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Department of Physics and Applied PhysicsPHYS.1410 Lecture 9 Danylov

Lecture 9

Chapter 7

Newton’s Third Law

Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI

Physics I


Today we are going to discuss:

Chapter 7:

Some leftover (Ch.6) Interacting Objects: Section 7.1 Analyzing Interacting Objects: Section 7.2 (skip Propulsion) Newton’s Third Law: Section 7.3 Ropes and Pulleys: Section 7.4

The static friction force has a maximum of sN = 50 N. The tension in the rope is only 30 N. So the pulling force is not big enough to overcome friction.

ConcepTest Will It move?A) moves to the left

B) moves to the right

C) the box does not move

D) moves down

E) moves up

A box of weight 100 N is at rest on a floor where s = 0.5. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move?

T=30Nm

Follow-up: What happens if the tension is 55 N?

NF Sfr NmgS 501005.0


DemonstrationTwo Interleaved Books

Simply lay the pages of two phone books on top of each other one by one before attempting to pull them apart.

FA Fperp

Fpull

It increases max friction

It pulls books apart

Ffr NxFfr

http://www.youtube.com/watch?v=AHq82D78Igg

Fpull

Ffr

When we apply FA force, we create two components Fpull and Fperp.

Since there is no motion, we have a static friction force Ffr equal to the Fpull.

As we increase Fpull, Fperp gets larger increasing (Ffr )max , which means that our goal moves

farther away. So the more we pull, the larger goal becomes But luckily Fpull grows faster

and with two tanks it can be done.


Newton’s 3rd Law

Chapter 7.Forces come from other objects


Newton’s Third Law of Motion

Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first.

“For every action force, there is an equal and opposite reaction force”NOTE: Action and Reaction forces act on different objects!

=


Let’s try to justify N.3rd law.

If you still don’t believe me, hit the nail with a glass hammer.It’s the force of the nail on the hammer that would cause the glass to shatter!

Is it true that a nail also exerts a force on a hammer (in return)?

When a hammer hits a nail, it exerts a forward force on the nail.

When a hammer hits a nail, the hammer stops very quickly

i.e. the hammer changes velocity, i.e. it accelerates (a) If the hammer accelerates (decelerates), there must be a force acting on the hammer from the “poor nail”, according to the N. 2nd law.

10 /0 /

Me


Convention how to denote these forces

SWWS FF

A key to the correct application of the third law is that the forces are exerted on different objects.

Make sure you don’t use them as if they were acting on the same object.

Helpful notation:

the second subscript is the object that the force is being exerted on; the first is the source.

Let’s come up with a convention how to denote these pairs of forces

FB on A

If object A exerts a force on object B, then object B exerts a force on object A. - The pair of forces, as shown, is called an action/reaction pair.

A BFBAOr simply

FA on BFAB Or simply

- i.e. If you want to describe a motion of a skater, you have to use a force FWS acting from the wall on the skater

- And if you want to describe the wall, use FSW .

=


Newton’s Third Law of Motion

Rocket propulsion can also be explained using Newton’s third law: hot gases from combustion spew out of the tail of the rocket at high speeds. The reaction force is what propels the rocket.


Acceleration Constraints

If two objects A and B move together, their accelerations are constrained to be equal: aA = aB. This equation is called an acceleration constraint.

Consider a car being towed by a truck. In this case, the acceleration constraint is

aCx = aTx = ax . Because the accelerations of both objects

are equal, we can drop the subscripts C and T and call both of them ax .

But, as A moves to the right in the x direction, B moves down in the y direction.

In this case, the acceleration constraint is aAx = aBy .


Internal forces cancel each other!!!!!!Cancel due to Newton’s Third Law


Two blocks problem

x

yF

Treat, m=m1+m2, as the system (one big block)

Fx maxmaF

21 mmFa

Given: F, m1, m2

Find: a(common acceleration)

F maApply N. 2nd law to m

x component of N. 2nd law

amm )( 21

m=m1+m2

We can now forget about the internal forces

ConcepTest Crazy Mosquito

A mosquito runs head-on into a truck. Which is true during the collision?

A) The magnitude of the mosquito’s acceleration is larger than that of the truck.

B) The magnitude of the truck’s acceleration is larger than that of the mosquito.

C) The magnitude of the mosquito’s acceleration is the same as that of the truck.

D) The truck accelerates but the mosquito does not.E) The mosquito accelerates but the truck does not.

Newton’s 2nd law:

Don’t confuse cause and effect! The same force can have very different effects.

FMT FTM Newton’s 3rd law:FMT=FTM=F aM >> aT

The same idea can be applied to an interaction of an apple and the Earth in the next slide.But you don’t have to read the next slide. Only if you want.

End of the class


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Ropes and Pulleys


Tension

If a flexible cord pulls an object, the cord is said to be under TENSION

Often in problems the mass of the string or rope is much less than the masses of the objects that it connects.

massless string approximation:

T1

T2

m

Let’s assume that the cord is a described object and apply N 2nd law

0

m=0

For problems in this book, you can assume that any strings or ropes are massless unless it explicitly states otherwise.

Tension is the same at any point of the rope

,

Tension is the same at any point of the massless rope

(Read this if you want)

ConcepTest Going Sledding

1

2

In case 1, the force F is pushing down(in addition to mg), so the normal force is larger. In case 2, the force Fis pulling up, against gravity, so the normal force is lessened. Recall that the frictional force is proportional to the normal force.

A) pushing her from behind

B) pulling her from the front

C) both are equivalent

D) it is impossible to move the sled

E) tell her to get out and walk

Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this?


Thank you

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