newton’s third law sec. 6.3 interaction forces ► objectives explain the meaning of interaction...

Newton’s Third LawNewton’s Third Law

Sec. 6.3Sec. 6.3Interaction ForcesInteraction Forces

►ObjectivesObjectives Explain the meaning of interaction pairs Explain the meaning of interaction pairs

(third law pairs) of forces and how they (third law pairs) of forces and how they are related by Newton’s third laware related by Newton’s third law

List the four fundamental forces and List the four fundamental forces and illustrate the environment in which each illustrate the environment in which each can be observed.can be observed.

Explain the tension in ropes and strings in Explain the tension in ropes and strings in terms of Newton’s third lawterms of Newton’s third law

1. The law of inertia. An object in motion remains in motion with constant velocity if the net force on the object is zero.

2. Force and acceleration. If the net force acting on an object of mass m is F, then the acceleration of the object is a = F/m. Or, F = ma.

3. Action and reaction. For every action there is an equal but opposite reaction.

Newton’s laws of motion

Action means force.

Newton’s 3rd Law

For every action force there is an equal and

opposite reaction force

(You cannot touch without being touched)

Action-reaction Pair

If object A exerts a force on object B, then object B exerts an equal and opposite force on object A. The pair of forces (due to one interaction), is called an action/reaction pair.

NOTA BENE: The action/reaction pair will never appear in the same free body diagram.

Summary: Interaction forcesSummary: Interaction forces(‘Third Law Pairs’)(‘Third Law Pairs’)

►Two forces that are in opposite Two forces that are in opposite directions and have equal magnitudedirections and have equal magnitude

►They never act on the same object!They never act on the same object!(Therefore never in same FBD!)(Therefore never in same FBD!)

►General Form is: General Form is:

FFA on B A on B = -F = -FB on AB on A

force of A on B

force of B on A

Newton’s third law

We will see that this is very hard to accept! It is just not common sense. That is why it took a great genius like Newton to figure it out.

Whenever one object exerts a force on another object, the second object exerts an equal but opposite force on the first object.

Forces always occur like this, in pairs.

For every action there is an equal but opposite reaction.

Action: tire pushes on roadReaction: road pushes on

tire

When you walk or run, what forces occur?

• At constant velocity the horizontal force is 0 and you continue to move because of inertia.

• To accelerate, you push backward against the floor; the reaction force, which is a friction force exerted by the floor on your foot, pushes you forward.

This reaction force may be hard to visualize, but imagine what would happen if you were on a frictionless surface – can’t accelerate!

Force accelerating bulletRecoil force

a = F/ma = F/m

A bug and a car collide. Which experiences the greater force?

• (a) bug

• (b) car• (c) neither, they both experience the same

magnitude of force(c) neither, they both experience the same magnitude of force

Consider hitting a baseball with a bat. If we call the force applied to the ball by the bat the action force,

identify the reaction force.

(a) the force applied to the bat by the hands

(b) the force applied to the bat by the ball

(c) the force the ball carries with it in flight

(d) the centrifugal force in the swing

(b) the force applied to the bat by the ball

Which vehicle exerts a greater force ― the tow truck or the car?

The truck pulls to the right. According to Newton’s third law, the car pulls to the left with an equal force. So how can they start moving, or accelerate?

A puzzle:

Resolution: Consider each part separately, and don’t forget that other forces are also acting.

Playing catch with a medicine ball

A throws the ball and B catches it. four forces

When A throws the ball he exerts a force on the ball (toward the right) and the ball exerts a force on him so he recoils (toward the left).

A B

When B catches the ball he exerts a force on the ball (toward the left to stop it) and the ball exerts a force on him so he is knocked back (toward the right).

► Newton’s third law for the throw

► Newton’s third law for the catch

Example – A collision

The force exerted by the ball on the toe (reaction) is equal to the force exerted by the toe on the ball.

Really hard to accept!

Which team will end up in the puddle?

But aren’t the forces equal but opposite !?

Resolution: Don’t forget that there are other forces acting.Each team exerts a force on the Earth, so the Earth exerts a force on the team (3rd law!). The net force on either team is toward the left.

