Transcript 02_lecture_ppt

Chapter 2
Lecture
Outline
1
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 2: Force
•Forces
•Vector Addition
•Newton’s First and Third Laws
•Gravity
•Contact Forces
•Tension
•Fundamental Forces
2
§2.1 Forces
Isaac Newton was the first to discover that the laws that
govern motions on the Earth also applied to celestial bodies.
Over the next few chapters we will study how bodies interact
with one another.
3
Simply, a force is a “push” or “pull” on an object.
4
How can a force be measured? One way is with a spring
scale.
By hanging masses on a
spring we find that the
spring stretchapplied
force.
The unit of force is the newton (N).
5
Vectors versus scalars:
A vector is a quantity that has both a magnitude and a
direction. A force is an example of a vector quantity.
A scalar is just a number (no direction). The mass of an
object is an example of a scalar quantity.
6
Notation:
Vector:

F or F
The magnitude of a vector:

F or F or F .
The direction of vector might be “35 south of east”; “20
above the +x-axis”; or….
Scalar: m (not bold face; no arrow)
7
• Important!
– When the symbol for a vector is written
without the arrow and in italics rather than
boldface (F), it stands for the magnitude of the
vector (which is a scalar).
8
Conceptual Example and Practice
Problems 2.1
• Is temperature a vector or scalar?
• When you deposit a paycheck, the
balance of your account “goes up.” When
you pay a bill, it “goes down.” Is the
balance of your account a vector quantity?
9
§2.2 Graphical Vector Addition
To graphically represent a vector,
draw a directed line segment.
The length of the line can be used to represent the vector’s
length or magnitude.
10
To add vectors graphically they must be placed “tip to tail”.
The result (F1 + F2) points from the tail of the first vector to
the tip of the second vector.
F2
F1
Fnet
For collinear vectors:
F1
F2
Fnet
11
12
• Important!
– A plus sign (+) between vectors indicates
vector addition, not ordinary addition. An
equals sign (=) between vector quantities
means that the vectors are identical in
magnitude and direction, not simply that their
magnitudes are equal.
– A common error is to draw the sum from the
tip of the second vector to the tail of the first.
13
Practice and HW
• Practice – p. 62 # 3, 5
• HW – p. 62 # 2, 4
14
Checkpoint 2.2
• What is the vector sum of a force of 20 N
north and a force of 50 N directed south?
15
Example 2.2
• Two draft horses, Sam and Bob, are dragging a
sled loaded with jugs of maple syrup. They pull
with horizontal forces of equal magnitude 1.50
kN (kilonewtons) on the front of the sled. The
force due to Sam is in the direction 15° north of
east, and the force due to Bob is 15° south of
east. Use the graphical method of vector
addition to find the magnitude and direction of
the sum of the forces exerted on the sled by the
two horses.
16
17
18
Practice Problem 2.2
• If Sam and Bob were to pull with forces of the
same magnitude as before but angle 30° north
and south of east, would the sum of the two
angles be larger, smaller, or the same
magnitude as before? Illustrate the sketch.
• If Sam pulls 10° north of east while Bob pulls 15°
south of east, is it still possible for the sum of the
two forces to be due east if their magnitudes are
not the same? Which force must have the larger
magnitude? Illustrate the sketch.
19
20
Practice and HW
• Practice – p. 62 # 7, 11
• HW – p. 62 # 8, 10, 12
21
§2.3 Vector Addition Using
Components
Vector Addition: Place the vectors tip to tail as before. A
vector may be moved any way you please provided that you
do not change its length nor rotate it. The resultant points
from the tail of the first vector to the tip of the second (A+B).
22
Problem Solving Strategy - Finding x and y
components of a Vector for Its Magnitude and
Direction
• Draw a right angle with the vector as the
hypotenuse and the other two sides parallel to
the x- and y-axes.
• Determine one of the angles in the triangle.
• Use trigonometric functions to find the
magnitudes of the components. Make sure your
calculator is in “degree mode” to evaluate
trigonometric functions of angle in degrees and
“radian mode” for angles in radians.
• Determine the correct algebraic sign for each
component.
23
Example: Vector A has a length of 5.00 meters and points
along the x-axis. Vector B has a length of 3.00 meters and
points 120 from the +x-axis. Compute A+B (=C).
y
B
C
120
A
x
24
Example continued:
y
sin  
opp
hyp
cos  
B
hyp
By
60
Bx
sin 60  
By
B
cos 60  
tan  
120
A
sin 
cos 

