Transcript Lecture #11 - the GMU ECE Department

CHAPTER 10
Force and Force-Related Parameters
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10-1
Outline
In this chapter we will
• explain what we mean by force
• discuss Newton’s laws in mechanics
• explain tendencies of unbalanced mechanical forces
• quantify the tendencies of a force acting at a distance
(moment), over a distance (work), and over a period
of time (impulse)
• discuss mechanical properties of materials
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10-2
What Is Force?
Force represents the interaction between
2 objects
Objects: car and person
Interaction: person pushing the car
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10-3
What Is Force?
Force exerted by your hand on
a lawn mower.
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Force exerted by bumper hitch
on the trailer.
10-4
Definition of Force
All forces are defined by their magnitudes, their
directions, and the point of applications
Examples to demonstrate the effect of magnitude, direction,
and the point of application of a force on the same object
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10-5
Tendencies of a Force
Examples to demonstrate the tendencies of a force
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10-6
Units of Force
Force is a derived parameter
force  mass  accelerati on
 ma
Units of force
kg  m
 m
1 N  1 kg  1 2   N  2
s
 s 
slug  ft
 ft 
1 lb f  1 slug  1 2   lb f  lb 
s2
 s 
1 lb f  4.448 N
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10-7
Types of Forces
Spring forces and Hooke’s Law
F  kx
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x
10-8
Example 10.1 – Spring Constant
Given: a spring as shown; dead weight is attached to
the end of the spring and the corresponding deflection is
measured.
Find: the spring constant
Results of the Experiments
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10-9
Example 10.1 – Spring Constant
Solution:
Spring constant k is determined by calculating the slope of a forcedeflection line
change in force
slope 
change in deflection

19.6  4.9N
k
 0.54 N/mm
36  9mm
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10-10
Types of Forces
• Friction forces
 dry friction
Fmax  N
weight
applied
force

viscous (fluid) friction
normal
force
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10-11
friction
force
Example 10.2 – Friction Force
Given: the coefficient of static friction between a book
and a desk surface is 0.6. The book weighs 20 N. A
horizontal force of 10 N is applied to the book.
Find: would the book move? If not, find the friction force
and magnitude of the horizontal force F to set the book
in motion.
weight
Solution:
applied
force
Fmax  N  0.620  12 newtons
normal
force
since Fmax > 10 N, the book would not move. It would
take a 12 N horizontal force to set the book in motion.
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10-12
friction
force
Newton’s Laws in Mechanics
• First law
 an object at rest remains at rest if no
unbalanced forces acting on it
 an object in motion with a constant velocity,
and if there are no unbalanced forces acting
on it, the object will continue to move with the
same velocity
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10-13
Newton’s Laws in Mechanics
Second law


 F  ma
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10-14
Newton’s Laws in Mechanics
• Third law
 For every action there is a reaction
 The forces of action and reaction have the same
magnitude and act along the same line, but they
have opposite directions
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10-15
Newton’s Laws in Mechanics
• Universal law of gravitation attraction

Two masses attract each other with a force that is
equal in magnitude and opposite in direction
Gm1m2
F
r2
G = 6.673 x10-11 m3/kg.s2

weight of an object is the force that is exerted on the
mass of the object by earth’s gravity
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10-16
Newton’s Laws in Mechanics
Weight of an object having a mass m
on the earth is
GM earthm
GM earth
W
, let g 
2
2
Rearth
Rearth
W  mg
Mearth = 5.97 x 1024 kg
Rearth = 6378 x 103 m
g = 9.8 m/s2
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10-17
Example 10.3 – Newton’s Laws
Given: an exploration vehicle with a mass of 250 kg on
Earth
Find: the mass of the vehicle on Moon (gMoon = 1.6 m/s2)
and on planet Mars (gMars = 3.7 m/s2); weight of vehicle
on Moon and Mars
Solution:
The mass of the vehicle is the same on Earth, Moon and Mars. The weight
of the vehicle is determined from W = mg,