A puzzle : Tug of War

String tension

String tension

Contact force Contact force

Resolution: Consider each part separately, and don’t forget that there are other forces acting.

So how can they start moving, or accelerate?

A puzzleHorse and Cart

The horse pulls the cart with a force A (to the right).

According to Newton, the cart pulls the horse with a force –A (to the left).

A small car is pushing a larger truck that has a dead battery. The mass of the truck is larger than the mass of thecar. Which of the following statements is true?A. The truck exerts a larger force on the car than the car exerts

on the truck.B. The truck exerts a force on the car but the car doesn’t exert a

force on the truck.C. The car exerts a force on the truck but the truck doesn’t exert

a force on the car.D. The car exerts a larger force on the truck than the truck

exerts on the car.E. The car exerts the same amount of force on the truck as the

truck exerts on the car.

A small car is pushing a larger truck that has a dead battery. The mass of the truck is larger than the mass of thecar. Which of the following statements is true?A. The truck exerts a larger force on the car than the car exerts

on the truck.B. The truck exerts a force on the car but the car doesn’t exert a

force on the truck.C. The car exerts a force on the truck but the truck doesn’t exert

a force on the car.D. The car exerts a larger force on the truck than the truck

exerts on the car.E. The car exerts the same amount of force on the truck as

the truck exerts on the car.

Newton’s Third Law PairsNewton’s Third Law Pairs

Example - analyzing interacting objects

A person pushes a large crate across a rough surface.

• Identify the objects that are systems of interest

• Draw free-body diagrams for each system of interest.

• Identify all action/reaction pairs with a dashed line.

Forces involved in pushing a crate – FBD of person and crate

Propulsion Force• The force label fp

shows that the static friction force on the person is acting as a propulsion force.

• This is a force that a system with an internal source of energy uses to drive itself forward.

Propulsion forces

Free Body Diagrams – Exercises

Draw a freebody diagram of each object in the interacting system.

Show action/reaction pair with red/orange dotted lines.

Draw force vectors in another color.

Label vectors with standard symbols.

Label action/reaction pairs FAonB , FBonA for example.

A fishing line of negligible mass lifts a fishupward at constant speed. The line and the fish are the system, the fishing pole is part of the environment. What, if anything, is wrong with the free-body diagrams?

A fishing line of negligible mass lifts a fishupward at constant speed. The line and the fish are the system, the fishing pole is part of the environment. What, if anything, is wrong with the free-body diagrams?

The gravitational force and the tension force are incorrectly identified as an action/reaction pair. The correct action reaction pair is…?

Action/reaction pairs are never on the same free body diagram.

Mass of line considered negligible so no weight force necessary.

Boxes A and B are sliding to the right across a frictionless table. The hand H is slowing them down. The mass of A is larger than the mass of B. Rank in order, from largest to smallest, the horizontal forces on A, B, and H. Ignore forces on H from objects not shown in the picture.

Draw a FBD for each object: A, B, and H.

Only consider horizontal forces in this problem.

FBA

A B C

FHBFBHFAB

Recognize any third law pairs?

Boxes A and B are sliding to the right across a frictionless table. The hand H is slowing them down. The mass of A is larger than the mass of B. Rank in order, from largest to smallest, the horizontal forces on A, B, and H.

FB on H = FH on B > FA on B = FB on A

from Newton’s 2nd and 3rd Laws

Problem-Solving Strategy: Interacting-Objects Problems

Challenge Problem:Two strong magnets each weigh 2 N

and are on opposite sides of the table. The table, by itself, has a weight of 20 N. The long range-range attractive force between the magnets keeps the lower magnet in place. The magnetic force on the lower magnet is 3 times its weight.

a. Draw a FBD for each magnet and table. Use dashed lines to connect all action/reaction pairs.

b. Find the magnitude of all forces in your FBD and list them in a table.

