Chapter 4:

Dynamics Newton’s Laws

What if we all jumped at once?

Newton’s 1st Law

• Objects with mass have Inertia: the

tendency to stay at rest (or moving!)

• The more mass an object has, the more

difficult it is to move it (or stop it!)

• Activity: penny/cardboard

LAB P. 76-79

Lab p. 76-79

Write out headings, purpose

procedure (choose one of a-h)

Observations: Copy out data

tables

Lab p. 76-79: choose on of A-H

A: Caitlyn/Susan

B: Carson

C: Lindsay,Miranda

D: Hailey/Brett

E: Steven^2,Kev,Zach

F: Elisabeth/Lalia

G: Noah

H: Liam/Captain Proton!

Email logger pro file to yourself - lab

Newton’s 2nd Law

• The force necessary to move (or stop)

objects depends on:

– mass

– acceleration

amFnet

• Ex 1: how much force is necessary to

accelerate a 80kg student at 10 m/s2?

280 10 800net

mF ma kg Ns

What do we mean “net” force?

• Net force is zero if there are no

unbalanced forces

• We usually do not notice forces until

they become unbalanced

• Ex. What are the forces acting on a

suction cup?

Free Body Diagrams

• The point of a FBD is to simplify the

dynamics involved

• We only point out the forces acting on the

body in question

• To get to the point, we draw the body as a…

point!

• The forces are drawn

pointing away from the

body

F1 F2

F3

More than one force?

• Ex 1: Cody applies a 50 N force to a

2.5kg book to slide it across the table.

Find the acceleration if there is a 45 N

friction force resisting this motion

netF ma

50 ( 45 )

2.5

netF N Na

m kg

20.2

5.2

5

sm

kg

N

Fa Ff

• Ex 2: Find the force necessary to

accelerate a 1200 kg rocket at 3.4 m/s2,

if there is a 250 N force of air

resistance. Fa

Fr

𝐹𝑛𝑒𝑡 = 𝑚𝑎

𝐹𝑎+𝐹𝑟= 𝑚𝑎

Exercises p. 81 #1-6

Weight

• Weight is the force of gravity acting on

an object

• Where g is the gravitational field strength at

some location in space

• Near the surface of the earth, each kg of mass

has a weight of 9.8 N

• Ex 1: what is the weight of a 86 kg student?

843gF N

m

Fg

mgF

Ex 2

• What is the weight of the same student,

while carrying a 55N bag of tricks?

843gF N

Ex 3

• What is the mass of a student with a

weight of 700N?

m

Fg

g

Fm

kgN

Nm

/8.9

700 kgm 71

Ex 3

• What is the weight of the same student

in orbit around the earth, where

g=9.2m/s2?

• 792N

Ex 4:

• What is the

weight of the

same student

on the moon,

whose

gravitational

field strength is

1/6 earth’s?

• 140N

Exercises p. 82-3 #1-5

Prep Lab p. 85

Percent Difference

• Comparing two numbers: subtract, then

divide

9.782 − 9.832

9.832

𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 − 𝐴𝑐𝑐𝑒𝑝𝑡𝑒𝑑

𝐴𝑐𝑐𝑒𝑝𝑡𝑒𝑑

Lab p. 76-79: choose on of A-H

A: Caitlyn/Susan

B: Carson

C: Lindsay,Miranda

D: Hailey/Brett

E: Steven^2,Kev,Zach

F: Elisabeth/Lalia

G: Noah

H: Liam/Captain Proton!

Email logger pro file to yourself - lab

Newton’s 3rd Law

• For every action there is an equal and

opposite reaction

• When you hit something, it hits back!

Normal Force

• When an object is in contact with a supporting

surface, it pushes down on that surface

• Newton’s 3rd Law states the surface pushes back

with an equal and opposite force

• This is normally (but not always!) equal to the

object’s weight

• We sometimes refer to Normal force as the

“apparent weight”

Simple case: object at rest

• Ex 3: What is the normal force acting on

the 2.5 kg book resting on your desk?

– What forces act on the book?

• Gravity and Normal force

– Free body diagram

– Apply 2nd law

FN

Fg

0netF ma

0g NF F

N gF F mg FN

Fg

2.5 9.8 24.5NNF kg Nkg

Extended object at rest

• Ex 3: what is the normal force acting on

your book as you lean on it with a 35 N

force?

– What forces act on the book?

• Gravity, Applied and Normal force

– Free body diagram

– Apply 2nd law

0netF ma

FN

Fa Fg

0netF ma

0g a NF F F

N g aF F F

FN

Fa Fg ( 24.5 ) ( 35 )NF N N

60NF N

Accelerating object

• Ex 4: find the apparent weight of a 50 kg

student accelerating upwards at 3.4 m/s2

– What forces act on the student?

