Transcript Physics 4

Physics
Chapter 4
Forces and the Laws of Motion
Forces and the Laws of Motion
 A force is defined as a push or pull exerted on
an object.
 Forces can cause objects to speed up, slow
down, or change direction as they move.
 Based on the definitions of velocity and
acceleration, a force exerted on an object
causes that object’s velocity to change; that is,
a force causes an acceleration.
Forces and the Laws of Motion
 How can you cause an object to move?
 You can push on it or you can pull on it. The
push or pull is a force that you exert on the
object.
 If you push harder on an object, you have a
greater effect on its motion.
 The direction in which force is exerted also
matters. If you push the book to the right, the
book moves towards right.
Forces and the Laws of Motion
 When considering how a force affects motion, it
is important to identify the object of interest.
This object is called the system.
 Everything around the object that exerts forces
on it is called the external world.
Forces and the Laws of Motion
 Think about the different ways in which you
could move a textbook.
 You could touch it directly and push or pull it,
or you could tie a string around it and pull on
the string. These are examples of contact forces.
 A contact force exists when an object from the
external world touches a system and thereby
exerts a force on it.
Forces and the Laws of Motion
 Contact Forces

Acts on an Object by Direct Contact
 Field Forces


Acts on an Object Without Direct Contact
Examples
• Magnetic Attraction and Repulsion
• Gravitational Forces
Forces and the Laws of Motion
 Agent



Origin of Force
Identifiable
May be animate or inanimate
 Examples?

Contact / Field Forces
Forces and the Laws of Motion
 Free Body Diagrams

Illustrate All Forces Acting on an Object
• Examples?
Newton’s Laws of Motion
Forces and the Laws of Motion
 Newton’s First Law of Motion

“Every Body Continues In Its State of Rest,
Or of Uniform Motion In A Straight Line,
Unless It is Compelled to Change That State
by Forces Impressed Upon It.”
Forces and the Laws of Motion
 Newton’s First Law of Motion


“Every Body Continues In Its State of Rest,
Or of Uniform Motion In A Straight Line,
Unless It is Compelled to Change That State
by Forces Impressed Upon It.”
aka “The Law of Inertia”
Forces and the Laws of Motion
 Inertia


The Tendency of an Object to Resist a
Change in Movement
Mass is a Measurement of Inertia
 Equilibrium

Net Force = 0
Forces and the Laws of Motion
 Force (F)




Energy exerted on an object
Possesses magnitude and direction
May be in contact with an object
May be distant from an object.
Newton’s Laws of Motion
Forces and the Laws of Motion
 Newton’s Second Law of Motion

“The Acceleration of a Body is Directly
Proportional to the Net Force Acting on the
Body and Inversely Proportional to the Mass
of the Body.”
Forces and the Laws of Motion
 Newton’s Second Law of Motion


The Greater the Force on the Object, the
Greater the Acceleration of the Object
The Greater the Mass of the Object, the
Weaker the Acceleration of the Object
Forces and the Laws of Motion
 Newton’s Second Law of Motion

Force is Related to…
• The Mass of the Object (m)
• The Acceleration of the Object (a)
F  ma
Forces and the Laws of Motion
 Measurement of Force

SI Units
• Mass - Kilogram (kg)
• Distance - Meter (m)
• Time - Second (s)
Forces and the Laws of Motion
 Measurement of Force

The Newton (N)
• One Newton can accelerate 1 kg to 1 m/s2
• N = 1 kg*m/s2
• (F=ma)
Newton’s Laws of Motion
Forces and the Laws of Motion
 Interaction Forces

Newton’s Third Law of Motion
• “When an Object Exerts a Force On a Second
Object, That Second Object Exerts an Equal and
Opposite Force On the First Object.”
Forces and the Laws of Motion
 Interaction Forces

Interaction Pair
• “Force Pair”
– Non-Force-Pairs?
• Two Forces in Opposite Directions with Equal
Magnitude
• FA ON B = -FB ON A
Forces and the Laws of Motion
 Interaction Forces

Fundamental Forces
• Gravity
• Electromagnetism
• Strong Nuclear Force
• Weak Nuclear Force
Forces and the Laws of Motion
 Combining Forces




Forces are Vectors
They May Strengthen Each Other
They May Cancel All or Part of Each Other
They Are Cumulative
Forces and the Laws of Motion
 Example Problem

Two horizontal forces of 225N and 165N are
exerted in the same direction on a crate.
What is the net force on the crate?
Forces and the Laws of Motion
 Solution
225N
165N
Forces and the Laws of Motion
 Solution
390N
Forces and the Laws of Motion
 Example Problem

