The Laws of Motion Physics 2053 Lecture Notes The Laws of Motion.

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The Laws of Motion Physics 2053 Lecture Notes The Laws of Motion

Transcript of The Laws of Motion Physics 2053 Lecture Notes The Laws of Motion.

The Laws of Motion

Physics 2053Lecture 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

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

“g” in terms of G

R

m

M

F

F

2R

MmGF

mgF

mgR

MmG

2

2R

GMg

g

Newton’s Second Law

The Laws of Motion

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

Friction:

Force of Kinetic Friction

Ffk m

mgNf kk

vN

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