On Moon : W  250 kg 1.6 m s   400 N
On Mars : W  250 kg 3.7 m s   925 N
On Earth : W  250 kg  9.8 m s 2  2450 N
2
2
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10-18
Example 10.4 – Newton’s Laws
Given: The space shuttle orbits the Earth at altitudes
ranging from 250 km to 965 km; mass of an astronaut is
70 kg on Earth
Find: The g value and the weight of the astronaut in orbit
Solution:
Space shuttle orbiting at an altitude of 250 km above Earth’s surface.
GM Earth
g
R2
3


m
11
24
 6.673 10

5
.
97

10
kg
2 
kg  s 

2


9
.
07
m
s
2
6378 103  250 103 m




W  70 kg  9.07 m s 2  635 N
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
 
10-19
Example 10.4 – Newton’s Laws
Solution (continued):
Space shuttle orbiting at an altitude of 965 km above the Earth’s surface.
GM Earth
g
R2
3


11 m
24
 6.673 10

5
.
97

10
kg
2 
kg  s 

2


7
.
38
m
s
2
3
3
6378 10  965 10 m




W  70 kg  7.38 m s 2  517 N

 
The near weightless conditions that you see on TV are created by the orbital
speed of the shuttle. For example, when the shuttle is orbiting at an altitude
of 935 km at a speed of 7744 m/s, it creates a normal acceleration of 8.2
m/s2. It is the difference between g and normal acceleration [(9.8 – 8.2)
m/s2] that creates the condition of weightlessness.
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10-20
Moment, Torque – Force Acting at a
Distance
Moment is the measure of the tendency of a force
acting about an axis or a point
hinge
Moment is a vector
Example – opening and closing a door
To open or close a door, we apply a pulling
or pushing force on the doorknob. This
force will make the door rotate about its
hinge. In mechanics, this tendency of
force is measured in terms of a moment of
a force about an axis or a point.
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10-21
Moment, Torque – Force Acting at a
Distance
Calculating moment of a force
Line of action
C
A
M A  d1 F
M B  d2 F
B
F
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MC  0
10-22
Example 10.5 – Moment, Torque
0.07 cos 35o
Given: the locations and forces
applied as shown
Find: the resulting moment of the
forces about point O
Solution:
0.1 cos 35o

M
O
 50 N 0.05 m   50 N 0.07 cos 35 m   100 N 0.1cos 35 m 
 13.55 N  m
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10-23
Example 10.6 – Moment, Torque
Given: two forces are applied as shown
Find: moment about points A, B, C, and D
Solution:
Note the two forces have equal magnitude and act in
opposite directions.




M
M
M
M
A
B
C
D
 100 N 0   100 N 0.1 m   10 N  m
 100 N 0.1 m   100 N 0  10 N  m
 100 N 0.25 m   100 N 0.15 m   10 N  m
 100 N 0.35 m   100 N 0.25 m   10 N  m
This pair of forces is called a couple.
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10-24
Internal Force
When an object is subjected to an external force,
internal forces are created inside the material to hold
the material and the components together.
An example of internal force. When you try to pull apart the
bar, inside the bar material, internal forces develop that keep
the bar together as one piece
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10-25
Reaction Force
Reaction forces are developed at the
supporting boundaries to keep the object held
in position as planned
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10-26
Reaction Force
Examples to demonstrate various support conditions and how
they influence the behavior of an object.
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10-27
Work – Force Acting Over a Distance
Mechanical work is defined as the component
of the force – that moves the object – times the
distance the object moves
work = F x d
units: N.m, J, lb.ft….
distance, d
force
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10-28
Work Done on a Car
d
Work done on the car by the force F is equal to
W1-2 = (F cos θ)(d)
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10-29
Mechanical Power
Power is the time rate of doing work. It
represents how fast you want to do the work.
work
power 
time
units: J/s (watts), N.m/s
lb.ft/s
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10-30
Example 10.7 – Work
Given: a box with weight of 100 N on the
ground as shown
Find: work required to lift the box 1.5 m
above the ground
Solution:
W  100 N1.5 m  150 N  m
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10-31
Example 10.7 – Power
Given: a box with weight of 100 N on the
ground as shown. We want to lift the box
in 3 seconds
Find: power required
Solution:
work 150 J
power 