• Gravity and Normal force

– Free body diagram

– Apply 2nd law

netF ma

FN

Fg

FN

Fg

maFF gN

maFF gN

mamgFN

4.3508.950 NF

NFN 660

Accelerating object

• Ex 4: find the acceleration of a 55 kg

student with a normal force of 1300N

– What forces act on the student?

• Gravity and Normal force

– Free body diagram

– Apply 2nd law

netF ma

FN

Fg

FN

Fg

maFF gN

m

FFa

gN

55

)8.9(551300

Na

28.13s

ma

Do Lab 4-3

• P. 85-86

Friction

• Once we know normal

force, we can calculate

friction

• Friction force can be

calculated as normal force

times “mu”

• Ff=μFN

• But how do we find mu?

• Lab 3-3 p. 55

FN

Fg

Ff Fa

Friction lab 3-3

• Conclusion: find the

average value of μ from

part 1

• Make a statement about

what factors affect friction

force

• As always, state sources of

error and possible

improvements

FN

Fg

Ff Fa

ReTest procedure

• Come in at lunch, do

corrections, get help

• Then book date to come in

another lunch to do retest

Friction

• We find friction force is proportional to the Normal force and a “stickiness factor” (AKA coefficient of friction)

Nf FF

• Ex: find the friction force acting on your

25N textbook as it slides across the table if

=0.55

• Applying 2nd law to find Normal force:

x y

Nf FF maFnet

maFF Ng

0 Ng FF

NFF gN 25

NFf 2555.0

NFf 14

Nf FF

• Ex: find the coefficient of friction if it takes

a 13 N force to slide your 2.2 kg book along

the table at a constant speed.

N

f

F

F

)8.9(2.2

13 60.0

Exercises

• P. 60 #1-4

• Lab p. 55 done!!

Friction on “static” objects

• If an object is not moving, forces must

be balanced

– friction force must be equal to the applied

horizontal force

• The maximum static friction force is

given by the previous formula, so we

have:

Nsf FF

• Where s is the coefficient of static friction, and

tends to be larger than k for kinetic friction

When does static become

kinetic? • Static friction increases

with applied force until it “breaks free”

– at this point, it starts moving and kinetic or “sliding” friction takes over

• Ex 6: for what applied force if s=0.65 does your textbook start moving?

– What will the acceleration be?

• We already have Normal force:

x y

Nf FF max NFN 25

NFf 2565.0max

NFf 16max

• Once it breaks free we have kinetic friction again:

x y

NFa 16

maFnet

m

FFa

fa

298.0s

ma

kg

NN

55.2

75.1325.16

Friction Lab

• Add a column to table 1 for coefficient of

friction

• Do questions #1, 3 (average) & 4 p. 56

• Do question #1 p. 58 (#2: sources of

error should be done in your conclusion)

Universal

Gravitation

• Newton’s most original contribution!

• All objects in the universe exert a gravitational pull on each other

• But why doesn’t the moon fall?

• Newton realized

objects don’t fall

to the Earth’s

surface if they

have a high

enough tangential

velocity

Universal gravitation:

• Fg=GMm/r2

– Where G is the

universal gravitational

constant

How do we find G?

Weigh the Earth!?

Mm

rFG

g

2

• Weigh the Earth!

Pendulum Method

• If only we had a large enough mass to get a

measurable force…

• Ex 1: Calculate G if you get a 4.6×10-6 N

force on a 1.5kg pendulum 150m away from

a mountain with a mass of 6.4×108 kg

Mm

rFG

g

2

Mm

rFG

g

2

26

8

4.6 10 150

1.5 6.4 10

N mG

kg kg

2

2101008.1kg

NmG

2

21010kg

NmG

Cavendish’s experiment

• To get a more

precise

measurement of G,

he used a torsional

pendulum

• Cavendish was able

to get a value of:

2

2111067.6kg

NmG

• Ex 2: What is the

Earth’s mass if 1kg

of mass has a

weight of 9.8N?

2r

GMmFg

mG

FrM

g

2

6 2

211

2

9.8 (6.38 10 )

1 6.67 10

N mM

Nmkgkg

kgM 241098.5

• Ex 3: Find the force

of attraction

between Kat (54 kg)

and Mikael (95 kg) if

they sit 4.0 m apart.

2r

GMmFg

2r

GMmFg

2

112

2

6.67 10 95 54

(4.0 )g

Nm kg kgkg

Fm

NFg8101.2

Start Exercises

• P. 67 question 6 a) through c)

• *part (a) consider what happens when we replace r with 2r?

21r

GMmF

222r

GMmF

What about g?

• How does gravitational field strength “g”

fit into all of this?

• Ex 4: find g at an

altitude of 130 km.

Careful!

2r

GMmmg

2

11 242

6 5 2

6.67 10 5.98 10

(6.38 10 1.3 10 )

Nm kgkg

gm m

kgNg 41.9

mgFg 2r

GMg

Ex. 3: find g on the moon.

