physics 207: lecture 8, pg 1 lecture 9 l goals describe friction in air (ch. 6) differentiate...

Physics 207: Lecture 8, Pg 1

Lecture 9

GoalsGoals

Describe Friction in Air (Ch. 6)

Differentiate between Newton’s 1st, 2nd and 3rd Laws

Use Newton’s 3rd Law in problem solving

Assignment: HW4, (Chap. 6 & 7, due 10/5)

1st Exam Thurs., Oct. 7th from 7:15-8:45 PM Chapters 1-7

in rooms 2103, 2141, & 2223 Chamberlin Hall


Friction in a viscous mediumDrag Force Quantified

With a cross sectional area, A (in m2), coefficient of drag of 1.0 (most objects), sea-level density of air, and velocity, v (m/s), the drag force is:

D = ½ C A v2 c A v2 in Newtons

c = ¼ kg/m3

In falling, when D = mg, then at terminal velocity Example: Bicycling at 10 m/s (22 m.p.h.), with projected

area of 0.5 m2 exerts a force of ~30 Newtons At low speeds air drag is proportional to v but at high

speeds it is v2

Minimizing drag is often important


Newton’s Third Law:

If object 1 exerts a force on object 2 (F2,1 ) then object 2

exerts an equal and opposite force on object 1 (F1,2)

F1,2 = -F2,1

IMPORTANT: Newton’s 3rd law concerns force pairs whichact on two different objects (not on the same object) !

For every “action” there is an equal and opposite “reaction”


Force Pairs vs. Free Body Diagrams

Consider the following two cases (a falling ball and ball on table),

Compare and contrast Free Body Diagram and

Action-Reaction Force Pair sketch


Free body diagrams

mg

mg

FB,T= N

Ball FallsFor Static Situation

N = mg


Force Pairs

1st and 2nd Laws Free-body diagram Relates force to acceleration3rd Law Action/reaction pairsShows how forces act between objects

FB,E = -mg FB,T= N

FE,B = mg

FB,E = -mg

FE,B = mg

FT,B= -N


Example (non-contact)

Consider the forces on an object undergoing projectile motion

FB,E = - mB g

EARTH

FE,B = mB g

FB,E = - mB g

FE,B = mB g

Question: By how much does g change at an altitude of 40 miles? (Radius of the Earth ~4000 mi)


Note on Gravitational Forces

Newton also recognized that gravity is an attractive, long-range force between any two objects.

When two objects with masses m1 and m2 are separated by distance r, each object “pulls” on the other with a force given by Newton’s law of gravity, as follows:


Cavendish’s Experiment

F = m1 g = G m1 m2 / r2

g = G m2 / r2

If we know big G, little g and r then will can find m2 the mass of the Earth!!!


Example (non-contact)

Consider the force on a satellite undergoing projectile motion 40 km above the surface of the earth:

FB,E = - mB g

EARTH

FE,B = mB g

FB,E = - mB g

FE,B = mB g

Compare: g = G m2 / 40002 g’ = G m2 / (4000+40)2

g / g’ = / (4000+40)2 / 40002 = 0.98


A conceptual question: A flying bird in a cage

You have a bird in a cage that is resting on your upward turned palm. The cage is completely sealed to the outside (at least while we run the experiment!). The bird is initially sitting at rest on the perch. It decides it needs a bit of exercise and starts to fly.

Question: How does the weight of the cage plus bird vary when the bird is flying up, when the bird is flying sideways, when the bird is flying down?

Follow up question:

So, what is holding the airplane up in the sky?


Static Friction with a bicycle wheel

You are pedaling hard and the bicycle is speeding up.

What is the direction of the frictional force?

You are breaking and the bicycle is slowing down

What is the direction of the frictional force?


Exercise Newton’s Third Law

A. greater than

B. equal to

C. less than

A fly is deformed by hitting the windshield of a speeding bus.

v

The force exerted by the bus on the fly is,

that exerted by the fly on the bus.


Exercise 2Newton’s Third Law

A. greater than

B. equal to

C. less than

A fly is deformed by hitting the windshield of a speeding bus.

v

The magnitude of the acceleration, due to this collision, of the bus is

that of the fly.

Same scenario but now we examine the accelerations


Exercise 3Newton’s 3rd Law

A. 2

B. 4

C. 6

D. Something else

a b

Two blocks are being pushed by a finger on a horizontal frictionless floor.

How many action-reaction force pairs are present in this exercise?


Force pairs on an Inclined plane

Forces on the block

Block weight is mg

NormalForce

Friction Force

“Normal” means perpendicular

mg cos

f

x

y

mg sin



Forces on the block (equilibrium case)

NormalForce

Friction Force

f= N

x

yForces on the plane



Forces on the block (non equilibrium case)

NormalForce

Friction Force

f

x

yForces on the plane


Example: Friction and Motion

A box of mass m1 = 1 kg is being pulled by a horizontal string having tension T = 40 N. It slides with friction

(k= 0.5) on top of a second box having mass m2 = 2 kg, which in turn slides on a smooth (frictionless) surface. What is the acceleration of the second box ?

1st Question: What is the force on mass 2 from mass 1?

(A) a = 0 N (B) a = 5 N (C) a = 20 N (D) can’t tell

mm22

T mm11slides with friction (k=0.5)

slides without frictiona = ?

v


ExampleSolution

First draw FBD of the top box:

m1

N1

m1g

T fk = KN1 = Km1g

v


Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2.

m1ff1,2 = Km1g = 5 N

m2

ff2,1 == -ff1,2

As we just saw, this force is due to friction:

ExampleSolution

ActionReaction

(A) a = 0 N (B) a = 5 N (C) a = 20 N (D) can’t tell


Now consider the FBD of box 2:

m2 ff2,1 = km1g

m2g

N2

m1g

ExampleSolution


Finally, solve Fx = ma in the horizontal direction:

m2 ff2,1 = Km1g

K m1g = m2a

ExampleSolution

= 2.5 m/s2

kg 2

N 5ga

2

k1

m

m


Home Exercise Friction and Motion, Replay

A box of mass m1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are s=1.5 and k= 0.5. The second box can slide freely (frictionless) on an smooth surface.

Compare the acceleration of box 1 to the acceleration of box 2 ?

mm22

T mm11

friction coefficients s=1.5 and k=0.5

slides without frictiona2

a1


Home Exercise Friction and Motion, Replay in the static case

A box of mass m1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are s=1.5 and k= 0.5. The second box can slide freely on an smooth surface (frictionless).

If no slippage then the maximum frictional force between 1 & 2 is

(A) 20 N (B) 15 N (C) 5 N (D) depends on T

mm22

T mm11friction coefficients

s=1.5 and k=0.5


a1


Home ExerciseFriction and Motion

fs = 10 N and the acceleration of box 1 is

Acceleration of box 2 equals that of box 1, with |a| = |T| / (m1+m2) and the frictional force f is m2a

(Notice that if T were in excess of 15 N then it would break free)

mm2 2

T mm11

friction coefficients s=1.5 and k=0.5


a1

TTm1 gg

NN

fS

fS S N = S m1 g = 1.5 x 1 kg x 10 m/s2

which is 15 N (so m2 can’t break free)


Recap

Wednesday: Review for exam

physics 207: lecture 8, pg 1 lecture 9 l goals describe friction in air (ch. 6) differentiate...

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