 50 watts
time
3s
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10-32
Pressure – Force Acting Over an Area
Pressure provides a measure of intensity of a
force acting over an area.
force
pressure 
area
weight
contact area
Units: Pa (N/m2), kPa, MPa, GPa
psi, ksi
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10-33
Pressure
An experiment demonstrating the concept of
pressure using bricks (a) and (b)
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10-34
Common Units of Pressure
1N
1 Pa  2
m
1 lb
1 psi 
1 in 2
1 lb
1 psf  2
1 ft
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1 kPa  1000 Pa
1 ksi  1000 psi
1 ksf  1000 psf
10-35
Pressure – Pascal’s Law
Pascal’s law states that for a fluid at rest,
pressure at a point is the same in all directions
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10-36
Pressure – Pascal’s Law
For a fluid at rest, the pressure increases with
the depth of fluid.
P  gh
P = fluid pressure at a point (Pa)
 = density of fluid (kg/m3)
g = acceleration due to gravity (m/s2)
h = height of fluid column (m)
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10-37
Pressure – Pascal’s Law
Buoyancy is the force that a fluid exerts on a
submerged object.
buoyancy force  FB  ρVg (N)
V  volume of the object (m 3 )
  density of fluid kg m 3 

g  accelerati on due to gravity 9.8 m s 2
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
10-38
Example 10.8 – Pressure
Given: a water of height h in a water tower
Find: develop a table showing the water
pressure in a pipeline located at the base of
the water tower relating to h; use the table to
determine h if a pressure of 50 psi is desired
Solution:
We will first develop the equation for pressure P as a function of height h
P  gh
 1 ft 2 
ft 
 lb 
 slug 
P 2   1.94 3  32.2 2 h ft 
2
in
ft
s
144
in







 0.4338h ft 
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10-39
Example 10.8 – Pressure
Solution (continued):
Substituting values of h from 10 ft to 240 ft yields the
pressure in the following table:
For a pressure of 50 psi, the water level in the tower
should be approximately 120 ft
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10-40
Pressure in Hydraulic Systems
In an enclosed fluid system, the pressure is nearly constant throughout
the system. The relationship between the forces F1 and F2 could be
established.
P1 
F1
A1
since P1  P2 ,
P2 
P1  P2 
F2
A2
F1 F2

A1 A2
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
F2 
10-41
A2
F1
A1
Pressure in Hydraulic System
Push to
raise load
To raise the load – push the
arm of the hand pump down –
fluid will be pushed into the
load cylinder, which in turn
creates a pressure that is
transmitted to the load
piston, and consequently the
load is raised.
To lower the load – open the
release valve. The fluid will
return to the reservoir and
the load is lowered.
Open to
lower load
reservoir
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10-42
Pressure in Hydraulic System
To raise the load – move the
control lever UP – fluid will
be pushed into the load
cylinder, which in turn
creates a pressure that is
transmitted to the load
piston, and consequently the
load is raised.
Gear or rotary
pump
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To lower the load – push the
control lever down. The fluid
will return to the reservoir
and the load is lowered.
10-43
Example 10.12 – Hydraulic System
Given: the hydraulic system
shown, g = 9.81 m/s2
Find: the load that can be
lifted by the hydraulic
system.
Solution:


F1  m1 g  100 kg  9.81 m/s 2  981 N
A2
 0.15 m 
981 N   8829 N
F2 
F1 
2
A1
 0.05 m 
2

F2  8829 N  m2 kg  9.81 m/s 2
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
 m2  900 kg
10-44
Example 10.12 – Hydraulic System
Solution (continued):
Alternate approach
A
 R2 
m1 g 
F2  2 F1  m2 g 
2
A1
 R1 
2
2
2


