physics 151: lecture 8, pg 1 physics 151: lecture 8 l reaminder: çhomework #3 : (problems from...

Physics 151: Lecture 8, Pg 1

Physics 151: Lecture 8Physics 151: Lecture 8

Reaminder:Homework #3 : (Problems from Chapter 5)

due Fri. (Sept. 22) by 5.00 PM

Today’s Topics :

Review of Newton’s Laws 1 and 2 - Ch. 5.1-4

Newton’s third law: action and reaction - Ch. 5.6


ReviewReviewNewton’s Laws 1 and 2Newton’s Laws 1 and 2

Isaac Newton (1643 - 1727) published Principia Mathematica in 1687. In this work, he 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

See text: 5.1-4


Newton’s Third Law: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

See text: 5-6

This is among the most abused concepts in physics.

REMEMBER: 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”


An ExampleAn Example

FB,E = - mB g

EARTH

FE,B = mB g

Consider the forces on an object undergoing projectile motion

FB,E = - mB g

FE,B = mB g


Normal ForcesNormal Forces

Certain forces act to keep an object in place. These have what ever force needed to balance all others

(until a breaking point).

FT,B

FB,T


Force PairsForce Pairs

Newton’s 3rd law concerns force pairs. Two members of a force pair cannot act on the same object.

Don’t confuse gravity (the force of the earth on an object) and normal forces. It’s an extra part of the problem.

FT,B

FB,T

FB,E = -mg

FE,B = mg



Consider the following two cases



The Free Body Diagrams

mg

mg

FB,T= N

Ball FallsFor Static Situation

N = mg



The action/reaction pair forces

FB,E = -mg FB,T= N

FE,B = mg

FB,E = -mg

FE,B = mg

FT,B= -N


Lecture 8, Lecture 8, Act 1Act 1Newton’s 3rd LawNewton’s 3rd Law

Two blocks are being pushed by a finger on a horizontal frictionless floor. How many action-reaction pairs of forces are present in this system?

(a) 2 (b) 4 (c) 6

a b


246

Lecture 8, Lecture 8, Act 1Act 1Solution:Solution:

a bFFa,f FFf,a FFb,a FFa,b

FFg,ag,a

FFa,ga,g

FFg,bg,b

FFb,gb,g

FFE,aE,a

FFa,Ea,E

FFE,bE,b

FFb,Eb,E

Is Fa,f = Fb,a? (A) YES (B) NO Fb,a = Fa,f [mb/(mb+ma)]


You are going to pull two blocks (mA=4 kg and mB=6 kg) at constant acceleration (a= 2.5 m/s2) on a horizontal frictionless floor, as shown below. The rope connecting the two blocks can stand tension of only 9.0 N. Would the rope break ?

(A) YES (B) CAN’T TELL (C) NO

Lecture 8, Lecture 8, Act 2Act 2

A Ba= 2.5 m/s2rope


What are the relevant forces ?

Lecture 8, Lecture 8, Act 2Act 2 Solution:Solution:

mAmB

Fapp

a= 2.5 m/s2

mA

rope

mB

T

-T T

-T

Fapp = a (mA + mB)Fapp = 2.5m/s2( 4kg+6kg) = 25 N

total mass !

aA = a = 2.5 m/s2

T = a mA T = 2.5m/s2 4kg = 10 NT > 9 N, rope will brakeANSWER (A)

Fapp - T = a mB T = 25N - 2.5m/s2 6kg=10NT > 9 N, rope will brake

a = 2.5 m/s2

Fapp

a = 2.5 m/s2

THE SAME ANSWER -> (A)


Exercise: Inclined planeExercise: Inclined plane

A block of mass m slides down a frictionless ramp that makes angle with respect to horizontal. What is its acceleration a ?

ma

See text: Example 5.7


Inclined plane...Inclined plane...

Define convenient axes parallel and perpendicular to plane:

Acceleration a is in x direction only.

ma

ii

jj



Inclined plane...Inclined plane...

Consider x and y components separately: ii: mg sin = ma a = g sin

jj: N - mg cos . N = mg cos

mgg

NN

mg sin

mg cos

maa

ii

jj


m


Angles of an Inclined planeAngles of an Inclined plane

a = g sin

mg

N


m


Free Body DiagramFree Body Diagram

A heavy sign is hung between two poles by a rope at each corner extending to the poles.

Eat at Bob’s

What are the forces on the sign ?


Free Body DiagramFree Body Diagram

Eat at Bob’s

T1

mg

T2

Add vectors :

x

yT1

T2

mg

Vertical (y):mg = T1sin1 + T2sin2

Horizontal (x) :T1cos1 = T2cos2


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)


Lecture 9, Lecture 9, ACT 1ACT 1Gravity 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 1ACT 1Gravity and Normal ForcesGravity and Normal Forces

The free body diagram is,

For the woman to accelerate upwards, the normal force on the woman must be A) greater than the force due to gravity acting on the womanNote, both of these forces act on the woman, they cannot be an action/reaction pair

N mg


Lecture 9, Lecture 9, ACT 1bACT 1bGravity 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 the woman exerts on the elevator.


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

The action/reaction force diagram for the woman and elevator is,

By Newton’s third law these must be (B) equal.

N = FW,EFE,W


Recap of today’s lectureRecap of today’s lecture

Newton’s 3 Laws

Free Body Diagrams

Action/Reaction Force pairs

Reading for Friday, Ch 5.7-8, pp. 123-139

Applications of Newton’s Laws and Friction

Homework #3 (due next Wed. / Sept. 21 by 11:59 pm)

physics 151: lecture 8, pg 1 physics 151: lecture 8 l reaminder: çhomework #3 : (problems from...

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