PHYS 100 Lecture 10, Slide 1

PHYS 100Lecture 10

FAB ?

M1 ?

PHYS 100 Lecture 10, Slide 2

Physics 212 Lecture 20

Music

Why?

He did a spectacular show at Krannert last fall

Legendary Cuban piano player Hasn’t been in the U.S. in seven years A real treat !!

Who is the Artist?

A) Arturo SandovalB) Tiempo LibreC) Chucho Valdes & Afro-Cuban MessengersD) Freddie Omar con su bandaE) Los Hombres Calientes

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PHYS 100 Lecture 10, Slide 3

CheckPoint 1

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A B C

T1 must pull the most total weight, T2 pulls the second most and T3 is only responsible for block 2m.

T1 is pulling all three, T2 is pulling the back two, and T3 is only pulling one.

The highest tension is T2 because it is pulling the heaviest object

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PHYS 100 Lecture 10, Slide 4

Follow-Up

What is the value of T2?

How do we start?

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Draw Free Body Diagram(s) and write down Newton’s Second Law

3mT2

T3

2mT3

T2 – T3 = 3ma

T3 = 2ma

Add eqns

T2 = 5ma

Note: even easier --- Draw FBD for 2m+3m !!

PHYS 100 Lecture 10, Slide 5

CheckPoint 2a

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A B C

The force needed to accelerate block R is greater than the force needed to accelerate block L. But neither of there forces is as great as the external force.

Since the blocks are acting on each other in equal or opposite forces. I believe that the magnitude of the total force on the right block is the same as the magnitude on the left.

Because L and R are accelerating at the same rate, they must have the same force acting upon them.

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PHYS 100 Lecture 10, Slide 6

CheckPoint 2a

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IT IS MOST IMPORTANT TO READ THE PROBLEM STATEMENT !!

What is FR ?? FR is the TOTAL FORCE ACTING ON THE RIGHT BLOCK

What is FL ?? FL is the TOTAL FORCE ACTING ON THE LEFT BLOCK

amFtotal�

�

= MR > ML

FR = MRa

FL = MLaFR > FL

FEXT = (ML +MR)a FEXT = FL + FR

FR ∫ FLonR FL ∫ FRonL

PHYS 100 Lecture 10, Slide 7

Follow-Up

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Suppose MR = 5 ML

What is FRonL in terms of FEXT ??

How do we start? Draw Free Body Diagram(s) and write down Newton’s Second Law

L + RFEXT FEXT = (ML + MR)a

LFRonL

FEXT

a = FEXT / (ML + MR)

FEXT – FRonL = MLa FRonL = FEXT - MLa

FRonL = FEXT (MR/(ML + MR))

PHYS 100 Lecture 10, Slide 8

Follow-Up

Suppose we add friction between floor and boxes and boxes slide together.How does FRonL change?

(A) FRonL increases B) FRonL decreases C) FRonL stays the same

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PHYS 100 Lecture 10, Slide 9

Follow-Up

Suppose we add friction between floor and boxes and boxes slide together.How does FRonL change?

(A) FRonL increases B) FRonL decreases C) FRonL stays the same

What is the confidence level of your answer?

A) PRETTY SURE IT’S RIGHT

B) PROBABLY RIGHT

C) UNSURE (GUESS?)

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PHYS 100 Lecture 10, Slide 10

Suppose we add friction between floor and boxes and boxes slide together.How does FRonL change?

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How do we start? Draw Free Body Diagram(s) and write down Newton’s Second Law

L + RFEXT

µµµµKN(ML+MR)g

N

FEXT – µµµµK(ML+MR)g= (ML+MR)a

a = (FEXT / (ML+MR)) -µµµµKg

acceleration decreases!

R

FLonR

µµµµKNRNRMRg

FLonR – µµµµKMRg= MRa

FLonR = MR((FEXT / (ML+MR)) –µµµµKg) + µµµµKMRg

FLonR = FEXT (MR/(ML + MR))

Force remains the same!

PHYS 100 Lecture 10, Slide 11

CheckPoint 2b

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A B C

the force going up minus the force going down is equal to ma, so rearranging the equation you get m(g+a)

to find the acceleration you take the accleration from the elevator minus the accleration of gravity put it into the f=ma equation

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PHYS 100 Lecture 10, Slide 12

CheckPoint 2b

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PROBLEM: Even though 65% chose A, several reasons were not correct !!

For example: “The acceleration of A is g + a so the force on be has to be M(a+g) “

The acceleration of A is a NOT a+g !!

Just DO the FBD !! This is the ONLY sure-fire way to answer the question

A

MgFBonA

FBonA –Mg = Ma FBonA = M(g+a)

Newton’s Third Law

FBonA = FAonB

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PHYS 100 Lecture 10, Slide 13

CheckPoint 3

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What’s wrong with A?

F =T = ma

What’s wrong with C?

T = M1(g-a)

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Strategies:A) We start w/FBD. There is a weight force

going down on M1. We know T1 = 10N. Therefore, F =T = ma. M1 = 1.91.

B) I assumed the acceleration of M2 applied for the acceleration for the system in general.So I made a free body diagram of M1 and summed the forces in the x direction then solved for m1, since we have all the values needed.

C) I took the acceleration of gravity and subtracted the acceleration of the block M2 and then divided the tension force 10 by the result and got 2.13.

PHYS 100 Lecture 10, Slide 14

CheckPoint 3

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A B C D E

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M1

Draw FBD T1 = 10 N

M1g

T1 – M1g = M1a

ga

TM

+=

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PHYS 100 Lecture 10, Slide 15

Calculation

How to start?

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We will do this one as a CheckPoint in a couple of weeks

NEXT WEEK: NEW CONTENT

Springs & Universal GravitationDo Prelecture & CheckPoint before Lecture