PHY 113 C Fall 2013 -- Lecture 5 19/10/2013

PHY 113 A General Physics I11 AM-12:15 PM MWF Olin 101

Plan for Lecture 5:

Chapter 5 – NEWTON’S GENIUS1. Concept of force

2. Relationship between force and acceleration

3. Net force

Note: today we are going to neglect friction.

PHY 113 C Fall 2013 -- Lecture 5 29/10/2013

PHY 113 C Fall 2013 -- Lecture 5 39/10/2013

Webassign question from Assignment #4

East

North

3

1

2

4

PHY 113 C Fall 2013 -- Lecture 5 49/10/2013

Webassign question from Assignment #4

Given: h,H,x,q Determine: vi

vi

x=

PHY 113 C Fall 2013 -- Lecture 5 59/10/2013

Problem solving steps1. Visualize problem – labeling

variables2. Determine which basic physical

principle(s) apply3. Write down the appropriate

equations using the variables defined in step 1.

4. Check whether you have the correct amount of information to solve the problem (same number of knowns and unknowns).

5. Solve the equations.6. Check whether your answer makes

sense (units, order of magnitude, etc.).

Two-dimensional motion with constant acceleration

2

221

221

cos tan

ˆ sin

ˆ cos

ˆ

ii

iiii

iii

iii

vxxg

xxyxy

gttvy

tvxt

g

qq

q

q

j

ir

ja

PHY 113 C Fall 2013 -- Lecture 5 69/10/2013

Isaac Newton, English physicist and mathematician (1642—1727)

http://www.newton.ac.uk/newton.html

1. In the absence of a net force, an object remains at constant velocity or at rest.

2. In the presence of a net force F, the motion of an object of mass m is described by the form F=ma.

3. F12 =– F21.

PHY 113 C Fall 2013 -- Lecture 5 79/10/2013

Detail: “Inertial” frame of reference

Strictly speaking, Newton’s laws work only in a reference frame (coordinate system) that is stationary or moving at constant velocity. In a non-inertial (accelerating) reference frame, there are some extra contributions.

iclicker question:Are we (here in Winston-Salem)

A. In an inertial frame of reference (exactly)B. In a non-inertial frame of reference

Newton’s laws are only approximately true in Winston-Salem, but the corrections are very small.

PHY 113 C Fall 2013 -- Lecture 5 89/10/2013

PHY 113 C Fall 2013 -- Lecture 5 99/10/2013

Example force – weight

1 lb = 4.45 N 1 N = 0.225 lb

Newton

PHY 113 C Fall 2013 -- Lecture 5 109/10/2013

Gravitational mass versus inertial mass –

The concept of mass (“inertial” mass m) is distinct from the concept of weight mg. In fact the gravitation “constant” g varies slightly throughout the globe. The weight of mass m on the surface of the earth is different from its weight measured on the surface of the moon or on the surface of Mars. (Stay tuned to Chapter 13 – “Universal Gravitation”.)

iclicker questionWill we always need to take gravity into account?

A. YesB. No

PHY 113 C Fall 2013 -- Lecture 5 119/10/2013

Isaac Newton, English physicist and mathematician (1642—1727)

http://www.newton.ac.uk/newton.html

1. In the absence of a net force, an object remains at constant velocity or at rest.

2. In the presence of a net force F, the motion of an object of mass m is described by the form F=ma.

3. F12 =– F21.

PHY 113 C Fall 2013 -- Lecture 5 129/10/2013

Examples

Fgravity=-mg j

Ftension= T j

F=Fnet=Ftension+Fgravity=(T-mg) j =0

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Examples

Fgravity=-mg j

Ftension= T j

F=Fnet=Ftension+Fgravity=(T-mg) j =0

Fgravity=-mg j

F=Fnet=Fgravity=(-mg) j = ma

a=-gj

PHY 113 C Fall 2013 -- Lecture 5 14

vi

x=9/10/2013

Two-dimensional motion with constant acceleration

2

221

221

cos tan

ˆ sin

ˆ cos

ˆ

ii

iiii

iii

iii

vxxg

xxyxy

gttvy

tvxt

g

qq

q

q

j

ir

ja

PHY 113 C Fall 2013 -- Lecture 5 159/10/2013

iclicker question:An object having a mass of 2 kg is moving at constant velocity of 3 m/s in the upward direction near the surface of the Earth. What is the net force acting on the object?

