The Laws of Motion Physics 2053 Lecture Notes The Laws of Motion.
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Transcript of The Laws of Motion Physics 2053 Lecture Notes The Laws of Motion.
The Laws of Motion
4-01 Force
4-02 Newton’s First Law
4-03 Newton’s Second Law
4-04 Newton’s Third Law
4-05 Applications of Newton’s Laws
4-06 Forces of Friction
Topics
The Laws of Motion
Types Range
Gravitational Unlimited
Electromagnetic Unlimited
Weak Nuclear 1012 m
Strong Nuclear 1015 m
Size
100
106
1020
1035
Force
The Laws of Motion
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.
Newton’s First Law
The Laws of Motion
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
0F
Newton’s First Law
Newton’s first law is often called the law of inertia.
The Laws of Motion
When you sit on a chair, the resultant force on you is
A) zero.
B) up.
C) down.
D) depending on your weight.
Newton’s First Law
The Laws of Motion
0F
maF
zz
yy
xx
maF
maF
maF
Units of ForceSystem Mass Acceleration Force SI kg m/s2 N = kg m/s2
British slug ft/s2 lb = slug ft/s2
Newton’s second law is the relation between acceleration and force. Acceleration is proportional to force and inversely proportional to mass.
Force is a vector, so F = ma is true along each coordinate axis.
Newton’s Second Law
The Laws of Motion
A man stands on a scaleinside a stationary elevator.
N
mg
Forces acting on the man
0 F
0 mgN
mgN
Reading on scale
Newton’s Second Law
The Laws of Motion
N
mg
v
When Moving Upward With Constant Velocity
am F
0m mgN
mgN
Forces acting on the man
Reading on scale
Newton’s Second Law
0a
The Laws of Motion
N
mg
a
When Moving Upward With Constant Acceleration
am F
am mgN
mamgN
agmN
Forces acting on the man
Reading on scale
Newton’s Second Law
The Laws of Motion
N
mg
a
When Moving Downward With Constant Acceleration
am F
am Nmg
mamgN
agmN
Forces acting on the man
Reading on scale
Newton’s Second Law
The Laws of Motion
A constant net force acts on an object. Describe the motion of the object.
A) constant acceleration
B) constant speed
C) constant velocity
D) increasing acceleration
Newton’s Second Law
The Laws of Motion
maF
F vo = 0
m
t = 5 s v = ?
maF tv
a
tv
mF
tΔ
vvm o
mtF
v
kg 5s 5N 20
F = 20 N
m = 5 kg
m/s 20
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 t.
The Laws of Motion
What average force is required to stop an 1100 kg car in 8.0 s if the car is travelling at 95 km/h?
ov
FNewton’s 2nd Law
maF
t
vvmF o
s 3600
h 1km 1
m 1000s 8.0
km/h 590kg 1100F
N 10x 6.3F 3
Newton’s Second Law (Problem)
v,v t, m,F o
atvv o
t
vva o
The Laws of Motion
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
Newton’s Second Law
The Laws of Motion
m
Fa
maF
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?
Newton’s Second Law (Problem)
Tmax
mg
m
Newton’s 2nd Law
maF
maxmax mamgT
m
mgTa max
max
gk 2125
m/s 8.9 gk 2125N 750,21 2
2max m/s 44.0a
g T, m,a
a
The Laws of Motion
How much tension must a rope withstand if it is used toaccelerate a 1200 kg car vertically upward at 0.80 m/s2.
mg
a
Newton’s 2nd Law
maF
mamgT
22 m/s 8.9m/s 0.80 kg 1200T
mgmaT gam
N 10x 3.1 4
Newton’s Second Law (Problem)
g a, m,T
T
The Laws of Motion
Gravitational Force:
Newton’s Second Law
Gravitational Force is the mutual force of attraction between any two objects in the Universe.
R
m
M
F
F2R
MmGF
Universal Gravitational Constant
2
211
kg
Nm 10 x 67.6G
The Laws of Motion
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.
Newton’s Second Law
The Laws of Motion
2R
MmGF
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
Newton’s Second Law
The Laws of Motion
2R
MmGF
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
D) 4F
Newton’s Second Law
The Laws of Motion
2R
MmGF
Mass is the measure of inertia of an object. In the SI system, mass is measured in kilograms.
Gravitational mass mg
mg = mi
Inertial mass mi
Newton’s Second Law
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.
The Laws of Motion
m
Weight = mg
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:
2m/s 8.9g
Newton’s Second Law
The Laws of Motion
Mass and weight
A) both measure the same thing.
B) are exactly equal.
C) are two different quantities.
D) are both measured in kilograms.
Newton’s Second Law
The Laws of Motion
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.
Newton’s Second Law
The Laws of Motion
F1
F2
21 FF
Action/Reaction Forces
Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first.
Newton’s Third Law
The Laws of Motion
N
mg
0 Fy
0mgN
mgN
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.
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.
m
Newton’s Third Law
The Laws of Motion
0FmgN
0 Fy
If an upward force F is applied to the block, the magnitude of the normal force is
FmgN
N
mg
F
Newton’s Third Law
The Laws of Motion
N
mg
0FmgN
F
0 Fy
If a downward force F is applied to the block, the magnitude of the normal force is m
FmgN
Newton’s Third Law
The Laws of Motion
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.
