Physics 151: Lecture 9, Pg 1 Physics 151: Lecture 9 l Announcements çHomework #3 (due this Fri....

Physics 151: Lecture 9, Pg 1

Physics 151: Lecture 9Physics 151: Lecture 9

Announcements

Homework #3 (due this Fri. 9/22/06, 5 PM)Homework #4 (due Mon. 10/2/06, 5 PM)

Review sessions:Sept. 25, Mon. 8.10 PM - 9:00 pm in P-36

Today’s Topics:Review of Newton’s Law 3 - Ch. 5.1-6Some applications of Newton’s laws - Ch5.7Friction - Ch5.8


ReviewReviewNewton’s Laws 1, 2, 3Newton’s Laws 1, 2, 3

Isaac Newton (1643 - 1727) proposed three “laws” of motion:

Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame.

Law 2: For any object, FFNET = FF = maa

Law 3: Forces occur in pairs: FFA ,B = - FFB ,A

(For every action there is an equal and opposite reaction.)

See text: Ch. 5


Lecture 9, Lecture 9, ACT 1ACT 1

A book is placed on a chair. Then a videocassette is placed on the book. The floor exerts a normal force

A. on all three.

B. only on the book.

C. only on the chair.

D. upwards on the chair and downwards on the

book.

E. only on the objects that you have defined to be

part of the system.


Lecture 9, Lecture 9, ACT 2ACT 2Gravity and Normal ForcesGravity and Normal Forces

A woman in an elevator is accelerating upwards

The normal force exerted by the elevator on the woman is,A) greater thanB) the same asC) less than

the force due to gravity acting on the woman


Lecture 9, Lecture 9, ACT 1bACT 1bGravity and Normal ForcesGravity and Normal Forces

A woman in an elevator is accelerating downwards

The normal force exerted by the elevator on the woman is,A) greater thanB) the same asC) less than

the force the woman exerts on the elevator.


Lecture 9, Lecture 9, ACT 1cACT 1cGravity and Normal ForcesGravity and Normal Forces

A woman in an elevator is downwards at constant speed

The tension on the rope holding the elevator is,A) greater thanB) the same asC) less than

the weight of the elevator and the woman ?


Example-1 with pulleyExample-1 with pulley

Two masses M1 and M2 are connected by a rope over the pulley as shown. Assume the pulley is massless and

frictionless.Assume the rope massless.

If M1 > M2 find :

Acceleration of M1 ?

Acceleration of M2 ?Tension on the rope ?

Free-body diagram for each object

M1

T2T1

M2

aAnimation

Video


Example-2 with pulleyExample-2 with pulley

A mass M is held in place by a force F. Find the tension in each segment of the rope and the magnitude of F.Assume the pulleys massless and

frictionless.Assume the rope massless.

M

T5

T4

T3T2

T1

F We use the 5 step method.Draw a picture: what are we looking for ?What physics idea are applicable ? Draw

a diagram and list known and unknown variables.

Newton’s 2nd law : F=ma

Free-body diagram for each object


Pulleys: continuedPulleys: continued FBD for all objects

M

T5

T4

T3T2

T1

F

T4

F=T1

T2

T3

T2 T3

T5

M

T5

Mg


Pulleys: finallyPulleys: finally

Step 3: Plan the solution (what are the relevant equations)F=ma , static (no acceleration: mass is held in place)

M

T5

Mg

T5=Mg

T2 T3

T5

T2+T3=T5

T4

F=T1

T2

T3

F=T1

T1+T2+T3=T4


Pulleys: really finally!Pulleys: really finally! Step 4: execute the plan (solve in terms of variables)

We have (from FBD):

T5=MgF=T1 T2+T3=T5 T1+T2+T3=T4

M

T5

T4

T3T2

T1

F

T2=T3T1=T3

T2=Mg/2

T2+T3=T5 gives T5=2T2=Mg

F=T1=Mg/2

T1=T2=T3=Mg/2 and T4=3Mg/2

T5=Mg and

Pulleys are massless and frictionless

Step 5: evaluate the answer (here, dimensions are OK and no numerical values)


New Topic: New Topic: FrictionFriction

What does it do?It opposes motion!

How do we characterize this in terms we have learned?Friction results in a force in a direction opposite to the

direction of motion!

maaFFAPPLIED

ffFRICTION mgg

NN

ii

j j

See text: 5.8


Friction...Friction...

Friction is caused by the “microscopic” interactions between the two surfaces:

See text: 6-1


Friction...Friction...

Force of friction acts to oppose motion:Parallel to surface.Perpendicular to NNormal force.

maaFF

ffF mgg

NN

ii

j j

See text: 6-1

See figure 6-1


Model for Sliding FrictionModel for Sliding Friction

The direction of the frictional force vector is perpendicular to the normal force vector NN.