A. 0 N.

B. 6 N in the upward direction.

C. 25.6 N in the upward direction.

D. 19.6 N in the downward direction.

PHY 113 C Fall 2013 -- Lecture 5 169/10/2013

Example: (assume surface is frictionless)

Suppose T measured

a observed & measured

T = m a

kg50m/s1.0N5

m/s 0.1 N;5

2

2

aTm

aT

PHY 113 C Fall 2013 -- Lecture 5 179/10/2013

Example of two dimensional motion on a frictionless horizontal surface

m

m

m

21

21

FFa

aFF

PHY 113 C Fall 2013 -- Lecture 5 189/10/2013

Example of two dimensional motion on a frictionless horizontal surface -- continued

mN 10.14

N100cos85285 2221

oFF

2m/s 8.33kg 3.0

N 10.14

m21 FF

a

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Example – support forces

appliedsupport

ˆ

FFyF

mgg

Fsupport acts in direction to surface

(in direction of surface “normal”)

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Another example of support force:

FFn

FFn

g

g

g

0ˆ0

j

FFn

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iclicker question:Why do we care about forces in equilibrium?

A. Normal people don’t care.B. Physicist care because they are weird.C. It might be important if the materials

responsible for the support forces are near their breaking point.

PHY 113 C Fall 2013 -- Lecture 5 229/10/2013

Example of forces in equilibrium

PHY 113 C Fall 2013 -- Lecture 5 239/10/2013

Example of forces in equilibrium -- continued

0sinsin :component ˆ0coscos :component ˆ

0

122

32211

2211

3

1

3

TTT

TT

NFT

ii

g

qq

qq

j

i

T

PHY 113 C Fall 2013 -- Lecture 5 249/10/2013

Example of forces in equilibrium -- continued

0sinsin 0coscos

:algebra using equations Solving

32211

2211

TTT

TTqq

qq

NTT

NTT

NTTT

TT

4.73coscos

4.97sinsin

coscos

122sinsincoscos

coscos

1

221

211

2

32

32211

22

1

221

qq

qqqq

qqqq

qq

PHY 113 C Fall 2013 -- Lecture 5 259/10/2013

Newton’s second lawF = m a

Types of forces:

Fundamental Approximate Empirical

Gravitational F=-mg j Friction

Electrical Support

Magnetic Elastic

Elementary

particles

PHY 113 C Fall 2013 -- Lecture 5 269/10/2013

2

2

dtdm

dtdmm rvaF

Examples (one dimension):

0FF (constant)

tmFtvxtx

sin)( 20

00 tFF sin0

2000 2

1)( tmFtvxtx

kxF tmkxtx cos)( 0

mbvFF 0

btebg

bgtv )(

PHY 113 C Fall 2013 -- Lecture 5 279/10/2013

Newton’s third law

F12 = - F21

1 2

FPT T

T=m1a a=FP/(m1+m2)

FP-T=m2a T=FP (m1/(m1+m2))

PHY 113 C Fall 2013 -- Lecture 5 289/10/2013

T-m1g=m1a T-m2g=-m2a => after some algebra:

a

gmmmma

12

12

g

mmmmT

12

212

PHY 113 C Fall 2013 -- Lecture 5 299/10/2013

Fgravity=-mg j

Fscale

Fscale + Fgravity = 0; Fscale = mg

Question: What if you step on the scale in the elevator?

Fscale + Fgravity = m a; Fscale = mg +m a

PHY 113 C Fall 2013 -- Lecture 5 309/10/2013

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ExerciseFor this exercise, you should ride the elevator from the first floor of Olin Physical Laboratory up to the second or third floor. While in the elevator, stand on the scale and record the scale readings to answer the following questions.

1. What is the scale reading when the elevator is stationary?

2. What is the scale reading when the elevator just starts moving upward?

3. What is the scale reading when the elevator is coming to a stop just before your destination floor?

x

<x

>x

PHY 113 C Fall 2013 -- Lecture 5 329/10/2013

mg

Ffloor = mg

mgFfloor = mg-T

T T

Mg

Mg=12 lbmg=14.5 lb

T=?

iclicker exerciseA. mgB. MgC. Not enough info

PHY 113 C Fall 2013 -- Lecture 5 339/10/2013

Example: 2-dimensional forcesA car of mass m is on an icy (frictionless) driveway, inclined at an angle t as shown. Determine its acceleration.

Conveniently tilted coordinate system:

qq

q

sinsin : Along

0cos : Along

gamamgx

mgny

x

x

PHY 113 C Fall 2013 -- Lecture 5 349/10/2013

vi

vf=0

tgvtvtgtvxtx

i

ii

sin)( sin)( :incline Along 2

21

qq

Another example: Note: we are using a tilted coordinate frame

i