Download Newton`s Second Law

The Laws of Motion
Physics 2053
Lecture Notes
The Laws of Motion
The Laws of Motion
Topics
4-01
4-02
4-03
4-04
4-05
4-06
Force
Newton’s First Law
Newton’s Second Law
Newton’s Third Law
Applications of Newton’s Laws
Forces of Friction
The Laws of Motion
Force
Types
Range
Size
Gravitational
Unlimited
100
Electromagnetic
Unlimited
106
Weak Nuclear
 10-12 m
1020
Strong Nuclear
 10-15 m
1035
The Laws of Motion
Newton’s First Law
A force is a push or pull.
An object at rest needs a force
to get it moving; a moving
object needs a force to change
its velocity.
The magnitude of a
force can be measured
using a spring scale.
The Laws of Motion
Newton’s First Law
Newton’s first law is often called the law of inertia.
Every object continues in its state of rest,
or of uniform velocity in a straight line,
as long as no net force acts on it.
If no external force acts
F  0
The Laws of Motion
Newton’s First Law
When you sit on a chair, the resultant force on you is
A) zero.
B) up.
C) down.
D) depending on your weight.
F  0
The Laws of Motion
Newton’s Second Law
Newton’s second law is the relation between acceleration
and force. Acceleration is proportional to force and
inversely proportional to mass.
 Fx  ma x
 F  ma
 Fy  ma y
 Fz  ma z
Force is a vector, so SF = ma is true along each coordinate axis.
Units of Force
System Mass
SI
kg
British
slug
Acceleration
Force
m/s2
N = kg m/s2
ft/s2
lb = slug ft/s2
The Laws of Motion
Newton’s Second Law
A man stands on a scale
inside a stationary elevator.
Forces acting on the man
F  0
N - mg  0
N
N  mg
Reading
on scale
mg
The Laws of Motion
Newton’s Second Law
When Moving Upward With
Constant Velocity
Forces acting on the man
 F  ma
v
a0
N - mg  m0
N
N  mg
Reading
on scale
mg
The Laws of Motion
Newton’s Second Law
When Moving Upward With
Constant Acceleration
Forces acting on the man
 F  ma
N - mg  ma
a
N
N  mg  ma
N  mg  a 
Reading
on scale
mg
The Laws of Motion
Newton’s Second Law
When Moving Downward With
Constant Acceleration
Forces acting on the man
 F  ma
mg - N  ma
a
N
N  mg - ma
N  mg - a 
Reading
on scale
mg
The Laws of Motion
Newton’s Second Law
A constant net force acts on an object. Describe the
motion of the object.
A) constant acceleration
B) constant speed
C) constant velocity
 F  ma
D) increasing acceleration
The Laws of Motion
Newton’s Second Law
A constant force F acts on a block of mass m. which is initially
at rest. Find the velocity of the block after time Dt.
F
vo = 0
Dt = 5 s
m
F  ma
v =?
F = 20 N
m = 5 kg
Dv
a
Dt
 v - vo 
 Dv 
F  m   m 

 Dt 
 Δt 
FDt
v
m
20 N5 s 

5 kg
 20 m/s
The Laws of Motion
Newton’s Second Law (Problem)
What average force is required to stop an 1100 kg car in 8.0 s
if the car is travelling at 95 km/h? Fm, t, v , v 
o
Newton’s 2nd Law
 F  ma
vo
v  vo  at
F
v - vo
a
t
 v - vo 
F  m

 t 
 0 - 95 km/h  1000 m  1 h 
F  1100 kg



8.0 s

 1 km  3600 s 
F  -3.6 x 103 N
The Laws of Motion
Newton’s Second Law
A net force F accelerates a mass m with an acceleration a.
If the same net force is applied to mass 2m, then the
acceleration will be
A) 4a.
B) 2a.
C) a/2
D) a/4
 F  ma
F

a
m
The Laws of Motion
Newton’s Second Law (Problem)
The cable supporting a 2,125 kg elevator has a maximum
strength of 21,750 N. What maximum upward
acceleration can it give the elevator without breaking?
am, T, g 
Newton’s 2nd Law
Tmax
 F  ma
Tmax - mg  mamax
a
m

