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

Options for Tomorrow

• Questions, bottle rocket (don’t need help)

How do forces affect motion? • If it takes a force to accelerate an object,

what happens if we change the force? Or

the mass?

• Write a hypothesis:

• If we increase _____, then acceleration

will _________

Lab p. 76-79

Write out headings, purpose

procedure (refer to text p. 76-9,

then choose one of a-h, specify

equipment used)

Observations: Copy out data

tables, eventually do graphs

Conclusion: state results,

errors, possible improvements

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

-choose stopwatch, iPad, phone or ticker tape

timer

A: Sammy 1.4 m/s2

B: Eric/Alexa 1.8 m/s2

C: Anna/Filippo 1.9 m/s2

D: Preston/Riley 3.7 m/s2

E: Harleen/Dayton 1.11 m/s2

F: Alexia/Camryn 0.66 m/s2

G: Jason/Darrell 0.40 m/s2

H: Ashera/Estela 0.2 m/s2

a = Rise/Run = (1.2-0.2)/*(2.75-1.85)

a = 1.0/0.9 = 1.1 m/s2

Lab p. 76-79: 3 graphs:

Graph 1: your own data velocity vs time to

find acceleration (don’t hand in)

Graph 2: (abcd) acceleration vs force

Graph 3a: (efgh) acceleration vs mass

Graph 3b: (efgh) acceleration vs 1/mass

Conclusions:

State results (proportional? Inversely

proportional?), including the meaning of

each slope (control variable?)

State sources of error and possible

improvements

Using our lab results:

• The acceleration of the cart seems to

be:

– Directly proportional to force

– Inversely proportional to mass

• Can you write a formula that shows how

these factors affect acceleration?

maFnet

• Ex 1: how much unbalanced or “net” force

is necessary to accelerate a 80kg student at

10 m/s2?

280 10 800netmF ma kg N

s

“Net” Force?

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: Griffin applies a 50 N force to a

2.5kg book to slide it right 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 horizontally

at 3.4 m/s2, if there is a 250 N force of

air resistance.

FT Fr

𝐹𝑛𝑒𝑡 = 𝑚𝑎

𝐹𝑇+𝐹𝑟= 𝑚𝑎

𝐹𝑇 = 𝑚𝑎 −𝐹𝑟

𝐹𝑇 = 1200 3.4 − (−250)

𝐹𝑇 = 4330𝑁

Exercises p. 81 #1-6

Start p. 90-91 #1-8

Dynamics Quiz #1

maF a

Fm

4.4

250 kg57

mgF )8.9(57 N560

NF 560

2

560

2

NF N280

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?

• 653N

Ex 4:

• What is the

weight of the

same student

on the moon,

whose

gravitational

field strength is

1/6 earth’s?

• 120N

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

Newton’s 3rd Law

• For every action force F there is an

equal and opposite reaction force R

• When you hit something, it hits back!

• Forces always come in pairs

Reaction Force R

• Newton’s 3rd Law states a supporting surface

pushes back with an equal and opposite force

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

object’s weight

• Because it acts perpendicular, we sometimes call

it “Normal force” FN

• We sometimes refer to Reaction force as the

“apparent weight”

R

Fg

How can we make you feel

weightless?

How would you feel if there was

no Reaction force acting on you?

Ex: object at rest

• What is the Reaction force acting on the

2.5 kg book resting on your desk?

– What forces act on the book?

• Gravity and Reaction 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 N

kg

Extended object at rest

• Ex 3: what is the Reaction force acting on

your book as you lean on it with a 35 N

force?

– What forces act on the book?

• Gravity, Applied and Reaction 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 Reaction force of 1300N

– What forces act on the student?

• Gravity and Reaction force

– Free body diagram

– Apply 2nd law

netF ma

FN

Fg

FN

Fg

maFR g

m

FRa

g

55

)8.9(551300

Na

28.13s

ma

Do Lab 4-3

• P. 85-86

What’s the difference?

Write a hypothesis about R

• What combination of 3 carts will

provide the best push? Why?

Prizes for the greatest distance (video evidence)

Friction

• Once we know Reaction

force, we can calculate

friction

• Friction force can be

calculated as Reaction

force times “mu”

• Ff=μR

R

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

R

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)

RFf

• 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

RFf maFnet

maRFg

0 RFg

NFR g 25

NFf 2555.0

NFf 14

RFf

• 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.

R

Ff

)8.9(2.2

13 60.0

RFf

• Ex: find the force necessary to slide your 2.2

kg wooden block along the table at a constant

speed if coefficient of friction is 0.3.

mgFf

8.9)2.2(3.0fF

NFf 5.6

RFf

• Ex: find the friction coefficient to accelerate

a top fuel dragster to 56 m/s in 1.0s

mgma

g

a =56/9.8 =5.7

Questions

• p. 81 #1-6

• Start #1-10 Ch Rev p. 90-91

• Quiz Thursday

• Bottle Rockets launch Wednesday?