If the same two horizontal forces of 225N
and 165N are exerted in opposite directions
on the crate. What is the net force on the
crate?
Forces and the Laws of Motion
 Solution
225N
165N
Forces and the Laws of Motion
 Solution
60N
Forces and the Laws of Motion
 Example Problem

If you hold a 1.0 lb object using 6.5N of
upward force, what is the net force on the
object? (1 lb = 0.454kg)
Forces and the Laws of Motion
 Solution

If you hold a 1.0 lb object using 6.5N of upward force,
what is the net force on the object? (1 lb = 0.454kg)
Fdown  mg  (0.454kg)(9.8m / s)  4.5N
Fup  Fdown  6.5 N  (4.5 N )  2.0 N
Forces and the Laws of Motion
 Problem

On Earth, a scale shows that Ben weighs
712N. What is Ben’s mass?
Forces and the Laws of Motion
 Solution


F = 712N
g = 9.8m/s2
F
712 N
m 
 72.7kg(160lb )
2
g 9.8m / s
Forces and the Laws of Motion
 Problem

The acceleration of gravity on the moon is
1.60m/s2. What would the scale indicate that
Ben weighs if he were on the moon and his
mass is the same as the previous question?
(72.7kg)
Forces and the Laws of Motion
 Solution


m = 72.7kg
a = 1.60m/s2
F  ma  (72.7kg)(1.60m / s )  116.3N
2
Forces and the Laws of Motion
 Problem

A 2.0kg mass (ma) and a 3.0kg mass (mb) are
attached to a cord that passes over a pulley.
What is the acceleration of the smaller mass?
Forces and the Laws of Motion
 Solution



ma = 2.0kg
mb = 3.0kg
g = 9.8m/s2
ma a  mb g  mb a  ma g
(ma  mb )a  (mb  ma ) g
 mb  ma 
 3.0kg  2.0kg 
 g  
9.8m / s 2  2.0m / s 2
a  
 2.0kg  3.0kg 
 ma  mb 
Forces and the Laws of Motion
 Problem

The maximum force a grocery sack can
stand and not rip is 250N. If Elizabeth loads
20.0kg of groceries into the sack and lifts the
sack from the floor to the table with an
acceleration of 6.0m/s2, will the sack hold?
Forces and the Laws of Motion
 Solution



a = 6.0m/s2
m = 20.0kg
Fmax = 250N
ma  Fnet  Fapplied  Fg
Fapplied  ma  mg  m(a  g )
Fapplied  (20.0kg)[(6.0m / s )  (9.8m / s )]  300N
2
2
Forces and the Laws of Motion
 Problem

The instruments attached to a weather
balloon have a mass of 5.0kg. The balloon is
released and exerts an upward force of 98N
on the instruments. What is the acceleration
of the balloon and the instruments?
Forces and the Laws of Motion
 Solution




m = 5.0kg
Fu = 98N
g = 9.8m/s2
a=?
ma  Fnet  Fapplied  Fg
ma  98 N  (49 N )  49 N up
 49 N
2
a
 9.8m / s up
5.0kg
Forces and the Laws of Motion
 Problem

The instruments attached to a weather
balloon have a mass of 5.0kg. The balloon is
released and exerts an upward force of 98N
on the instruments. After the balloon has
accelerated for 10.0s, the instruments are
released. What is the velocity of the
instruments at the moment of their release?
Forces and the Laws of Motion
 Solution




m = 5.0kg
Fu = 98N
g = 9.8m/s2 down
a = 9.8m/s2 up
v  at  (9.8m / s )(10.0s)  98m / s up
2
Forces and the Laws of Motion
 Problem

The instruments attached to a weather
balloon have a mass of 5.0kg. The balloon is
released and exerts an upward force of 98N
on the instruments. After the balloon has
accelerated for 10.0s, the instruments are
released. What net force acts on the
instruments after their release?
Forces and the Laws of Motion
 Solution





m = 5.0kg
Fu = 98N
g = 9.8m/s2 down
a = 9.8m/s2 up
v0 = +98m/s
Fg  mg  W  49 N
Forces and the Laws of Motion
 Problem

The instruments attached to a weather
balloon have a mass of 5.0kg. The balloon is
released and exerts an upward force of 98N
on the instruments. After the balloon has
accelerated for 10.0s, the instruments are
released. When does the instrument’s
velocity first become downward?
Forces and the Laws of Motion
 Solution





m = 5.0kg
Fu = 98N
g = 9.8m/s2 down
a = 9.8m/s2 up
v0 = +98m/s
v  v0  gt
v0
v0  (98m / s)
t 
 10.0s after release
2
g (9.8m / s )
Forces and the Laws of Motion
 Problem

