4. Newton's Laws

4. Newton's Laws

1. The Wrong Question

2. Newton’s 1st & 2nd laws

3. Forces

4. The Force of Gravity

5. Using Newton’s 2nd Law

6. Newton’s 3rd Law

What forces govern the motion of the sailboard?

Ans:

• Human body: contact force

• Wind & water: fluid pressure

• Gravity: action-at-a-distance

4.1. The Wrong Question

Q: Why do things move?

Aristotle (~350BC) : Because some forces are acting on them.

Galileo (~1600) :

Wrong question ( things move until stopped ).

The right question:

Why do moving things change direction?

Newton: Because some forces are acting on them.

If a ball is released here …

… it always rises to its starting height …

… so this ball should roll forever.

4.2. Newton’s 1st & 2nd laws

Net force determines motion.

Newton’s 1st law of motion:

A body at rest or in uniform motion remains so

unless acted on by a nonzero net force.

Fnet 0 v

Fnet = 0 v = const

GOT IT? 4.1.

On a horizontal tabletop is a curved barrier that exerts a force on a ball,

guiding its motion in a circular path.

After the ball leaves the barrier, which of the dashed paths shown does it follow?

Newton’s 2nd Law

Newton’s 2nd law of motion:The rate of change of momentum is equal to the net force.

Momentum: mp v (quantity of motion)

d

d t

pf

For a constant mass: m a

dm

d t v

d m dm

d t d t

vv

dmd t

v

f

Mass, Inertia, & Force

Loaded truck has

greater mass,

more inertia,

less acceleration.

Inertia: resistance to changes in motion.

1st law = law of inertia

known knownf m a unknown unknownm a

unknown known

known unknown

m a

m a Operational

definition of mass

1 N (Newton) force required to give a mass of 1 Kg an acceleration of 1 m / s2.

1 N = 1 Kg m / s2.

1 Dyne = 1 g cm / s2.

Example 4.1. Accelerating Car

A 1200 kg car accelerates constantly in a straight line from rest to 20 m/s in 7.8 s.

a)What is the net force on it?

b)What is the net force on it if it rounds a bend 85 m in radius at v = 20 m/s?

f m a vm

t

20 /1200

7.8

m skg

s

3.1 kN

f m a2v

mr

220 /

120085

m skg

m 5.6 kN

Inertial Reference Frames

Inertial Reference Frames: reference frame in which Newton’s 1st law works.

Examples of non-inertial reference frames:

• Accelerating planes.

• Car rounding a curve.

• Merry-go-round.

• Earth (rotating).

Examples of inertial reference frames:

Newton: Distant “fixed” stars.

Einstein: General relativity.

4.3. Forces

Examples of forces:

• Pushes & pulls.

• Car collided with truck & stopped.

• Moon circles Earth.

• Person sitting on chair.

• Climber on rope.

(action-at-a distance)

(contact force)

(tension force)

Fc = Fg

T = Fg

The Fundamental Forces

The fundamental forces:

• Gravity: large scale phenomena

• Electroweak force

• Electromagnetic force: everyday phenomena

• Weak (nuclear) force

• Strong (nuclear) force 1

1025

1036

1038

4.4. The Force of Gravity

m in f = m a is the inertia mass (same everywhere).

Weight = force of gravity on mass: mw g

For a mass of 65 kg,

weight on Earth = (65 kg) (9.8 m/s2 ) = 640 N.

weight on Moon = (65 kg) (1.6 m/s2 ) = 100 N.

weight in outer-space = (65 kg) (0 m/s2 ) = 0 N.

Caution:

In daily use, weight is often measured in units of mass.

E.g., a person “weights” 65 kg.

Strictly speaking, m in w = m g is the gravitational mass.

Galileo experiment: mI = mG (coincidence)

Einstein: mI = mG (exact: gravity is geometry)

Weightlessness

These astonauts feel “weightless” because

they are in freefall.

Their weight is about 93% of that on surface.

same a

4.5. Using Newton’s 2nd Law

Tactics 4.1. Free-Body Diagram

1.Identify object of interest & forces on it.

2.Object dot.

3.Draw forces on object as vectors at dot.

Example 4.3. Elevator

A 740 kg elevator accelerates upward at 1.1 m/s2.

Find the tension force on the cable (of negligible mass).

g m T F a

y yT m g m a

2 2740 9.8 / 1.1 /yT kg m s m s 8.1 kN

GOT IT? 4.4.

How’s T compared to w = m g if the elevator

a)moves upward starting at rest.

b)decelerates to stop while moving upward.

c)starts moving downward, accelerating from rest.

d)slows to stop while moving downward.

e)moves upward with constant speed.

greater

greater

equal

less

less

Conceptual Example 4.1. At the Equator

When you stand on a scale, the scale reading shows the force it pushes up to support you.If you stand on a scale at Earth’s equator, is the reading greater or less than your weight?

Ans:You’re in uniform circular motion so that the net force on you is centripetal.

2vm a m

R

scaleF W m a W

scaleW F

FscaleW = m g

m a

Parabolic Flight

4.6. Newton’s 3rd Law

Newton’s 3rd law: Action = Reaction

Forces of 3rd law pair act on different objects.

Hence, they don’t cancel each other out.

Horse-Cart dilemma

Example 4.4. Pushing Books

2 books lie on frictionless horizontal surface.

You push with force F on books of mass m1.

which in turn on book of mass m2.

What force does the 2nd book exert on the 1st ?

1 2m m

F

aAcceleration of books:

12 2mF aNet force on 2nd book:

21 12F F

2

1 2

m

m m

F

Force exerted by 2nd book on the 1st

2

1 2

m

m m

F

GOT IT? 4.5

Is the net force on the larger block

(a) greater than 2 N , (b) equal to 2N or (c) less than 2 N ?

Normal force n : contact force acting normal to contact surface.

3rd law also applies to non-contact forces such as gravity.

Not a 3rd law pair

Normal force Fnet 0

Measuring Force

Force can be measured using Newton’s 3rd law.

Ideal spring (Hooke’s law): spf k x k = spring constant

Works only in inertial frame.

fsp = 0

fsp = fwall < 0

fsp = fwall > 0

Example 4.5. Helicopter Ride

Helicopter rises vertically.

A 35 kg bag sits in it on a spring scale of k = 3.4 kN/m.

By how much the spring compresses

(a)when helicopter is at rest

(b) when it’s accelerating upward at 1.9 m/s2.

sp g m F F a

k x m g m a

m a gx

k

(a) 235 0 9.8 /

3400 /

kg m sx

N m

10 cm

(b) 2 235 1.9 / 9.8 /

3400 /

kg m s m sx

N m

12 cm

GOT IT? 4.6

Continued from Example 4.5:

(a)Would the answer to (a) change if the helicopter were

moving upward with constant speed?

(b)Would the answer to (b) change if the helicopter were

moving downward but still accelerating upward?

Example 4.5:

Helicopter rises vertically.

A 35 kg bag sits in it on a spring scale of k = 3.4 kN/m.

By how much the spring compresses

(a)when helicopter is at rest

(b) when it’s accelerating upward at 1.9 m/s2.

No

No