Physics 1501: Lecture 7 Today ’ s Agenda

Physics 1501: Lecture 7, Pg 1

Physics 1501: Lecture 7Physics 1501: Lecture 7TodayToday’’s Agendas Agenda

Announcements:

Physics Learning Center P201:

» Schedule posted on the door

Homeworks

» Lowest HW will be dropped

» HW 02: due next Monday

Today’s topics:

» Newton’s laws

» Applications of Newton’s laws


DynamicsDynamics


ReviewReviewNewtonNewton’’s Laws 1 and 2s 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

Force and mass


Lecture 7,Lecture 7, ACT 1ACT 1NewtonNewton’’s Second Laws Second Law

A constant force is exerted on a cart that is initially at rest on an air table. The force acts for a short period of time and gives the cart a certain final speed.

Air Track

CartForce

For a second shot, I can apply a force only half as large (I’m getting tired). To reach the same final speed, for how long must I apply the force ?

A) 4 x as long B) 2 x as long C) Same timeD) 1/2 as long E) 1/4 x as long


Lecture 7,Lecture 7, ACT 2ACT 2NewtonNewton’’s Second Laws Second Law

I push with a force of 2 Newtons on a cart that is initially at rest on an air table with no air. I push for a second. Because there is no air, the cart stops after I finish pushing. It has traveled a certain distance.

Air Track

CartForce

For a second shot, I push just as hard but keep pushing for 2 seconds. The distance the cart moves the second time versus the first is,

A) 4 x as long B) 2 x as long C) Same D) 1/2 as long E) 1/4 x as long


Lecture 7Lecture 7, , ACT 3ACT 3NewtonNewton’’s Second Laws Second Law

A constant force is exerted on a cart that is initially at rest on an air table. This force is applied for a short period of time and the cart acquires a certain final speed, which I call vf1.

Air Track

Cart Force

I repeat the experiment, but this time the cart is already moving with constant speed when I start applying the force. After exerting the same constant force for the same time interval, the cart’s final speed is,

A) vf1 B) 2vf1 C) vf12

D) cannot be determined from the information given.


NewtonNewton’’s Third Law: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

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

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) !


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


Lecture 7, Lecture 7, ACT 4ACT 4NewtonNewton’’s Third Laws Third Law

A fly gets smushed onto the windshield of a speeding bus.

The force exerted by the bus on the fly is,A) greater thanB) the same asC) less thanthat exerted by the fly on the bus.


Lecture 7, Lecture 7, ACT 5ACT 5NewtonNewton’’s Third Laws Third Law

A fly gets smushed onto the windshield of a speeding bus.

The acceleration due to this collision of the bus is,A) greater thanB) the same asC) less thanthat of the fly.


Newton's Third Law...Newton's Third Law...

FFA ,B = - FFB ,A an example,

FFm,w FFw,m

FFm,f

FFf,m


Example of Bad ThinkingExample of Bad Thinking

Since FFm,b = -FFb,m why isn’t FFnet = 0, and aa = 0 ?

a ??a ??FFm,b FFb,m

ice


Example of Good ThinkingExample of Good Thinking Consider only the box only the box as the system!

Free Body Diagram

FFm,b FFb,m

ice


Example of Good ThinkingExample of Good Thinking Consider only the box only the box as the system!

Free Body Diagram

aaboxbox

FFb,m

FgFN

abox = Fb,m/mbox


Free Body DiagramFree Body DiagramA 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

T2

Add vectors

mg

T1

T2mg


Free Body DiagramFree Body Diagram

Eat at Bob’s

T1 T2

mg

Vertical :mg = T1sin1 + T2sin2

Horizontal :T1cos1 = T2cos2


Normal Forces and String TensionNormal Forces and String Tension

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


Lecture 7, Lecture 7, Act 6Act 6NewtonNewton’’s 3rd Laws 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



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 7, Lecture 7, ACT 7ACT 7Gravity 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 thanthe force due to gravity acting on the woman


Lecture 7, Lecture 7, ACT 7bACT 7bGravity 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 thanthe force the woman exerts on the elevator.


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


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


Angles of an Inclined planeAngles of an Inclined plane

ma = mg sin

mgN

Physics 1501: Lecture 7 Today ’ s Agenda

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