opp
adj
x
 B y  B sin 60   3 . 00 m  sin 60   2 . 60 m
 Bx
B
adj
 B x   B cos 60    3 . 00 m cos 60    1 . 50 m
and Ax = 5.00 m and Ay = 0.00 m
25
Example continued:
The components of C:
C x  A x  B x  5 . 00 m    1.50 m   3.50 m
C y  A y  B y  0 . 00 m  2.60 m  2.60 m
y
The length of C is:
C
C  C 
Cy = 2.60 m

Cx = 3.50 m
The direction of C is:

x
tan  
2
m    2 . 60 m 
2
2
 4 . 36 m
Cy
Cx
  tan
3 . 50
2
Cx  Cy
1

2 . 60 m
 0 . 7429
3.50 m
0 . 7429   36 . 6  From the +x-axis
26
Problem Solving Strategy: Finding the Magnitude
and Direction of a Vector from Its x and y
Components
• Sketch the vector on a set of x- and y-axes in the correct
quadrant, according to the signs of the components.
• Draw a right triangle with the vector as the hypotenuse
and the other two sides parallel to the x- and y-axes.
• In the right triangle, choose which of the unknown angles
you want to determine.
• Use the inverse tangent to find the angle.
• tan  = Fy/Fx and  = tan-1 Fy/Fx
• Interpret the angle: specify whether it is the angle belwo
the horizontal, or the angle west of south, or the abgle
clockwise from the negative y axis, etc.
• Use the Pythagorean theorem to find the magnitude of
the vector.
F = √(Fx2 + Fy2)
27
Example 2.3
• Suppose you are standing on the floor doing your daily
exercises. For one exercise, you lift your arms up and
out until they are horizontal. In this position, assume that
the deltoid muscle exerts a force of 270 N at an angle of
15° above the horizontal on the humerus. What are the
x- and y-components of the force?
28
29
Practice Problem 2.3
• While you are tilling the garden, you exert
a force on the handles of the tiller that has
components Fx = +85 N and Fy = -132 N.
The x-axis is horizontal and the y-axis
points up. What are the magnitude and
direction of this force?
30
31
Problem Solving Strategy – Adding
Vectors Using Components
• C = A + B if and only if Cx = Ax + Bx and Cy = Ay + By
• Find the x- and y-components of each vector being
added.
• Add the x-components (with their algebraic signs) of the
vectors to find the x-component of the sum (if the signs
are not correct, then the sum will not be correct.)
• Add the y-components (with their algebraic signs) of the
vectors to find the y-component of the sum.
• If necessary, use the x- and y-components of the sum to
find the magnitude and direction of the sum.
32
Example 2.4
• In a traction apparatus, three cords pull on
the central pulley, each with magnitude
22.0 N, in the direction shown in the Fig.
2.12. What is the sum of the forces
exerted on the central pulley by the three
cords? Give the magnitude and direction
of the sum.
33
34
35
36
Practice Problem 2.4
• The pulleys are moved, after which F1 and
F2 are at an angle of 30.0° above the xaxis and F3 is 60.0° below the x-axis.
– What is the sum of these three forces in
component form?
– What is the magnitude of the sum?
– At what angle with the horizontal is the sum?
37
38
Practice and HW
• Practice – p. 63 # 17, 19, 21, 23
• HW – p. 63 # 18, 22, 24
39
§2.4 Newton’s First Law
Newton’s 1st Law (The Law of Inertia):
If no net force acts on an object, then its speed and direction
of motion do not change.
Inertia is a measure of an object’s resistance to changes
in its motion. It does not mean resistance to the
continuation of motion (or the tendency to come to rest.)
The net force is the vector sum of all the forces acting on
a body.
F net 
F
i
i
 F1  F 2  F3  
40
If the object is at rest, it remains at rest (speed = 0).
If the object is in motion, it continues to move in a straight
line with the same speed.
No force is required to keep a body in straight line motion
when effects such as friction are negligible.
41
42
Conceptual Example 2.5
• The task of shoveling newly fallen snow
from the driveway can be thought of as a
struggle against the inertia of the snow.
Without the application of a net force, the
snow remains at rest on the ground.
However, in an important way the inertia of
the snow makes it easier to shovel.
Explain.
43
44
Conceptual Practice Problem 2.5
• A college student stands on a subway car,
holding on to an overhead strap. As the
train starts to pull out of the station, she
feels thrust toward the rear of the car; as
the train comes to a stop at the next
station, she feels thrust forward. Explain
the role played by inertia in this situation.