Challenge Problem

↓ (Wg) E,U ↓ (Wg) E,T ↓ (Wg) E,L

↑ N T,U ↑ N S,T ↑ LM U,L

↓ LM L,U ↓ N U.T ↓ N T,L

- ↑ N L,T -

Upper Table Lower

Find Third Law Pairs

↓ (Wg) E,U ↓ (Wg) E,T ↓ (Wg) E,L

↑ N T,U ↑ N S,T ↑ LM U,L

↓ LM L,U ↓ N U.T ↓ N T,L

- ↑ N L,T -

Upper Table Lower

FBDs for EOC 8

Challenge Problem: Answer

Ranking Task – Pushing blocks

Block 1 has a mass of m, block 2 has a mass of 2m, block 3 has a mass of 3m. The surface is frictionless.

Rank these blocks on the basis of the net force on each of them, from greatest to least. If the net force on each block is the same, state that explicitly.

Ranking Task – Pushing blocks

Answer: 3 2 1

Reason: ΣF = ma. Acceleration is equal for all blocks.

Ranking Problem ExampleBlock 1 has a mass of 1 kg, block

2 has a mass of 2 kg, block 3 has a mass of 3 kg. The surface is frictionless.

a. Draw a fbd for each block. Use dashed lines to connect all action/ reaction pairs.

b. How much force does the 2-kg block exert on the 3-kg block?

c. How much force does the 2-kg block exert on the 1-kg block?

Ranking Problem Example – Answer:

b. How much force does the 2-kg block exert on the 3-kg block? – 6N

c. How much force does the 2-kg block exert on the 1-kg block? – 10N

Newton’s Third LawNewton’s Third Law

Acceleration constraint

• An acceleration constraint is a well-defined relationship between the acceleration of 2 (or more) objects.

• In the case shown, we can assume ac =aT = ax

Is there an acceleration constraint in this situation? If so, what is it? The pulley is considered to be

massless and frictionless.

Answer: Acceleration constraint is: aA = -aB

The actual signs may not be known until the problem is solved, but the relationship is known

from the start.

Exercises

17

answers

a2kg = -.5a1kg

Newton’s Third LawNewton’s Third Law

Page 163, 2nd Ed (Found in Chapter 5, 1st ed)

Interacting systems problem (EOC #35)

A rope attached to a 20 kg wooden sled pulls the sled up a 200 snow-covered hill. A 10 kg wooden box rides on top of the sled. If the tension in the rope steadily increases, at what value of tension will the box slip?

Interacting systems problem (EOC #35)

Find the max tension in the rope, so the box does not slip.

What are the objects of interest?What kind of axes for the FBD for each?Acceleration constraints?Draw FBDs, with 3rd law pairs connected with dashed lines.

0.06

Box: 3 forces

Sled: 6 forces

0.06

• I suggest starting with the equations for the sled, since the unknown of interest is found there.

• Identify quantities in sled equations that you can find by solving box equations.

• Solve box equations.

• Return to sled equations with newfound booty.

• Plug and chug.

Interacting systems problem (EOC #35)

A rope attached to a 20 kg wooden sled pulls the sled up a 200 snow-covered hill. A 10 kg wooden box rides on top of the sled. If the tension in the rope steadily increases, at what value of tension will the box slip?

Answer: 155 N.

The Massless String Approximation

A horizontal forces only fbd for the string:

● TBonS

TAonS

ΣF = TBonS – TAonS = ma. If string is accelerating to the right

TBonS = TAonS + ma

The Massless String Approximation

Often in physics and engineering problems the mass of the string or rope is much less than the masses of the objects that it connects. In such cases, we can adopt the following massless string approximation:

This allows the objects A and B to be analyzed as if they exert forces directly on each other.

PulleysIf we assume that the string is massless and the pulley is both massless and frictionless, no net force is needed to turn the pulley. TAonB and TBonA act “as if” they are an action/reaction pair, even though they are not acting in opposite directions.

Pulleys

• In this case the Newton’s 3rd law action/reaction pair point in the same direction!

Tm on 100kg

T 100kg on m

A. Equal to B. Greater thanC. Less than

All three 50 kg blocks are at rest. Is the tension in rope 2 greater than, less than, or equal to the tension in rope 1?

A. Equal to B. Greater thanC. Less than

All three 50 kg blocks are at rest. Is the tension in rope 2 greater than, less than, or equal to the tension in rope 1?

In the (moving) figure to the right, is the tension in the string greater than, less than, or equal to the weight ofblock B?

