PHYS 100: Lecture 5 - University of Illinois at Urbana ... · PHYS 100: Lecture 5 ... Who is the...

Physics 100 Lecture 5, Slide 1

PHYS 100: Lecture 5

NEWTON’S SECOND LAW

NMg

T

θ

u

v

tot iF F ma= =∑� �

�


Music

Who is the Artist?

A) David Benoit

B) Oscar Peterson

C) Thelonius Monk

D) Dr. John

E) Mose Allison

SORRY.. DON’T HAVE MY IPOD TODAY!

My theme for today was to have been Mardi Gras!

Which of the above piano players is most closely associated with New Orleans?

classics

BB


How To Learn• Lectures

– Study the Lecture slides after lecture. Ask yourself: What was the Point of the lecture? Try to understand the logic of what was asked.

• Homework– Do Interactive Examples first: Use as much HELP as you need !

• Do the Follow-Ups: they test whether you understand what you just did!– Your initial score on the delayed feedback questions is your first indicator of your understanding. If you do not answer those correctly, try to understand why!

• Discussion– Study the solutions after discussion. What was the point of asking those questions?

KEYS FOR PROBLEM SOLVING

AVOID EQUATION HUNTING

THINK FIRST !!• What concepts apply to the problem?• What are the relevant variables?

• Which are known? Which are unknown?• Develop a plan (first in English)• Write down equations for relevant variables• Solve


THE BIG IDEAS

1. An object accelerates only when the total force on it is non-zero.2. The weight of an object is always equal to Mg3. The Normal Force and Tension forces must be determined from F=ma

NOTE: THE BIG IDEAS ARE ALWAYS GIVEN IN THE LAST SLIDE


Thrown BallA ball is thrown straight up in the air. Take up to be the positive direction

What are the velocity (v) and acceleration (a) of the ball at its highest point?

Just before the ball reaches its highest point: what is its velocity?

(A) v = 0a > 0

(B) v = 0a < 0

(C) v = 0a = 0

(D) v ∫ 0a = 0

(E) v ∫ 0a ∫ 0

v

v1

Just after the ball reaches its highest point: what is its velocity? v2

Just as the ball reaches its highest point: what is its acceleration?

dt

vda

�

�

≡12

vvva��

−=∆∝v2

+

-v1

Just as the ball reaches its highest point: what is its velocity? v = 0

=

a

+

BB


CheckPoint 1A ball is thrown straight up in the air.

How many forces are acting on the ball at its highest point?

You said:

• The ball is not moving at its highest point so there are no forces acting on the ball at that moment.

• weight is the only force acting on the ball

• the gravitational force and the force exerted to throw it into the air

(A) 0 (B) 1 (C) 2 (D) 3

0

10

20

30

40

50

60

70

A B C D E

(E) Greater than 3

We know a = g throughout the flight of the ball ! The net force at all times then must be given by:F = ma = mg (the weight !!)

BB


Pulling a block

WHAT DO YOU KNOW??

In both cases shown, forces of equal magnitude, F, are exerted on identical blocks supported by a frictionless floor. In Case I, the force is applied at an angle θ wrt horizontal while in Case II, the force is applied horizontally.

Compare aI to aII:

(A) aI < aII

(B) aI = aII

(C) aI > aII

M M

Case I Case II

FFθ

amFFitot

�

��

==∑

WHAT DO YOU DO FIRST?

Draw Free Body Diagram !!

M M

FFθ

NI NIIMg Mg

APPLY NEWTON’S SECOND LAW IN HORIZONTAL

DIRECTION

IMaF =θcos II

MaF =

θcosM

Fa

I=

M

Fa

II=θcos

IIIaa =

i) ΣFxi = max

ii) ΣFyi = may

BB


Pulling a block

WHAT DO YOU KNOW??

In both cases shown, forces of equal magnitude, F, are exerted on identical blocks supported by a frictionless floor. In Case I, the force is applied at an angle θ wrt horizontal while in Case II, the force is applied horizontally.

Compare NI to NII:

(A) NI < NII

(B) NI = NII

(C) NI > NII

M M

Case I Case II

FFθ

amFFitot

�

��

==∑



M M

FFθ

NI NIIMg Mg

APPLY NEWTON’S SECOND LAW IN VERTICAL

DIRECTION

MgNFI

=+θsin MgNII

=

θsinFMgNI

−= θsinFNNIII

−=

BB


CheckPoint 2

0

10

20

30

40

50

60

70

A B C

You said:

• Case 1 has gravity acting against it at angle theta while case 2 is acting at a 90 degree angle. In case 1 part of the force is used to counteract that of gravity while case 2 all of the force is used to move the box

• Acceleration = Force/mass, so since the blocks have the same mass and each block has the same force acting upon it, the acceleration is equal.