45
46
An object is in translational equilibrium if the net force on
it is zero.
47
Free Body Diagrams:
•Must be drawn for problems when forces are involved.
•Must be large so that they are readable.
•Draw an idealization of the body in question (a dot, a
box,…). You will need one free body diagram for each
body in the problem that will provide useful information
for you to solve the given problem.
•Indicate only the forces acting on the body. Label the
forces appropriately. Do not include the forces that this
body exerts on any other body.
48
Free Body Diagrams (continued):
•A coordinate system is a must.
•Do not include fictitious forces. Remember that ma is itself
not a force!
•You may indicate the direction of the body’s acceleration or
direction of motion if you wish, but it must be done well off to
the side of the free body diagram.
•For an object in equilibrium,
SFx = 0
SFy = 0
SFz = 0
49
Example 2.6
• A red-tail hawk that weighs 8 N is gliding
due north at constant speed. What is the
force acting on the hawk due to the air?
Draw a FBD for the hawk.
50
51
Practice Problem 2.6
• An 80-N crate of apples sits at rest on the
horizontal bed of a parked pickup truck.
What is the for C exerted on the crate by
the bed of the pickup truck? Draw the
FBD for the crate.
52
53
Example 2.7
• The forces on an airplane in flight heading
eastward are as follows: gravity = 16.0 kN
downward; lift = 1.8 kN upward; thrust =
1.8 kN east; and drag = 0.8 kN west.
What is the net force on the plane?
(Thrust, drag, and lift are forces exerted on
the plane)
54
55
56
Practice Problem 2.8
• Find the net force on the airplane if the
forces are G = 16.0 kN down; L = 15.5 kN
up; Th = 1.2 kN north; D = 1.2 kN south.
57
58
Example 2.8
• To slide a chest that
weighs 750 N across
the floor at constant
velocity, you must
push it horizontally
with a force of 450 N.
Find the contact force
that the floor exerts
on the chest.
59
60
Practice Problem 2.8
• Suppose the same chest is at rest. You
push it horizontally with a force of 110 N
but it does not budge. What is the contact
force on the chest due to the floor during
the time you are pushing it?
61
62
Practice and HW
• Practice – p. 63 # 26, 27, 31
• HW – p. 63 # 28, 29, 32
63
§2.5 Newton’s Third Law
Newton’s 3rd Law:
When 2 bodies interact, the forces on the bodies from each
other are always equal in magnitude and opposite in
direction. Or, forces come in pairs.
Mathematically: F 21   F12
64
Newton’s 3rd Law
• We call two the two forces an interaction
pair; each force is the interaction partner
of the other.
• Interaction partners act on different objects
– the two objects that are interacting.
65
Example: Consider a box resting on a table.
F1
(a) If F1 is the force of the Earth
on the box, what is the interaction
partner of this force?
The force of the box on the Earth.
66
Example continued:
F2
(b) If F2 is the force of the box on the
table, what is the interaction partner
of this force?
The force of the table on the box.
67
Conceptual Example 2.9
• Earth exerts a gravitational force on an
orbiting communications satellite. What is
the interaction partner of this force?
68
69
Practice Problem 2.9
• In Example 2.8, the contact force exerted
on the chest by the floor was 870 N,
directed 59° above the leftward horizontal
(-x axis). Describe the interaction partner
of this force – in other words what object
exerts it on what other object? What are
the magnitude and direction of the
interaction partner?
70
71
• Do not assume that Newton’s third law in
involved every time two forces happen to
be equal and opposite.
• Remember, to be a third law pair, the
forces must act on different objects.
72
External forces:
Any force on a system from a body outside of the system.
F
Pulling a box
across the floor
73
Internal forces:
Force between bodies of a system.
Fext
Pulling 2 boxes across the floor
where the two boxes are attached
to each other by a rope.
74
Practice and HW
• Practice – p. 64 # 37, 41
• HW – p. 64 # 38, 43
75
§2.6 Gravity
Gravity is the force between two masses. Gravity is a longrange or field force. No contact is needed between the
bodies. The force of gravity is always attractive!
F 
GM 1 M
r
M1
r is the distance between the two masses
M1 and M2 and G = 6.6710-11 Nm2/kg2.
2
2
F21
F12
M2
F 21   F12
r
76
Let M1 = mass of the Earth.
 GM
F 
2
r