Donnie and Candice are skating. Donnie
pushes Candice, whose mass is 40.0-kg, with
a force of 5.0 N. What is Candice’s resulting
acceleration?
Forces and the Laws of Motion
 Solution


mj = 40.0kg
F = 5.0N
F  ma
F
a
m
F
5.0 N
2
a 
 0.12m / s
m 40.0kg
Forces and the Laws of Motion
 Problem

Your new motorcycle weighs 2450 N. What is
its mass in kilograms?
Forces and the Laws of Motion
 Solution

F = 2450N
Fg  mg
F
m
g
F
2450 N
m 

250
kg
2
g 9.8m / s
Forces and the Laws of Motion
 Problem

A 873-kg (1930-lb) dragster, starting from
rest, attains a speed of 26.3m/s (58.9 mph) in
0.59 s. What is the magnitude of the average
net force on the dragster during this time?
Forces and the Laws of Motion
 Solution




m = 873kg
v f  vi 26.3m / s  0m / s
vf = 26.3m/s a 

 44.58m / s 2
Dt
0.59s
vi = 0m/s
Dt = 0.59s
F  ma  (873kg)(44.58m / s )  3.9 x10 N
2
4
Forces and the Laws of Motion
 Problem

A 873-kg (1930-lb) dragster, starting from
rest, attains a speed of 26.3m/s (58.9 mph) in
0.59 s. Assume that the driver has a mass of
68kg. What horizontal force does the seat
exert on the driver?
Forces and the Laws of Motion
 Solution





mc = 873kg
md = 68kg
vf = 26.3m/s
vi = 0m/s
Dt = 0.59s
a
v f  vi
Dt

26.3m / s  0m / s
 44.58m / s 2
0.59s
F  ma  (68kg)(44.58m / s )  3.1x10 N
2
3
Forces and the Laws of Motion
 Problem

A raindrop, with mass 2.45 mg, falls to the
ground. As it is falling, what magnitude of
force does it exert on Earth?
Forces and the Laws of Motion
 Solution

m = 2.45x10-6 kg
6
5
Fg  mg  (2.45x10 kg)(9.8m / s )  2.4 x10 N
2
Forces and the Laws of Motion
 Problem

Male lions and human sprinters can both
accelerate at about 10.0 m/s2. If a typical
lion has a mass of 170 kg and a typical
sprinter’s mass is 75 kg, what is the
difference in the force exerted on the ground
during a race between these two species?
Forces and the Laws of Motion
 Solution



ml = 170 kg
mh = 75kg
a = 10.0m/s2
Flion  Fhuman
mliona  mhumana
(170kg)(10.0m / s 2 )  (75kg)(10.0m / s 2 )  950 N
Forces and the Laws of Motion
 Homework

Problems
• Pages 151 – 152
– 10 (1210N 62o above the 1520N Force)
– 12 (4N, 3N)
– 23 (a, 6.00s b, 72.0m c, 63.6m/s)
– 25 (a, 2.3m b, 1.8m)
Forces and the Laws of Motion
 Types of Forces

Weight (Fg)
• Long Range Force Exerted by Gravity on the
Object
• Toward the Center of the Earth

Normal (FN)
• Exerted by Surface on the Object
• Perpendicular to Surface
Forces and the Laws of Motion
 The Normal Force


The normal force is the perpendicular
contact force exerted by a surface on another
object.
The normal force is important when
calculating resistance.
Forces and the Laws of Motion
 Types of Forces

Friction (Ff)
• Opposes Horizontal Motion
• Parallel to Surface

Tension (FT)
• Pull or Push Exerted by Secondary Object
• Parallel to Secondary Object
Forces and the Laws of Motion
 Friction

Drag (Fdrag)
• Friction of Medium
on an Object

Terminal Velocity
• Fdrag = Fg
Forces and the Laws of Motion
 Forces of Ropes and Strings


The force exerted by a string or rope is
called tension.
At any point in a rope, the tension forces are
pulling equally in both directions.
Forces and the Laws of Motion
 Mass & Weight

On Earth
• Fg = mg
– g is the acceleration of Gravity (9.8
m/s2)
• Mass = Weight?
Forces and the Laws of Motion
 Apparent Weight

Fg +/- Any Other Forces
 Weightlessness
Apparent Weight = 0
 Elevator Example...