Newton’s Third Law
The Laws of Motion
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.
Newton’s Third Law
The Laws of Motion
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.
Newton’s Third Law
The Laws of Motion
F
vo = 20 m/s
mv = 0
maF xa2vv 2
o2
F = 10 Nm = 5 kg
x
F2
vvmx
2o
2
N 102
m/s 20m/s 0kg 5 22
m 100
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.
Newton’s 2nd Law
x2
vva
2o
2
x2
vvmF
2o
2
The Laws of Motion
m1
m2
N
m1g
T
T
m2g
0Fy
amF 1x
amT 1
maF
amTgm 22
amgmT 22
amgmam 221
21
2mmgm
a
Forces on m1
Forces on m2
0gmN 1 gmN 1
Find the acceleration ofthe two block system
Solving Problems with Newton’s Laws
a
The Laws of Motion
m1
m2T
T
m1gm2g
amgmT 11
amF 1
gmamT 11
amTgm 22
amF 2
amgmT 22
Mass 1 Mass 2
amgmgmam 2211
21
12mm
gmma
Find the acceleration ofthe two block system
Solving Problems with Newton’s Laws
The Laws of Motion
N
mg cos()
mg sin()
0Fy
y
x
amFx
mg
0cos mgN
sin ga
amsin mg cos mgN
Solving Problems with Newton’s Laws
The Laws of Motion
A pair of fuzzy dice is hanging by a string from your rear-view mirror. While you are accelerating from astoplight to 24 m/s in 6.0 s, what angle does the string make with the vertical?
Problem
mg
TF
a
The acceleration of the diceatvv o
t
vva o
s 0.6
0m/s 24 2m/s 0.4
Newton’s 2nd Law
maFx
0Fy
masinFT 0mgcosFT
mgcosFT
masinFT mgcosFT g
atan
ga
tan 1
2
21
m/s 8.9
m/s 0.4tan
o2.22
The Laws of Motion
T1 T2 Fm 3m2m
maF
a m3m2mF
m6F
a
Three mass system - find acceleration
a m6
Solving Problems with Newton’s Laws
The Laws of Motion
T1 T2 Fm 3m2m
maF
a m3TF 2
m6F
a
ma3FT2
m6F
m3FT2 2F
T2
Three mass system - find T2
Solving Problems with Newton’s Laws
The Laws of Motion
T1 T2 Fm 3m2m
maF
a mT1
m6F
mT1 6F
m6F
a
Three mass system - find T1
Solving Problems with Newton’s Laws
The Laws of Motion
T1T2 F
m 3m2m
maF
a m2TT 12
ma2TT 21
m6F
m22F
T1
2F
T2
6F
T1
m6F
a
Three mass system - find T1Or
Solving Problems with Newton’s Laws
The Laws of Motion
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.
Forces of Friction
The Laws of Motion
Friction:Force of Static Friction
F
fs m
mg 0F
Ffs
Nf ss
Nf ss(max)
0v N
Forces of Friction
The Laws of Motion
Steel on steel 0.74 0.57Aluminum on steel 0.61 0.47Copper on steel 0.53 0.36Rubber on concrete 1.0 0.8Wood on wood 0.25-0.5 0.2Glass on glass 0.94 0.4Waxed wood on wet snow 0.14 0.1Waxed wood on dry snow ------ 0.04Metal on metal (lubricated) 0.15 0.06Ice on ice 0.1 0.03Teflon on Teflon 0.04 0.04
Coefficients of Friction
k s
Forces of Friction
The Laws of Motion
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?
N
mg
fF
a
Newton’s 2nd Law
maFxmaFf
0Fy
0mgN mgN
Frictional force
NF sf
mamgs
ga
s 2
2
m/s .89
m/s 0.2 20.0
Forces of Friction (Problem)
mgs
The Laws of Motion
The force that keeps you from sliding on an icy sidewalk is
A) weight.
B) kinetic friction.
C) static friction.
D) normal force.
Forces of Friction
The Laws of Motion
m
mg
N
F
0F y
f
N
mg
F
0mgsinFN
sinFmgN
The normal force
x
y
f
k
Pulling a block
Pulling a block with constant speed
Forces of Friction
The Laws of Motion
m
mg
N F
f
0F x
f
N F
mg
0fcosF The frictional force
x
y
cosFf
k
Pulling a block with constant speed
Forces of Friction
The Laws of Motion
m
mg
N F
f
Nf k
f
N F
mg
The coefficient of friction
x
y
k
Nf
μ k
Pulling a block with constant speed
Forces of Friction
The Laws of Motion
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?
fF
gF
NF
0Fx 0sinFF gf
0Fy 0cosFF gN
Nsf FF
sinFF gf cosFF gN
cosFsinF gsg
stan s1tan 8.0tan 1 o39
Forces of Friction (Problem)
The Laws of Motion
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.
fFNewton’s 2nd Law
maFx
gF
NF
Nsf FF maFf
mamgs
ga
s
2at
tvx2
o
2t
x2a
2gt
x2
22 s 12m/s 8.9
m 10002 4.1
Forces of Friction (Problem)
mgF sf
The Laws of Motion
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.
Forces of Friction
The Laws of Motion
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
Newton’s First Law
The Laws of Motion