E

M 2

Here F = the force the Earth exerts on mass M2. This is the
force known as weight, w.
 GM E
w
 r 2
 E

 M 2  gM 2 .


where g 
GM
2
rE
E
 9 . 8 N/kg
M E  5 . 97  10
24
kg
rE  6370 km
Near the surface
of the Earth
77
78
Note that
g
F
m
is the gravitational force per unit mass.
This is called the gravitational field
strength. It is often referred to as the
acceleration due to gravity.
What is the direction of g?
What is the direction of w?
79
Example: What is the weight of a 100 kg astronaut on the
surface of the Earth (force of the Earth on the astronaut)?
How about in low Earth orbit? This is an orbit about 300 km
above the surface of the Earth.
On Earth:
w  mg  980 N
 GM E
In low Earth orbit: w  mg ( ro )  m 
2


R

r
o
 E

  890 N


Their weight is reduced by about 10%.
The astronaut is NOT weightless!
80
Example 2.10
• When you are in a
commercial airliner
cruising at an altitude
of 6.4 km (21000 ft),
by what percentage
has your weight (as
well as the weight of
the airplane) changed
compared with your
weight on the ground?
81
82
Practice Problem 2.10
• After an automobile collision, one driver
claims that the gravitational force between
the two cars caused the collision.
Estimate the magnitude of the gravitational
force exerted by one car on another when
they are driving side-by-side in parallel
lanes and comment on the driver’s claim.
83
84
Example 2.11
• In most countries other than the US,
produce is sold in mass units (grams or
kilograms) rather than in force units
(pounds or newtons). The scale still
measures a force, but is calibrated to
show the mass of the produce instead of
the weight. What is the weight of 350 g of
fresh figs, in newtons and in pounds?
85
Practice Problem 2.11
• What would be those figs weigh on the
surface of the Moon, where g = 1.62 N/kg?
86
87
Practice and HW
• Practice – pp. 64-65 # 45, 47, 51
• HW – pp. 64-65 # 46, 48
88
§2.7 Contact Forces
Contact forces: these forces arise because of an
interaction between the atoms in the surfaces in contact.
89
Normal force: this force acts in the direction perpendicular
to the contact surface.
N
Normal force
of the ground
on the box
w
N
Normal force
of the ramp on
the box
w
90
• When the surface is not horizontal
between two objects, the normal force is
not equal to the weight force.
• Normal forces are always perpendicular.
• Weight force is always straight downward.
91
Example: Consider a box on a table.
y
N
FBD for box
x
w
Apply
Newton’s
2nd law
F
y
 N w0
So that N  w  mg
This just says the magnitude of the
normal force equals the magnitude
of the weight; they are not Newton’s
third law interaction partners.
92
Friction: a contact force parallel to the contact surfaces.
Static friction acts to prevent objects from sliding.
Kinetic friction acts to make sliding objects slow down.
93
Static Friction:
The force of static friction is modeled as
fs  s N .
where s is the coefficient of static friction and N is the
normal force.
94
Kinetic Friction:
The force of kinetic friction is modeled as f k   k N .
where k is the coefficient of kinetic friction and N is the
normal force.
95
Example (text problem 2.97): A box full of books rests on a
wooden floor. The normal force the floor exerts on the box is
250 N.
(a) You push horizontally on the box with a force of 120
N, but it refuses to budge. What can you say about the
coefficient of friction between the box and the floor?
y
FBD for
box
N
F
x
fs
w
Apply
Newton’s
2nd Law
(1)  F y  N  w  0
( 2 ) Fx  F  f s  0
96
Example continued:
From (2):
F  fs  s N
 s 
F
 0 . 48
N
This is the minimum value of s, so s > 0.48.
(b) If you must push horizontally on the box with 150 N force
to start it sliding, what is the coefficient of static friction?
Again from (2):
F  fs  s N
 s 
F
 0 . 60
N
97
Example continued:
(c) Once the box is sliding, you only have to push with a
force of 120 N to keep it sliding. What is the coefficient of
kinetic friction?
y
N
FBD for
box
F
x
fk
Apply
Newton’s
2nd Law
(1)  F y  N  w  0
( 2 ) Fx  F  f k  0
w
From 2:
F  fk  k N
k 
F
N