2r

GMg

2

11 222

6 2

6.67 10 7.35 10

(1.74 10 )

Nm kgkg

gm

kgNg 62.1

Ex 4: Find the Sun’s

gravitational field here

2r

GMg

2

11 302

11 2

6.67 10 1.98 10

(1.50 10 )

Nm kgkg

gm

kgNg 0059.0

Do you feel lighter? • How much difference would a 85kg

student feel from noon to midnight?

mgFg

NFg 5.0

85 0.0059gNF kgkg

Solve for Mass

• Ex 1: Find the mass of the Earth

required to exert a force of 2.0x1020 N

on the Moon

2r

GMmFg

Gm

rFM

g

2

kgM 24100.6

Gravitational Definitions • This is where most people mix up

problems

• Mass, Force or Field?

– 25N

– Weight

– 5kg

– 9.8N/kg

– Apparent Weight

– Gravitational acceleration

– 25lb!?

– 1.63m/s2

Did it really

happen like that?

• Mostly true story but...

• 2nd law actually written in terms of "impulse":

Impulse

defined as a

change in

momentum

We use momentum to analyze collisions…

MOMENTUM IS THE PRODUCT

OF MASS AND VELOCITY

p=m·v

What is the momentum of a 120 kg

rugby player running at 11 m/s?

p=mv

p=120kg · 11 m/s

p=1320 kg ·m/s

More angular momentum

• http://www.youtube.com/watch?v=vj5XFK5p38c&fe

ature=youtube_gdata_player

http://www.youtube.com/watch?v=

M1fTD69CYtA&feature=related

Second Law Revisited?

Newton originally wrote his second law in this

form:

Rearrange by

multiplying both

sides by Δt:

Impulse

If t is small, F can be very large

Can we find the average

force Venus applies to this

57g tennis ball?

She returns a 46m/s serve

at 35m/s and the ball only

contacts the racquet for

6.5ms

#1-7 p. 89-90

Start with #1,2,5,7

FIAQD

Ex: #2 find Impulse

tF

sN 001.055

#1-7 p. 89-90

Start with #1,2,5,7

FIAQD

Ex: #5 find Impulse

vm

s

ms

mkg 6.125.2510

Principle of Momentum Conservation:

Total momentum

before is equal to

total momentum

after the collision

This is true for all

closed, isolated

systems

No

one allowed

in or out

No net

external

force

Chabal: timing is everything

Can we use momentum to analyze

this collision?

Momentum is conserved so p before

must equal p after

Rokocoko has a

mass of 105 kg and

runs at 7.5 m/s into

92 kg Betson

How fast are they

moving after the

collision?

Tm

umv 11

Most common physics 11 collision: moving object

collides with stationary one; they move off

together

Ex: a student with a mass of 105 kg runs at 7.5

m/s into a 92 kg classmate, find v

Same steps every time!

Variables

m1=

u1=

m2=

u2=

v'=

pi=pf

m1u1= (m1 + m2)v?

0 = m1v1+ m2v2?

• Explosions

For an explosion? Ex: bottle rocket

Variables

m1=0.320 kg

v1=?

m2=0.850 kg

v2=-25 m/s

pi=pf

0 = m1v1+ m2v2

m1v1 =- m2v2

v1=- m2v2/m1

v1=-0.850kg(-25m/s)/0.320kg

v1= 66 m/s

Practice Problems

• Finish q's 1-7p.89-90

• Start Ch 4 review p. 90

• Force quiz tomorrow

• Momentum Quiz Friday

• Test next week

• Start bottle rocket design

Ex: question 3

Variables

m1=0.060 kg

v1=600 m/s

m2=3.0 kg

V2=? m/s

pi=pf

0 = m1v1+ m2v2

m2v2 =- m1v1

v2=- m1v1/m2

v2=-0.060kg(600m/s)/3.0kg

v2= -12 m/s

Ex: question 6

Variables

m1=1500 kg

u1=44 m/s

m2=1000 kg

u2=-22 m/s

mT=2500 kg

v=? m/s

pi=pf

m1u1+ m2u2 = mTv

(m1u1+ m2u2)/mT= v

v=18 m/s

Ex: question 13

Variables

m1=0.250 kg

u1=5.0 m/s

m2=0.300 kg

u2=2.0m/s

mT=0.550 kg

v=? m/s

p=mv

m1u1+ m2u2 = mTv

(m1u1+ m2u2)/mT= v

v=3.4 m/s

Elastic vs. Inelastic?

• An elastic collision is

one where

mechanical energy is

conserved

• This can be

represented by a

bouncing ball

elastic inelastic completely

inelastic

Elasticity when bouncing

Chapter Review

• Finish p. 89-90 #1-7 Finish Chapter Review p. 90 q's 1-18

Momentum Quiz Friday

Chapter 4 Test next week

Bonus! Find the velocity of the

wreckage after this horrible

collision :(

Chapter Review

Finish p. 91 #10-18

Start “Test Yourself” p. 92

Momentum Quiz Today