Tmax - mg 21,750 N - 2125 kg 9.8 m/s
amax 

m
2125 kg
2

mg
amax  0.44 m/s 2
The Laws of Motion
Newton’s Second Law (Problem)
How much tension must a rope withstand if it is used to
accelerate a 1200 kg car vertically upward at 0.80 m/s2.
Tm, a, g 
Newton’s 2nd Law
 F  ma
T
T - mg  ma
a
mg
T  ma  mg  ma  g 

T  1200 kg 0.80 m/s 2  9.8 m/s 2

 1.3 x 104 N
The Laws of Motion
Newton’s Second Law
Gravitational Force:
Gravitational Force is the mutual force of attraction between
any two objects in the Universe.
m
F
F
R
FG
Mm
R2
M
Universal Gravitational Constant
2
Nm
G  6.67 x 10 -11
kg 2
The Laws of Motion
Newton’s Second Law
The gravitational force between two objects is
proportional to
A) the distance between the two objects.
B) the square of the distance between the two objects.
C) the product of each objects mass.
D) the square of the product of each objects mass.
FG
Mm
R2
The Laws of Motion
Newton’s Second Law
Two objects attract each other gravitationally. If the
distance between their centers is cut in half, the
gravitational force
A) is cut to one fourth.
B) is cut in half.
C) doubles.
D) quadruples
FG
Mm
R2
The Laws of Motion
Newton’s Second Law
Two objects, with masses m1 and m2, are originally a
distance r apart. The magnitude of the gravitational force
between them is F. The masses are changed to 2m1 and
2m2, and the distance is changed to 4r. What is the
magnitude of the new gravitational force?
A) F/16
B) F/4
C) 16F
FG
Mm
R2
D) 4F
The Laws of Motion
Newton’s Second Law
“g” in terms of G
g
F
m
FG
Mm
R2
F
F  mg
R
Mm
G
 mg
2
R
M
g
GM
R2
The Laws of Motion
Newton’s Second Law
Mass is the measure of inertia of an object. In the SI system,
mass is measured in kilograms.
Mass is not weight:
Mass is a property of an object. Weight is the force exerted on
that object by gravity.
If you go to the moon, whose gravitational acceleration is
about 1/6 g, you will weigh much less. Your mass, however,
will be the same.
Gravitational mass
mg
mg = mi
Inertial mass
mi
The Laws of Motion
Newton’s Second Law
Weight is the force exerted on an object by gravity. Close to
the surface of the Earth, where the gravitational force is
nearly constant, the weight is:
m
g  9.8 m/s 2
Weight = mg
The Laws of Motion
Newton’s Second Law
Mass and weight
A) both measure the same thing.
B) are exactly equal.
C) are two different quantities.
D) are both measured in kilograms.
The Laws of Motion
Newton’s Second Law
A stone is thrown straight up. At the top of its path, the
net force acting on it is
A) equal to its weight.
B) greater than its weight.
C) greater than zero, but less than its weight.
D) instantaneously equal to zero.
The Laws of Motion
Newton’s Third Law
Whenever one object exerts a force on a second object,
the second exerts an equal force in the opposite direction
on the first.
F1
F1  -F2
F2
Action/Reaction Forces
The Laws of Motion
Newton’s Third Law
An object at rest must have no net
force on it. If it is sitting on a table,
the object exerts a downward force
mg on the surface of the table.
m
N
mg
The surface of the table exerts
an upward force on the block,
called the normal force. It is
exactly as large as needed to
balance the force from the
object.
 Fy  0
N - mg  0
N  mg
The Laws of Motion
Newton’s Third Law
If an upward force F is applied
to the block, the magnitude of
the normal force is
 Fy  0
F
N
mg
N - mg  F  0
N  mg - F
The Laws of Motion
Newton’s Third Law
If a downward force F is applied
to the block, the magnitude of
the normal force is
 Fy  0
F
m
N
mg
N - mg - F  0
N  mg  F
The Laws of Motion
Newton’s Third Law
Action-reaction forces
A) sometimes act on the same object.
B) always act on the same object.
C) may be at right angles.
D) always act on different objects.
The Laws of Motion
Newton’s Third Law
A 40,000 kg truck collides with a 1500 lb car and causes
a lot of damage to the car.
A) the force on the truck is greater then the force on the car.
B) the force on the truck is equal to the force on the car.
C) the force on the truck is smaller than the force on the car.
D) the truck did not slow down during the collision.
The Laws of Motion
Newton’s Third Law
A golf club hits a golf ball with a force of 2,400 N.
The force the golf ball exerts on the club is
A) slightly less than 2400 N.
B) exactly 2400 N.
C) slightly more than 2400 N.
D) close to 0 N.
The Laws of Motion
Applications of Newton’s Laws
A block of mass m moving with a speed vo is brought to rest
by a constant force F. Find the distance the block moves.
vo = 20 m/s
v=0
F
m
F = -10 N
m = 5 kg
Dx
v 2  vo2  2aDx
Newton’s 2nd Law
F  ma
 v 2 - v o2 