120 N
250 N
 0 . 48
98
Consider a box of mass m that is at rest on an incline.
Its FBD is:
There is one longrange force acting on
the box: gravity.
y
FRB
w
x
There is one contact
force acting on the
box from the ramp.
If the net force acting on the box is zero, then the contact
force from the ramp must have the same magnitude as the
weight force, but be in the opposite direction.
99
The force FRB can be resolved into components that are
perpendicular and parallel to the ramp.
y
FRB
N
The perpendicular component is
what we call the normal force.
fs
w
x
The parallel component is the
static friction force.
100
Example: Let the box on the ramp have a mass 2.5 kg. If
the angle between the incline and the horizontal is 25, what
are the magnitudes of the weight force, normal force, and
static friction force acting on the box?
y
FRB
Apply
Newton’s
2nd Law
N
fs

w
F
F
y
 N  w cos   0
x
 w sin   f s  0
x
w  mg   2 . 5 kg 9 . 8 N/kg
The forces are:
  24 . 5 N
N  w cos    24 . 5 N  cos 25   22 . 2 N
f s  w sin    24 . 5 N  sin 25   9 . 31 N
101
Example 2.12
• Example 2.8 involved sliding a 750-N
chest to the right at constant velocity by
pushing it with a horizontal force of 450 N.
We found that the contact force on the
chest due to the floor had components Cx
= -450 N and Cy = 750 N, where the x-axis
points to the right and the y-axis points up.
What is the coefficient of kinetic friction for
the chest-floor surface?
102
103
Practice Problem 2.12
• Suppose the same chest is at rest. You
push to the right with a force of 110 N but
the chest does not budge. What are the
normal and frictional forces on the chest
due to the floor while you are pushing?
Explain why you do not need to know the
coefficient of static friction to answer the
question.
104
Conceptual Example 2.13
• A horse pulls a sleigh to the right at constant velocity on level
ground. The horse exerts a horizontal force Fsh on the sleigh. a)
Draw 3 FBDs, one for the horse, one for the sleigh, and one for the
horse-sleigh system. b) To make the sleigh increase its velocity,
there must be a nonzero net force to the right acting on the sleigh.
Suppose the horse pulls harder. According to Newton’s third law,
the sleigh always pulls back on the horse with the same magnitude
as the force with which the horse pulls on the sleigh. Does this
mean that no matter how hard it pulls, the horse can never make the
net force on the sleigh nonzero? Explain. c) Identify the interaction
partner for each force acting on the sleigh.
105
106
107
108
Practice Problem 2.13
• A car is moving north and speeding up to pass a
truck on a level road. The combined contact
force exerted on the road by all four tires has
vertical component 11.0 kN downward and
horizontal component 3.3 kN southward. The
drag force exerted on the car by the air is 1.2 kN
southward. A) Draw the FBD for the car. B)
What is the weight of the car? C) What is the
net force acting on the car?
109
Equilibrium on an Inclined Plane
• An object in equilibrium is not accelerating (or
rotating).
• Let the y-axis be in the direction of the normal
force.
• Let the x-axis be along the incline, with positive
being in the direction that the object would
move.
• In other words, tilt your axes to simplify the
math.
110
Example 2.14
• A new safe is being delivered to the Corner Book Store.
It is to be placed in the wall at a height of 1.5 m above
the floor. The delivery people have a portable ramp,
which they plan to use to help them push the safe up an
into position. The mass of the safe is 510 kg, the
coefficient of static friction is 0.42 and the coefficient of
kinetic friction is 0.33. The ramp forms an angle of 15°
above the horizontal. A) How hard do the movers have
to push to start the safe moving up the incline? Assume
that they push in a direction parallel to the incline. B) To
slide the safe up at a constant speed, with what
magnitude force must the movers push?
111
112
113
Practice Problem 2.14
• During the seventh-inning stretch of a baseball
game, groundskeepers drags mats across the
infield dirt to smooth it. A groundskeeper is
pulling a mat at a constant velocity by applying a
force of 120 N at an angle of 22° above the
horizontal. The coefficient of kinetic friction
between the mat and the ground is 0.60. Find a)
the magnitude of the frictional force between the
dirt and the mat and b) the weight of the mat.
114
115
Practice and HW
• Practice – pp. 65-66 # 59, 61
• HW – pp. 64-65 # 58, 68
116
§2.8 Tension
This is the force transmitted through a “rope” from one end
to the other.
An ideal cord has zero mass, does not stretch, and the
tension is the same throughout the cord.
117
Example (text problem 2.79): A pulley is hung from the ceiling by a
rope. A block of mass M is suspended by another rope that
passes over the pulley and is attached to the wall. The rope
fastened to the wall makes a right angle with the wall. Neglect the
masses of the rope and the pulley. Find the tension in the rope
from which the pulley hangs and the angle .
y
T
FDB for the
mass M
x
w
Apply Newton’s 2nd
Law to the mass M.
F
y
T w0
T  w  Mg
118
Example continued:
Apply Newton’s 2nd Law:
FBD for the pulley:
F
F
y
F
T
x
 F cos   T  0
y
 F sin   T  0
 T  F cos   F sin 