The magnitude of the frictional force vector |ffK| is proportional to the magnitude of the normal force |N N |.

|ffK| = K | N N | ( = K|mg g | in the previous example)

The “heavier” something is, the greater the friction will be...makes sense!

The constant K is called the “coefficient of kinetic friction”. (This friction is called Kinetic Friction)

See text: 6-1


Model...Model...

Dynamics:

i : F KN = m a

j : N = mg

so F Kmg = m a

maaFF

mgg

NN

ii

j j

K mg

See text: 6-1


Lecture 9, Lecture 9, ACT 4ACT 4

In a game of shuffleboard (played on a horizontal surface), a puck is given an initial speed of 6.0 m/s. It slides a distance of 9.0 m before coming to rest. What is the coefficient of kinetic friction between the puck and the surface ?

A. 0.20B. 0.18C. 0.15D. 0.13E. 0.27


Lecture 9, Lecture 9, ACT 4ACT 4 A box of mass m1 = 1 kg is being pulled by a

horizontal string having tension T = 40 N. It slides with friction (k= .5) on top of a second box having mass m2 = 2 kg, which in turn slides on an ice rink (frictionless).

What is the acceleration of the second box ?(A) a = 0 m/s2 (B) a = 2.5 m/s2 (C) a = 10 m/s2

mm2 2

T mm11slides with friction (k=0.5)

slides without frictiona = ?


Lecture 9, Lecture 9, ACT 3ACT 3Two-body dynamicsTwo-body dynamics

A block of mass m, when placed on a rough inclined plane ( > 0) and given a brief push, keeps moving down the plane with constant speed.If a similar block (same ) of mass 2m were placed on the

same incline and given a brief push, it would:

(a)(a) stop

(b) (b) accelerate

(c) (c) move with constant speed

m


Lecture 9, Lecture 9, ACT 3ACT 3SolutionSolution

Draw FBD and find the total force in the x-direction

ii

jj

mg

N

KN

FTOT,X = mg sin Kmg cos

= ma = 0 (first case)

Doubling the mass will simplydouble both terms…net forcewill still be zero !

Speed will still be constant !Speed will still be constant !

Animation


Static Friction...Static Friction...

FF

mgg

NN

ii

j j

fS

See text: Ch 5.8

So far we have considered friction acting when something moves.We also know that it acts in un-moving, “static” systems:

In these cases, the force provided by friction will depend on the forces applied on the system.



Just like in the sliding case except a = 0.

i : F fS = 0

j : N = mg

FF

mgg

NN

ii

j j

fS

While the block is static: fS F ! (unlike kinetic friction)

See text: Ch 5.8



FF

mgg

NN

ii

j j

fS

The maximum possible force that the friction between two objects can provide is fMAX = SN, where s is the “coefficient of static friction”.So fS S N.

As one increases F, fS gets bigger until fS = SN and the object “breaks loose” and starts to move.

See text: Ch 5.8



S is discovered by increasing FF until the block starts to slide:

i : FMAX SN = 0

j : N = mg

S FMAX / mg

FFMAX

mgg

NN

ii

j j

Smg

See text: Ch 5.8


Additional comments on Friction:Additional comments on Friction:

See text: 6-1

Since f = N , the force of friction does not depend on the area of the surfaces in contact.

By definition, it must be true that S > K for any system (think about it...).

Animation


ExampleExampleProblem 5.40 from the bookProblem 5.40 from the book

Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of 0.350, the masses are m1 = 4.00 kg, m2 = 1.00kg and m3 = 2.00kg.

a) What is the magnitude and direction of acceleration on the three blocks ?

b) What is the tension on the two cords ?

m1

T1m2

m3


m1

T1m2

m3

m1m2 m3

m2g

T23T12

m1g

T12

m3g

T23

T12

T12 T23

T23

T12 - m1g = - m1a T23 - m3g = m3a

k m2ga

a

a

-T12 + T23 + k m2g = - m2a

SOLUTION: T12 = = 30.0 N , T23 = 24.2 N , a = 2.31 m/s2 left for m2


Recap of today’s lectureRecap of today’s lecture

Newton’s Third Law Friction Reading for Wed., Ch 6.1, pp. 151-158

Newton’s Laws and Circular Motion

Homework #3 (due Fri. 9/22/06, 5 PM)Homework #4 (due Mon. 10/2/06, 5 PM)

Review sessions:Sep. 25, Mon. 8.10 - 9:00 pm in P-38

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Physics 151: Lecture 9, Pg 1 Physics 151: Lecture 9 l Announcements çHomework #3 (due this Fri....

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