Forces and the Laws of Motion
 Static and Kinetic Friction

Kinetic friction
Exerted on one surface by another when the
two surfaces rub against each other because
one or both of them are moving.

Static friction
Exerted on one surface by another when there
is no motion between the two surfaces.
Forces and the Laws of Motion
 Static and Kinetic Friction


You might push harder and harder, as shown in
the figure below, but if the couch still does not
move, the force of friction must be getting larger.
This is because the static friction force acts in
response to other forces.
Forces and the Laws of Motion
 Static and Kinetic Friction



Push hard enough and the couch will begin to
move.
There is a limit to how large the static friction
force can be.
Once your force is greater than this maximum
static friction, the couch begins moving and kinetic
friction begins to act on it instead of static friction.
Forces and the Laws of Motion
 Static and Kinetic Friction


Frictional force depends on the materials that the
surfaces are made of.
The normal force between the two objects also
matters. The harder one object is pushed against
the other, the greater the force of friction that
results.
Forces and the Laws of Motion
 Static and Kinetic Friction
Forces and the Laws of Motion
 Static and Kinetic Friction

The slope of this line, μk, is called the coefficient of kinetic
friction between the two surfaces and relates the
frictional force to the normal force, as shown below.
Kinetic Friction Force
f k  k FN
Forces and the Laws of Motion
 Static and Kinetic Friction

The static friction force is less than or equal to the
product of the coefficient of the static friction and the
normal force.
Static Friction Force
f s   s FN
Forces and the Laws of Motion
 Static and Kinetic Friction


The forces themselves, fs and FN, are at right
angles to each other.
Although all the listed coefficients are less than
1.0, this does not mean that they must always be
less than 1.0.
Forces and the Laws of Motion

Force…
• Is Not Transferred to the Object
• Is Not Needed to Keep an Object Moving
• Is Not Related to Inertia
Forces and the Laws of Motion
Problem
 If you use a horizontal force of 30.0N to
slide a 12.0kg wooden crate across a floor
at a constant velocity, what is the
coefficient of kinetic friction between the
crate and the floor?
Forces and the Laws of Motion
Solution


fk = 30.0N
m = 12.0kg
f k  k FN
fk
k 
FN
fk
fk
30.0 N
k 


 0.26
2
FN mg (12.0kg)(9.8m / s )
Forces and the Laws of Motion
Problem
 A force of 40.0N accelerates a 5.0kg block
at 6.0 m/s2 along a horizontal surface.
How large is the frictional force?
Forces and the Laws of Motion
Solution



Fapplied = 40.0N
m = 5.0kg
a = 6.0m/s2
Facc  Fapplied  f k
f k  Fapplied  Facc
f k  40.0 N  (5.0kg)(6.0m / s )  10.0 N
2
Forces and the Laws of Motion
Problem
 A force of 40.0N accelerates a 5.0kg block
at 6.0 m/s2 along a horizontal surface.
What is the coefficient of friction?
Forces and the Laws of Motion
f k  k FN
Solution



Fapplied = 40.0N
m = 5.0kg
a = 6.0m/s2
fk
fk
k 

FN mg
10.0 N
k 

0
.
20
2
(5.0kg)(9.8m / s )
Forces and the Laws of Motion
Problem

You are driving a 2500.0kg car at a constant
speed of 14.0m/s along a wet, but straight, level
road. As you approach an intersection, the
traffic light turns red. You slam on the brakes.
The car’s wheels lock, the tires begin skidding,
and the car slides to a halt in a distance of
25.0m. What is the coefficient of kinetic friction
between your tires and the wet road?
Forces and the Laws of Motion
Solution




d = 25.0m
m = 2500.0kg
vi = 14.0m/s
vf = 0m/s
2
f k  k FN  ma
m(v f  vi )
2
k mg 
2
2d
2
vi
(14.0m / s)
k 


0
.
40
2
2dg 2(25.0m)(9.8m / s )
Forces and the Laws of Motion
Problem
 A street lamp weighs 150N. It is
supported by two wires that form an
angle of 120.0° with each other. The
tensions in the wires are equal. What is
the tension in each wire supporting the
street lamp?
Forces and the Laws of Motion
Solution


W = 150.0N
q = 120o ?
Fg  2T sin q
T
Fg
2 sin q
150 N
T

150
N
o
2(sin 30.0 )
Forces and the Laws of Motion
 Homework

Problems
• Pages 153 – 154
– 38 (0.436)
– 41 (1.4m/s2 down the aisle)
– 42 (1.0m/s2)
– 43 (15.9N)