R2 
15 cm 
100 kg   900 kg
m2 
m1 
2
2
R1 
5 cm 
The second approach is preferred. We can see clearly the relationship
between m1 and m2.
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10-45
Stress
Stress provides a measure of the intensity of a force
acting over an area.
force
stress 
area
units: Pa (N/m2), kPa, MPa, GPa
psi, ksi
force component normal to area
normal stress 
area
force component parallel to area
shear stress 
area
Normal stress is often called pressure
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10-46
Strain
• Strain is a description of the deformation of a
body in terms of the relative displacement of
the particles in the body.
 That is, how much the body stretches.
• Deformation, or strain, is typically caused by
the application of an external force.
Engineering Fundamentals, By Saeed Moaveni, Fourth Edition,
Copyrighted 2011
10-47
Mechanical Properties of Materials
• Modulus of elasticity
a measure of how easily material stretches
• Modulus of Rigidity (also called shear
modulus)
a measure of how easily material twists
• Tensile strength
• Compressive strength
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10-48
Mechanical Properties of Materials
Modulus of elasticity, E – measures how easily a material
will stretch when pulled (subject to a tensile force) or how
well a material will shorten when pushed (subject to
compressive force).
When subjected to the same
force, which piece of material
will stretch more?
Esteel > Ealuminum > Erubber
L
L
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10-49
L
Tensile Test of Metal Specimen
Tensile test set up
Original
specimen
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Final
specimen
10-50
Stress-Strain Diagram
σe = elastic stress
(σY)u = upper-yield
stress
(σY)l = lower-yield
stress
σY = ultimate stress
σf = fracture stress
Stress-strain diagram for a mild steel sample
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10-51
Modulus of Elasticity
Hooke' s law
F
x
  E 
E
A
L
solve for E yields,
FL
E
Ax
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x
10-52
Other Mechanical Properties of
Materials
• Modulus of rigidity or shear modulus
 measures how easily a material can be
twisted or sheared
 is used to select materials for shafts or
rods subjected to twisting torques
 value is determined using a torsional test
machine
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10-53
Other Mechanical Properties of
Materials
• Tensile strength or ultimate strength
 maximum tensile force per unit of original
cross-sectional area of specimen
• Compressive strength
 maximum compressive force per unit of
cross-sectional area of specimen
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10-54
Modulus of Elasticity and Shear
Modulus of Selected Materials
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10-55
Strength of Selected Materials
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10-56
Strength of Selected Materials
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10-57
Example 10.13 – Material Properties
Given: a structural member and a load of
4000 N distributed uniformly over the crosssectional area of the member
Find: select a material to carry the load safely
Solution:
We will consider the strength of the material as the design factor in this
example.

4000 N
 16 MPa
0.05 m0.005 m
Aluminum alloy or structural steel material with yield strength of 50 MPa
and 200 MPa, respectively, could carry the load safely.
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10-58
Summary
• You should have a good understanding of
force and its common units.
• You should know the types of forces.
• You should understand the tendency of an
unbalanced force, to rotate or to translate.
• You should know that the application of forces
can lengthen, shorten, bend, and twist objects.
• You should be familiar with Newton’s laws in
mechanics.
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10-59
Summary
• You should understand pressure and should know
 Pascal’s law for static fluids
 relationship between fluid pressure and depth
of fluid
 how hydraulic systems work
 fluid properties: viscosity, bulk modulus of
compressibility
• You should know the mechanical properties of
materials: modulus of elasticity, shear modulus,
tensile and compressive strength
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10-60
Summary
• You should know how a material behaves under applied
forces
• You should understand the effect of an applied force on
an object
 stress: intensity of a force acting within the material
 Internal forces: forces created inside the material to
hold the material and the components together
 moment: force acting at a distance
 work: force acting over a distance
 linear impulse: force acting over a period of time
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10-61