F  m
 2 Dx 



m v 2 - v o2
Dx 
2F

v 2 - v o2
a
2 Dx

5 kg 0 m/s 2 - 20 m/s 2

2- 10 N 

 100 m
The Laws of Motion
Solving Problems with Newton’s Laws
 Fy  0
Find the acceleration of
the two block system
N - m1g  0  N  m1g
 Fx  m1a
Forces on m1
N
T
m1
T  m1a
 F  ma
m 2g - T  m 2a
T
m1g
m2
m2g
Forces on m2
T  m 2g - m 2a
a
m1a  m 2g - m 2a
m 2g
a
m1  m 2
The Laws of Motion
Solving Problems with Newton’s Laws
Find the acceleration of
the two block system
Mass 1
Mass 2
 F  m1a
 F  m 2a
T - m1g  m1a
m 2g - T  m 2a
T  m1a  m1g
T  m 2g - m 2a
T
T
m2
m1
m1a  m1g  m 2g - m 2a

m 2 - m1 g
a
m1  m 2
m1g
m2g
The Laws of Motion
 Fy  0
Solving Problems with Newton’s Laws
y
N
N - mg cosq   0
 Fx  ma
mg sinq   ma
N  mg cosq 
mg sin(q)
mg cos(q)
q
a  g sinq 
x
mg
q
The Laws of Motion
Problem
A pair of fuzzy dice is hanging by a string from your
rear-view mirror. While you are accelerating from a
stoplight to 24 m/s in 6.0 s, what angle does the string
make with the vertical?
 Fx  ma
FT sin q  ma
 Fy  0
FT cos q - mg  0
FT cos q  mg
a
FT sin q  ma
 tan q 
FT cos q  mg
g
q
q
The acceleration of the dice
v  vo  at
24 m/s - 0
v - vo
 4.0 m/s 2

a
6.0 s
t
Newton’s 2nd Law
FT
a
mg
-1  a 
q  tan  
g
2

4
.
0
m/s
o
-1 


22
.
2
q  tan
 9.8 m/s 2 


The Laws of Motion
Solving Problems with Newton’s Laws
Three mass system - find acceleration
m
T1
2m
T2
F
3m
 F  ma
F  m  2m  3m  a  6m  a
F
a
6m
The Laws of Motion
Solving Problems with Newton’s Laws
Three mass system - find T2
m
T1
2m
T2
F
3m
F
a
6m
 F  ma
F - T2  3m  a
T2  F - 3ma
 F 
T2  F - 3m

 6m 
 T2 
F
2
The Laws of Motion
Solving Problems with Newton’s Laws
Three mass system - find T1
T1
m
T2
2m
F
F
a
6m
3m
 F  ma
T1  m  a
 F 
T1  m

 6m 
F

6
The Laws of Motion
Solving Problems with Newton’s Laws
Or
Three mass system - find T1
m
T1
T2
2m
F
3m
a
F
6m
T2 
F
2
 F  ma
T2 - T1  2m  a
T1  T2 - 2ma
F
 F 
T1  - 2m

