5. Using Newton’s Laws

Newton’s Third Law

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Newton’s Third Law

Law of Action and Reaction

Forces always occur in equal and opposite pairs

AB

Aon BF

A acts on BB acts on A

Bon AF

Bon A Aon BF F

4

Example 3

As noted earlier, Newton’s 2nd law applies to all macroscopic objects. In particular, it appliesto each box separatelyand to bothboxes together.

5

Example 3

2,1 1AF F m a

1,2 2F m a

1,2 2,1F F

2nd Law for m1 2nd Law for m2

3rd Law

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Example 3

1AF F m a 2F m a

1,2 2,1F F F

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Example 3

1 2

AFa

m m

2

1 2A

mF F

m m

The accelerationis exactly what one expects fora mass m1+m2

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Using Newton’s Laws

netF ma

General MethodDetermine the object, or objects, of

interest.Determine real forces acting on

each object.For each object, find the net force.Insert the net force into the 2nd law

and solve:

Multiple Objects

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Example – Saving a Climber

Newton’s 2nd lawapplies toeach climber

For this example, we assume:no frictionrope does not stretchrope of negligible mass

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Example (2)

Forces on climber Steve• Gravity• Normal force• Tension in rope

Forces on climber Paul• Gravity• Tension in rope

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Example (3)

S 1 S SnF m g T m a

P 2 P Pm g T m a

Steve Paul

1. Choose coordinatesystem for eachclimber.

2. Sum forces foreach and apply 2nd law.

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Example (4) Steve

S 1 S SnF m g T m a

S S

1

S

S

S

S

S1

S

ˆ

ˆ

ˆ

0

(sin cos )

0

ˆ

ˆ

ˆ

nnF F

m g m g i

i

j

T

j

jT

i

S SS Sˆ0ia ja

Steve

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Example (5) Paul

P 2 P Pm g T m a

PP p

2 2

P

P P

0

0ˆ ˆ

ˆ ˆm g m gi

i jT

j

T

'y

Note: for Paul, we have chosen a frame of reference with x pointing down.

P PP Pˆ0ia ja

Paul

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Example (6)

p 2 P Pm g T m a

Steve

Paul

As usual, we equatecomponents. But for this problem only the x components are relevant:

S 1 S S0 sinm g T m a

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Example (7)

1 2

S P

T T

a

T

a a

By assumption:

p PTm g am

Steve

Paul

S S0 sin Tm ag m Therefore,

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Example (8)

S P

S P

(1 sin )m m

T gm m

S P

S P

sinm ma g

m m

Acceleration:

Rope tension:

Circular Motion

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Circular Motion

r

v

is a unit vector that pointsfrom the center of the circle to the object.

r

v

r

is the velocity of theobject. As the objectmoves around the circle, thedirection of the velocity changes as does thedirection of the unit vector .r

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Circular Motion

2va

r

The accelerationis directed towards thecenter of the circle, thatis, it is centripetal, andits magnitude is

r

v

r2

ˆv

a rr

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Example 5.7 – Loop-the-Loop!

What is the minimum speed needed to guaranteethat a roller-coaster car stays on the track at the top of the loop?

Identify forces on car• Gravity• Normal force from track

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Example 5.7 (2)

We have just seen that to move in a circle, anobject must have a centripetal acceleration.

According to the 2nd law, the acceleration is causedby a net force.

netF ma

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Example 5.7 (3)

Since the acceleration is in the same direction asthe net force, it follows that the net force must be centripetal, that is, directed towards the center of the loop.

What are these forces?

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Example 5.7 (4)

Presumably, they must be the two forces we have identified: the weight and the normal force.

gn F ma

As usual, we needto set up a coordinatesystem.

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Example 5.7 (5)

Coordinate systemTake +y to be downwards

Take +x to the right

gn F ma

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Example 5.7 (6)

At the top of the loop, the normal force andthe gravitational force point downwards andtowards the center of thecircle. Therefore, in they direction

2

an g

v

m

r

m

m

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Example 5.7 (7)

Solving for v we get ( / )v nr m gr At the minimum speed the caris on the verge of leaving thetracks at the top of the loop.

This occurs when the normal force, n, is zero!

Friction

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The Nature of Friction

Friction is an electrical force between the molecules of surfaces in contact.

Unlike gravity, however, friction is a very complicated force to describe accurately.

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The Nature of Friction

But, for many everyday situations, such as dragging an object along a floor, we can describe frictional forces using simple, approximate,

expressions.

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Frictional Forces

Static Friction – This is the frictional force between surfaces that are at rest relative to each other. The maximum static frictional force is found to be

fs = μs nwhere n is the magnitude of the normal force. μs is called the coefficient of static friction.

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Frictional Forces

Kinetic Friction – This is the frictional force between surfaces that are moving relative to each other. Its value is found to be

fk = μk nwhere n is the magnitude of the normal force. μk is called the coefficient of kinetic friction.

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Frictional Forces

It is found that as the applied force increases so does the opposing frictional force until a maximum value is reached. When the applied force exceeds the maximum frictional force the object accelerates.

During acceleration the frictional force decreases and remains constant when the motion is constant.

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Frictional Forces

Constant speed

Time

Fri

ctio

nal f

orce

Maximum frictional force

Accelerating

At rest

Frictional force remains equal to and opposite the applied force.

μs n

μk n

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Friction in Action

Without friction it wouldbe impossible to walk ormake a vehicle move. As you push against

the ground, the groundpushes you forwards!

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Example – Dragging a Box

What rope tension is needed to move the box atconstant velocity, assuming a coefficient of kineticfriction μk between box and floor?

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Example – Dragging a Box

Draw free-body diagram for box.

The magnitude of the kinetic friction force is

fk = μk n

n

kf

w

T

y

x

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Example – Dragging a Box

The motion is constant, so theforces cancel: – fk + T cos= 0 (x-dir.)–mg + n + T sin = 0 (y-dir.)

n

kf

w

T

y

x

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Example – Dragging a Box

The magnitude of the tension is therefore:

n

kf

w

T

y

x

cos sink

k

mgT

μ μ

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Summary

Big Idea: A net force causes changes in motion. How to Apply:

Find all real forces on a body, sum them, and apply Newton’s 2nd and 3rd laws.

Frictional forceIncreases until object moves, then reduces and

remains constant when motion is constant.