• It takes a greater acceleration to push the object up a incline than to push it straight across a surface.

In both cases shown, forces of equal magnitude, F, are exerted on identical blocks. The forces are directed parallel to the frictionless surfaces

Compare aI to aII:

(A) aI < aII

(B) aI = aII

(C) aI > aII

BB


CheckPoint 2


Compare aI to aII:

(A) aI < aII

(B) aI = aII

(C) aI > aII



Same kind of forces, but differences in orientation will cause differences in magnitude !

APPLY NEWTON’S SECOND LAW IN DIRECTION OF

ACCELERATION

IIMaF =

NIIMg

Mgθ

IMaMgF =− θsin

θsingM

Fa

I−=

M

Fa

II=

θsingaaIII

−=

x

ya


Follow-Up


Compare NI to NII:

(A) NI < NII

(B) NI = NII

(C) NI > NII



APPLY NEWTON’S SECOND LAW PERPENDICULAR TO

THE PLANE

NIIMg

Mgθ

0cos =− θMgNI

θcosIII

NN =

0=− MgNII

θcosMgNI

= MgNII

=

BB


Elevators

Both elevators shown carry identical crates and are moving down. In Case I, the elevator is moving at constant velocity v, while in Case II the elevator is moving at constant velocity 2v.

Compare NI to NII:

(A) NI < NII (B) NI = NII (C) NI > NII

2v


Draw Free Body Diagram !!2v

NIIMgNIMg


DIRECTION(a = 0 !!)

0=− MgNII0=− MgN

I

IIINN =

MgNI

= MgNII

=

BB


CheckPoint 3

0

10

20

30

40

50

60

A B C

You said:

• The acceleration is positive in case two where it is negative in case one, the overall force of case two is greater.

• I would say the normal forces are equal in both cases because its pushing against identical crates.

• N1 has more force opposing the fall, this is why it is slowing down. N2 has less force opposing the fall, this is why it speeds up

Both elevators shown carry identical crates and are moving down. In Case I, the elevator is slowing down, while in Case II the elevator is speeding up.

Compare NI to NII:


BB


CheckPoint 3

Both elevators shown carry identical crates and are moving down. In Case I, the elevator is slowing down, while in Case II the elevator is speeding up.

Compare NI to NII:




NIIMgNIMg

WHAT IS ACCELERATION ?? aI aII


DIRECTION(take direction of a to be

positive)

IIIIMaNMg =−

IIMaMgN =−

)(II

agMN += )(IIII

agMN −=

IIINN >


Follow Up

Both elevators shown carry identical crates and are moving up. In Case I, the elevator is slowing down, while in Case II the elevator is speeding up.

Compare NI to NII:

(B) NI = NII (C) NI > NII



WHAT IS ACCELERATION ?? aI aII


DIRECTION(take direction of a to be

positive)

IIIIMaMgN =−

IMaNMg =−

)(II

agMN −= )(IIII

agMN +=

(A) NI < NII

IIINN <

v v

NIIMgNIMg

v v

BB


CheckPoint 4

0

20

40

60

80

100

A B C

You said:

• In case I, the tension equals to the mass of the object (except the signs are different). In case II, the x and the y components of the tensions have to be considered in order to find the value of TL or TR.

• The similarity in angles calls for the tension to be equal.

• the sum of TL and TR will equal T since both cases are at equilibrium, therefore T must be greater than TL

In both cases shown identical signs are suspended from the ceiling. One string is used in Case I, while two strings, each making an angle of θ = 20o with the horizontal, are used in Case II.

Compare T to TL:

(A) T < TL (B) T = TL (C) T > TL

LET’S DO THE EXPERIMENT !!

BB


CheckPoint 4

In both cases shown identical signs are suspended from the ceiling. One string is used in Case I, while two strings, each making an angle of θ = 20o with the horizontal, are used in Case II.

Compare T to TL:

(A) T < TL (B) T = TL (C) T > TL


Draw Free Body Diagram !!T

Mg Mg

APPLY NEWTON’S SECOND LAW IN VERTICAL DIRECTION

(Symmetry of Case II insures ax = 0)(i.e., TL = TR)

MgT = MgTL

=θsin2

θsin2

MgT

L=

θsin2

TT

L=

TL > T if sinθ < 1/2 i.e., θ < 30o

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