2
 6m 
 T1 
F
6
The Laws of Motion
Forces of Friction
You are standing in a moving bus, facing forward, and
you suddenly fall forward. You can imply from this that
the bus's
A) velocity decreased.
B) velocity increased.
C) speed remained the same, but it's turning to the right.
D) speed remained the same, but it's turning to the left.
The Laws of Motion
Forces of Friction
Friction:
Force of Static Friction
v0
N
F
fs
F  0
m
mg
fs   s N
fs  F
fs(max)   s N
The Laws of Motion
Forces of Friction
Friction:
Force of Kinetic Friction
v
N
F
fk
m
mg
fk   k N
The Laws of Motion
Forces of Friction
Coefficients of Friction
s
Steel on steel
0.74
Aluminum on steel
0.61
Copper on steel
0.53
Rubber on concrete
1.0
Wood on wood
0.25-0.5
Glass on glass
0.94
Waxed wood on wet snow 0.14
Waxed wood on dry snow -----Metal on metal (lubricated) 0.15
Ice on ice
0.1
Teflon on Teflon
0.04
k
0.57
0.47
0.36
0.8
0.2
0.4
0.1
0.04
0.06
0.03
0.04
The Laws of Motion
Forces of Friction (Problem)
Suppose that you are standing on a train accelerating
at 2.0 m/s2. What minimum coefficient of static friction
must exist between your feet and the floor if you are not
to slide?
 Fy  0
Frictional force
nd
Newton’s 2 Law
N - mg  0
Ff   s N   smg
 Fx  ma
N  mg
Ff  ma
a
smg  ma
N
2
a
2.0 m/s
s 
 0.20

Ff
2
g
9.8 m/s
mg
The Laws of Motion
Forces of Friction
The force that keeps you from sliding on an icy sidewalk is
A) weight.
B) kinetic friction.
C) static friction.
D) normal force.
The Laws of Motion
Forces of Friction
Pulling a block with constant speed
Pulling a block
N
k
f
F
q
m
mg
The normal force
y
N
q
f
 Fy  0
F
x
mg
N  F sinq  - mg  0
N  mg - F sinq 
The Laws of Motion
Forces of Friction
Pulling a block with constant speed
N
k
f
F
q
m
y
N
F
q
f
x
mg
mg
 Fx  0
The frictional force
F cosq  - f  0
f  F cosq 
The Laws of Motion
Forces of Friction
Pulling a block with constant speed
N
k
f
q
m
y
F
F
q
f
x
mg
mg
The coefficient of friction
N
f  k N
f
μk 
N
The Laws of Motion
Forces of Friction (Problem)
The coefficient of static friction between hard rubber and
normal street pavement is about 0.80. On how steep a hill
(maximum angle) can you leave a car parked?F
N
Ff
 Fy  0
 Fx  0
Ff - Fg sin q  0 FN - Fg cos q  0
Ff  Fg sin q
FN  Fg cos q
q
q
Ff  s FN
Fg sin q   s Fg cos q
tan q  s  q  tan -1 s   tan -1 0.8
Fg
 39o
The Laws of Motion
Forces of Friction (Problem)
Drag-race tires in contact with an asphalt surface have a
very high coefficient of static friction. Assuming a constant
acceleration and no slipping of tires, estimate the coefficient
of static friction needed for a drag racer to cover 1.0 km
in 12 s, starting from rest.
FN
Newton’s 2nd Law
at 2
Ff
Dx  v o t 
 Fx  ma
2
2x
Ff  ma
Fg
a 2
Ff  s FN
t
smg  ma
Ff   smg
21000 m 
2x 
a
 1.4
 2
s 
2
2
9.8 m/s 12 s 
g
gt


The Laws of Motion
Forces of Friction
A brick and a feather fall to the earth at their respective
terminal velocities. Which object experiences the greater
force of air friction?
A) the feather
B) the brick
C) Neither, both experience the same amount of air friction.
D) It cannot be determined because there
is not enough information given.
The Laws of Motion
Newton’s First Law
Which of Newton's laws best explains why motorists
should buckle-up?
A) the first law
B) the second law
C) the third law
D) the law of gravitation
The Laws of Motion