• Gallery Walk Friday

Bottle Rockets

• For marks, submit:

– Video of launch to gsjokvist@gedu.sd73.bc.ca

– Paper with

• Group names

• Diagram of bottle design

• Estimate acceleration from ratio of time of thrust to

time remaining until maximum height

• Calculation of thrust force from F=ma

Why does a Bugatti Veyron have

the same drag coefficient as a

Cadillac Escalade?

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:

RF sf

• 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”

– Ex. Find the friction

force acting on your

textbook as it rests on

your desk*

RF sf

• Ex 6: for what applied force if s=0.65 does your textbook start moving? – What will the acceleration be?

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)

• Finish up to #6 p. 81

– F=ma gives μR=ma

Law of

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

Gravity PhET Lab

• Use the simulation to graph the

relationship between F and one of the

other variables

• Specify dependent, independent, and

control variables

• Ex: F dependent, r independent, M1 and

M2 control

Law of 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’s “Weigh

the Earth” Experiment

• Find G if Cavendish calculated the force

between the 2.5kg mass and the 0.55kg

mass (when their centers are 10.0 cm apart)

to be:

NFg

9102.9

2r

GMmFg

Mm

rFG

g

2

kgkg

mN

55.05.2

1.0102.929

2

2111069.6kg

NmG

• Finally! What is the Earth’s mass?

• Note: 1kg of mass has a weight of 9.8N

and Earth’s radius is 6380 km

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 Luke (74

kg) and his

lightsaber (0.5 kg) if

they are 1.5 m

apart. 2r

GMmFg

2r

GMmFg

2

11

5.1

1)74(1067.6 gF

nNFg 2.2

• Try #5-6 p. 66-67

**What happens to force if

r is doubled? Tripled?

2r

GMmFg

Start Exercises

• P. 66-67 questions 5-6

• *part 5 consider what happens when we replace 1 Earth radius with 2 Earth radii?

211 er

GMmF

222 er

GMmF

14

1F

What about g?

• How does gravitational field strength “g”

fit into all of this?

Ex. 3: find g on the Earth.

g=𝐺𝑀

𝑟2

g=6.67𝐸−11(5.98𝐸24)

(6.38𝐸6)2

g=9.8𝑚

𝑠2

Ex. 3: find g on the moon.

2

11 222

6 2

6.67 10 7.35 10

(1.74 10 )

Nm kgkg

gm

kgNg 62.1

2r

GMg

Ex. 3: find g on Saturn.

2r

GMg

27

2611

1083.5

1068.51067.6

g

kgNg /1.11

Ex.: find g on the Sun.

2r

GMg

g=6.67𝐸−11(1.98𝐸30)

(6.95𝐸8)2

g=273𝑚/𝑠2

Start #1-5 p. 82-83

% Difference = (A-B)/A x 100%

• Ex 4: find g at an

altitude of 130 km.

Careful!

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

2r

GMg

Ex 4: Find the Sun’s

gravitational field here (8 1/3

light minutes away)

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 kg

kg

• Ex 1: Find the mass of the

Earth required to exert a force

of 2.0x1020 N on the Moon,

(1.28 light seconds away)

2r

GMmFg

Gm

rFM

g

2

kgM 24100.6

2211

2820

1035.71067.6

1084.3100.2

Black hole Sun?

• Find the radius of the event horizon of a

Black hole Sun if g=1.5x1013 N/kg

2r

GMg r =

𝐺𝑀

𝑔

r = 6.67x10_11(1.98x1030)

1.5x1013 =2950m

Start Ch. Rev • P. 90 #1-10

Black hole Sun?

• Find the gravitational field 2.5 km from a

Black hole Sun

2r

GMg 2

3011

2500

)1098.1(1067.6

g

kgNg /10113.2 13

Spaghettification factor?

Gravitational field at your head: 2.501 km away

2r

GMg

23011

2501

)1098.1(1067.6

g

kgNg /10111.2 13

Difference:

kgNFS /10111.2113.2.. 13

kgNFS /102.. 10

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

Search: “Gravity and orbits PhET” • Can we transfer our understanding of

the concept of proportionality to other

situations?

• How does gravity change when we:

– Increase the Sun’s mass?

– Decrease orbital radius?

Search: “Gravity PhET” 1. Calculate force for default, compare

2. What is the smallest gravitational force we

can get for this simulation? Calculate:

a) Masses of 1 kg each, distance between

centres of mass of …

b) Compare to force from simulation of 10-12 N

3. Largest?

a) Masses of 1000 kg each, distance between

centres of mass of …

F=𝐺𝑀𝑚

𝑟2

F=

6.67𝐸−11(50)(200)

42

• Chapter 3 Review

• #1-9 p. 66-67

• Test Yourself

• #1-10 p. 69-70

• Chapter 4 Review

• #1-10 p. 90-91

• Test Yourself

• #1-6 p. 92, MC #1-14 p. 92-94

Chapter Review

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

Test Yourself p. 92-92 1-18

Make your own questions

Watch some videos

Review sheet

Dynamics Test MC Friday/ WR Monday?

Chapter Review

Finish p. 91 #10-18

Start “Test Yourself” p. 92

Momentum Quiz Today

Review Project: make an app on:

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