x
T
This statement is true
only when  = 45 and
F 
2T 
2 Mg
119
Example 2.15
• Figure 2.37 shows the
bowstring of a bow and
arrow just before it is
released. The archer is
pulling back on the midpoint of the bowstring with
a horizontal force of 162
N. What is the tension in
the bowstring?
120
121
Practice Problem 2.15
• Jorge decides to rig up a tightrope in the backyard so his children
can develop a good sense of balance. For safety reasons, he
positions a horizontal cable only 0.60 m above the ground. If the
6.00-m long cable sags by 0.12 m from its taut horizontal position
when Denisha (weight 250 N) is standing on the middle of it, what is
the tension in the cable. Ignore the weight of the cable.
122
The Ideal Pulley
• A pulley can change the direction of the force
exerted by a cord under tension.
• An ideal pulley has no mass and no friction.
• An ideal pulley exerts no forces on the cord that
are tangent to the cord – it is not pulling in either
direction along the cord.
• As a result, the tension of an ideal cord that runs
through an ideal pulley is the same on both
sides of the pulley.
123
124
Example 2.16
• A 1804-N engine is
hauled upward at
constant speed. What
are the tensions in the
three ropes labeled A, B,
and C? Assume the
ropes and the pulleys
labeled L and R are
ideal.
125
126
Practice Problem 2.16
• Consider the entire collection of ropes,
pulleys, and the engine to be a single
system. Draw the FBD for this system and
show that the net force is zero. [Hint:
Remember that only forces exerted by
objects external to the system are included
in the FBD.]
127
128
Practice and HW
• Practice – p. 66 # 71, 72, 75
• HW – p. 66 # 70, 80
129
§2.9 Fundamental Forces
The four fundamental forces of nature are:
•Gravity which is the force between two masses; it is the
weakest of the four.
•Strong Force which helps to bind atomic nuclei together;
it is the strongest of the four.
•Weak Force plays a role in some nuclear reactions.
•Electromagnetic is the force that acts between charged
particles.
130
Practice and HW
• Practice – None
• HW – p. 67 # 82, 84
131
Summary
•Newton’s First and Third Law’s
•Free Body Diagrams
•Adding Vectors
•Contact Forces Versus Long-Range Forces
•Different Forces (friction, gravity, normal, tension)
132
What is the net force acting on the object shown below?
y
15 N
15 N
x
10 N
a. 40 N
b. 0 N
c. 10 N down
d. 10 N up
133
The gravitational field strength of the Moon is about 1/6
that of Earth. If the mass and weight of an astronaut, as
measured on Earth, are m and w respectively, what will
they be on the Moon?
a. m , w
b.
1
m, w
6
c . m,
1
w
6
d.
1
6
m,
1
w
6
134