A. Equal toB. Greater thanC. Less than

In the figure to the right, is the tension in the string greater than, less than, or equal to the weight ofblock B?

A. Equal toB. Greater thanC. Less than

Interacting systems problem (EOC #40)

A 4.0 kg box (m) is on a frictionless 350 incline. It is connected via a massless string over a massless, frictionless pulley to a hanging 2.0 kg mass (M). When the box is released:

a. Which way will it go George?

b. What is the tension in the string?

350

4.0 kg

Interacting systems problem (EOC #40)

350

4.0 kg

a. Which way will it go? Even if you have no clue, follow the plan!

What are the objects of interest?What kind of axes for the FBD for each?Acceleration constraints??Draw FBDs, with 3rd law pairs connected with dashed lines.

Interacting systems problem (EOC #40)

How do you figure out which way the system will move, once m is released from rest?

massless string approx. allows us to join the tensions as an “as if” interaction pair

Interacting systems problem (EOC #40)

Interacting systems problem (EOC #40)

a = - 0.48 m/s2, T = 21 N.

Which way is the system moving?

How does the tension compare to the tension in the string while the box was being held?

Greater than, less than, equal to?

EOC # 33

The coefficient of static friction is 0.60 between the two blocks in the figure. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force F causes both blocks to slide 5 meters, starting from rest. Determine the minimum amount of time in which the motion can be completed without the upper block slipping.

EOC # 33The coefficient of static

friction is 0.60 between the two blocks in the figure. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force F causes both blocks to slide 5 meters, starting from rest. Determine the minimum amount of time in which the motion can be completed without the upper block slipping.

EOC # 33

amax = 3.27 m/s2

tmin = 1.75 s (time is important in this one)

EOC # 46

Find an expression for F, the magnitude of the horizontal force for which m1 does not slide up or down the wedge. This expression should be in terms of m1, m2 , θ, and any known constants, such as g. All surfaces are frictionless.

EOC # 46.

EOC # 46.

F = (m1 + m2) g tan θ

EXAMPLE 7.6 Comparing two tensions

QUESTION:

EXAMPLE 7.6 Comparing two tensions

EXAMPLE 7.6 Comparing two tensions

EXAMPLE 7.6 Comparing two tensions

EXAMPLE 7.6 Comparing two tensions

The spring force ---another example of 3rd law

Suppose the spring is stretched beyond its equilibrium length by a length x.

The force on the mass m1 is F1 = +kx.

(k = Hooke’s constant)

The force on the mass m2 is F2 = kx.

( + means to the right; - means to the left.)

The forces are equal but opposite.

Example

One end of a spring is attached to a wall. When the other end is pulled with a force of 50 N, the spring is stretched by 3 cm. What force would be required to stretch the spring by 5.5 cm?

Answer: 91.7 N

Hooke’s law: The strength of a spring force is proportional to the displacement (extension or compression). F = k x where k is called Hooke’s constant for the spring.

Forces obey Newton’s third law.

We’ll consider two examples:• The force of universal gravitation• The spring force

Universal Gravitation --- an example of Newton’s third law

ˆ

ˆ

2221

2

1221

1

nF

nF

rmGm

rmGm

The Earth pulls the apple down (“action”).

The apple pulls the Earth up (“reaction”).

The two forces are equal (but opposite).

When does a scientific theory become accepted as true?

The force on the 1 kg mass is +3.3 x 10-

10 N.The force on the 5 kg mass is –3.3 x 10-

10 N.( + means to the right, i.e., increasing x)

1-2- kg s m 31110676 .G

For a laboratory measurement, the gravitational force is really very weak.

Henry Cavendish, 1798 : first measurement of G

What makes g?

2

2

RGM

g

RGMm

mg

gravity of force weight

9.81 m/s2

Weighing the EarthCalculate the mass of the Earth.

The force of gravity on m is, by definition, its weight,

By Newton’s theory of universal gravitation,

mgF

2RGMm

F

kg .

.

).(.

24

22

10026

kg11E676

6E46819G

gRM

the mass of the Earth, relying on the Cavendish measurement

Quiz Question

The planet is pulled toward the moon (and vice versa).Calculate the gravitational force on the planet.