extent.
It is possible to change the size and/or shape of the object by
applying external forces.
Internal forces resist the deformation.
Section 12.4
39
Definitions Associated With Deformation
Stress
Is proportional to the force causing the deformation
It is the external force acting on the object per unit cross-sectional
area.
Strain
Is the result of a stress
Is a measure of the degree of deformation
Section 12.4
40
Elastic ModulusThe elastic modulus is the constant of proportionality between the
stress and the strain. For sufficiently small stresses, the stress is directly proportional to the
stress.
It depends on the material being deformed.
It also depends on the nature of the deformation.
The elastic modulus, in general, relates what is done to a solid object to
how that object responds.
Various types of deformation have unique elastic moduli.
stressElastic modulus
strain
Section 12.4
(12.5)
41
Three Types of Moduli
Young’s Modulus
Measures the resistance of a solid to a change in its length
Shear Modulus
Measures the resistance of motion of the planes within a solid
parallel to each other
Bulk Modulus
Measures the resistance of solids or liquids to changes in their
volume
Section 12.4
42
Young’s Modulus
The bar is stretched by an amount L
under the action of the force F.
The tensile stress is the ratio of the
magnitude of the external force to the
cross-sectional area A.
The tension strain is the ratio of the
change in length to the original length.
Young’s modulus, Y, is the ratio of
those two ratios:
Units are N / m2
tensile stress /tensile strain / i
F AY
L L
Section 12.4
(12.6)
43
• If you can't see the image above, please install Shockwave Flash Player.• If this active figure can’t auto-play, please click right button, then click play.
12.11: Young's ModulusIn this animation, a long bar clamped at one end is stretched under the action of a force F.
44
Stress vs. Strain Curve
Experiments show that for certain
stresses, the stress is directly
proportional to the strain.
This is the elastic behavior part
of the curve.
The elastic limit is the maximum
stress that can be applied to the
substance before it becomes
permanently deformed.
Section 12.4
45
Stress vs. Strain Curve, cont
When the stress exceeds the elastic limit, the substance will be
permanently deformed.
The curve is no longer a straight line.
With additional stress, the material ultimately breaks.
Section 12.4
46
Shear Modulus
Another type of deformation
occurs when a force acts parallel to
one of its faces while the opposite
face is held fixed by another force.
This is called a shear stress.
For small deformations, no change
in volume occurs with this
deformation.
A good first approximation
Section 12.4
47
Shear Modulus, cont.The shear strain is x / h. x is the horizontal distance the sheared
face moves.
h is the height of the object.
The shear stress is F / A. F is the tangential force.
A is the area of the face being sheared.
The shear modulus is the ratio of the
shear stress to the shear strain.
Units are N / m2Section 12.4
shear stress /shear strain /
F AS
x h
(12.7)
48
• If you can't see the image above, please install Shockwave Flash Player.• If this active figure can’t auto-play, please click right button, then click play.
12.13: Shear ModulusIn the initial conditions of this animation, we see shear deformation in which a rectangular block is distorted by two forces of equal magnitude but opposite directions applied to two parallel faces. The lower left shows a book under shear stress.
49
Bulk Modulus
Another type of deformation
occurs when a force of uniform
magnitude is applied
perpendicularly over the entire
surface of the object.
The object will undergo a
change in volume, but not in
shape.
Section 12.4
50
• If you can't see the image above, please install Shockwave Flash Player.• If this active figure can’t auto-play, please click right button, then click play.
12.14: Bulk ModulusWhen a solid is under uniform pressure as shown here, it undergoes a change in volume but no change in shape. This cube is compressed on all sides by forces normal to its six faces.
51
Bulk Modulus
The volume stress is defined as the ratio of the magnitude of the
total force, F, exerted on the surface to the area, A, of the surface.
This is also called the pressure.
The volume strain is the ratio of the change in volume to the
original volume.
Section 12.4
52
Bulk Modulus, cont.
The bulk modulus is the ratio of the volume stress to the volume
strain.
The negative indicates that an increase in pressure will result in a
decrease in volume.
volume stress /
volume strain / /i i
F A PB
V V V V
Section 12.4
(12.8)
53
Compressibility
The compressibility is the inverse of the bulk modulus.
It may be used instead of the bulk modulus.
Section 12.4
54
Moduli and Types of Materials
Both solids and liquids have a bulk modulus.
Liquids cannot sustain a shearing stress or a tensile stress.
If a shearing force or a tensile force is applied to a liquid, the liquid
will flow in response.
Section 12.4
55
Moduli Values
Section 12.4
56
Prestressed Concrete
If the stress on a solid object exceeds a certain value, the object fractures.
Concrete is normally very brittle when it is cast in thin sections.
The slab tends to sag and crack at unsupported areas.
The slab can be strengthened by the use of steel rods to reinforce the concrete. The concrete is stronger under compression than under tension.
Section 12.4
57
Pre-stressed Concrete, cont.
A significant increase in shear strength is achieved if the
reinforced concrete is pre-stressed.
As the concrete is being poured, the steel rods are held under
tension by external forces.
These external forces are released after the concrete cures.
This results in a permanent tension in the steel and hence a
compressive stress on the concrete.
This permits the concrete to support a much heavier load.
Section 12.4
58
Quick Quiz 12.4
For the three parts of this Quick Quiz, choose from the following choices the correct answer for the elastic modulus that describes the relationship between stress and strain for the system of interest, Which is in italics: (a) Young’s modulus (b) shear modulus (c) bulk modulus (d) none of those choices (i) A block of iron is sliding across horizontal floor. The friction force between the block and the floor causes the block to deform. (ii) A trapeze artist swings through a circular arc. At the bottom of the swing, the wires supporting the trapeze are longer than when the trapeze artist simply hangs from the trapeze due to the increased tension in them.
59
Quick Quiz 12.4
(iii) A space- craft carries a steel sphere to a planet on which atmospheric pressure is much higher than on the Earth. The higher pressure causes the radius of the sphere to decrease.
Answer: (i) (b) (ii) (a) (iii) (c)
60
Example 12.5
In Example 8.2, We analyzed a cable used to support an actor as he swung onto the stage. Now suppose the tension in the cable is 940 N as the actor reaches the lowest point. What diameter should a 10-m-long steel cable have if We do not Want it to stretch more than 0.50 cm under these conditions?
Solution:
310 2
2 2 2
(940N) (10m)2 3.5 10 m 3.5 mm
(20 10 N/m )(0.005 0 m)
i
i
FLA
Y L
FLAd r
Y L
d
61
Example 12.6
A solid brass sphere is initially surrounded by air, and the air pressure exerted on it is 1.0 × 105 N/m2 (normal atmospheric pressure). The sphere is lowered into the ocean to a depth Where the pressure is 2.0 × 107 N/m2. The volume of the sphere in air is 0.50 m3. By how much does this volume change once the sphere is submerged?
Solution:
B
PVV i
34
210
25273
m 106.1
N/m 101.6)N/m 100.1N/m 100.2)(m 50.0(
V
62
Analysis Model – Rigid Object in Equilibrium
A rigid object in equilibrium exhibits no translational or angular
acceleration.
The net external force acting on the object is zero:
This is the condition for translational equilibrium.
The net external torque acting on the object is zero:
This is the conditional for rotational equilibirium.
ext 0F
ext 0 τ
Summary
(12.1)
(12.2)
