i dNewton’s Laws - continued

Friction Inclined Planes N T LFriction, Inclined Planes, N.T.L.

TWO types of Frictionyp

Static – Friction that keeps an object at rest and prevents it from movingKinetic – Friction that acts during motion

Force of Friction

FF αThe Force of Friction is directly related to the F N l

Nf FF

µ

α

= alityproportion ofconstant Force Normal.

Mostly due to the fact that BOTH are surface Nssf FF µ

µ=

= frictionoft coefficienThe coefficient of f i i i i lthat BOTH are surface

forcesNkkf FF µ=

friction is a unitless constant that is specific to the

t i l t dmaterial type and usually less than one.

Note: Friction ONLY depends on the MATERIALS sliding against each other, NOT on surface area.

ExamplepA 1500 N crate is being pushed

across a level floor at aa) What is the coefficient of kinetic f i ti b t th t d thacross a level floor at a

constant speed by a force F of 600 N at an angle of 20° below the horizontal as shown in the

friction between the crate and the floor?

the horizontal as shown in the figure.

FNFa

82563)20(600

= Nkf

NFFF

FF

θ

µN

20Fay

2117051500)20( i600

1500sin

82.563)20(cos600cos

+

+=+=

====

aayN

aaxf

NF

FmgFF

NFFF

θ

θ

Fax

331021.170582.563

21.17051500)20(sin600

==

=+=

k

N NF

µµ

mgFf

331.0=kµ

ExamplepIf the 600 N force is instead pulling the

block at an angle of 20° above the horizontal as shown in the figure

FN

20

Fa

Fayhorizontal as shown in the figure, what will be the acceleration of the crate. Assume that the coefficient of friction is the same as found in (a)

20

Fax

friction is the same as found in (a)

mgFf

maFFmaFNet =

)i(cos maFF

maFF

Na

fax

=−

=−

θθµθ

1.153)20sin6001500(331.020cos600)sin(cos

amaFmgF aa

=−−=−− θµθ

2/883.01.15357.4288.563

smaa

=

=−

Inclines

FNFf

θθ

θθ

θcosmg θ

Ti

θsinmg

mg θTips•Rotate Axis•Break weight into components•Write equations of motion orθsinmg •Write equations of motion or equilibrium•Solve

ExamplepMasses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley. As shown in the diagram, m1 is held at rest on the floor and m2 rests on a fixed incline of angle 40 degrees. The masses are released from rest, and m2 slides

F

1.00 m down the incline in 4 seconds. Determine (a) The acceleration of each mass (b) The coefficient of kinetic friction and (c) the tension in the string.

gmamTamgmTmaFNET

1111 +=→=−=FNT

Fm gcos40

m g

T

Ff

40 amTFgm f 22 )(sin =+−θm2gcos40

m1

m2g40

m2gsin40

m1g

Example maFNET =pgmamTamgmT 1111 +=→=−

amTFgm f 22 )(sin =+−θ

2

2

)4(2101

21

a

attvx ox

+=

+= NT 7.39)8.9(4)125(.4 =+=

θsin amTFgm =

2/125.0

)4(201

sma

a

=

+

µθ

θ

θ

sin

)(sin

sin

2112

2112

22

amgmamFgm

amgmamFgm

amTFgm

Nk

f

f

=−−−

=+−−

=−−235.0

57.67125.12.395.07.56

=−−−

=kµ

θµθθµθ

µθ

cossincossin

sin

22112

21122

2112

gmamgmamgmamgmamgmgm

amgmamFgm

k

k

Nk

=−−−=−−−

θθµ

cossin

2

2112

gmamgmamgm

k−−−

=

Newton’s Third Law“For every action there is an EQUAL and

OPPOSITE reactionOPPOSITE reaction.This law focuses on action/reaction pairs (forces)They NEVER cancel outThey NEVER cancel out

All you do is SWITCH the wording!you do s S C t e o d g•PERSON on WALL•WALL on PERSON

N.T.LThis figure shows the force during a collision between a truck and a train. You can clearly see the forces are EQUAL and OPPOSITE. To help you understand the law better, look at this situation from the point of view of Newton’s Second Law.

FF

TrainTrainTruckTruck

TrainTruck

aMAmFF

==

TrainTrainTruckTruck

There is a balance between the mass and acceleration. One object usually h LARGE MASS d SMALL ACCELERATION hil h h hhas a LARGE MASS and a SMALL ACCELERATION, while the other has a SMALL MASS (comparatively) and a LARGE ACCELERATION.

N.T.L ExamplespAction: HAMMER HITS NAILReaction: NAIL HITS HAMMERReaction: NAIL HITS HAMMER

Action: Earth pulls on YOUReaction: YOU pull on the earth

Newton’s Law of GravitationWhat causes YOU to be pulled down? THE EARTH….or

more specifically…the EARTH’S MASS. Anything thatmore specifically…the EARTH S MASS. Anything that has MASS has a gravitational pull towards it.

MmFgαgWhat the proportionality above is saying is that for there to be asaying is that for there to be a FORCE DUE TO GRAVITY on something there must be at least 2 masses involved where one ismasses involved, where one is larger than the other.

N.L.o.G.As you move AWAY from the earth, your DISTANCE increases and your FORCE DUE TO GRAVITY decrease. This is a special INVERSE relationship called an Inverse-Square.

2

1Fg α 2rg

The “r” stands for SEPARATION DISTANCE and is the distance between the CENTERS OF MASS of the 2 objects. We us the symbol “r” as it symbolizes the radius. Gravitation is closely related to circular motion as you will discover later.

N.L.o.G – Putting it all togetherg g

221

rmmFg α

Constant nalGravitatio UniversalGalityproportion ofconstant G

r

==

22271067.6

mmkg

NmxG = −

221

rmmGFg =

ththhthiUF

earth eLEAVING th areyou when thisUse

earthon theareyou when thisUse

221 →=

→=

rmmGF

mgF

g

g

N.L.o.GLet’s set the 2 equations equal to each other since they BOTH represent your weight or force due to gravity

thLEAVING thhthiU

earth on the areyou when thisUse

21 →

→=

mmGF

mgFg

p y g g y

eartheLEAVING thareyou when thisUse221 →=

rGFg

Mm SOLVE FOR g!

MGg

rMmGmg

=

= 2

SOLVE FOR g!

226

2427

/81.9)1097.5)(1067.6( smxxg ==−

kgxMr

Gg

−==

=

6

24

2

1097.5Earth theof Mass

26 /81.9)1037.6(

smx

g

mxr −== 61037.6Earth theof radius

An interesting friction/calc problem…!F

g p

Suppose you had a 30 kg box that

FN

Suppose you had a 30- kg box that is moving at a constant speed until it hits a patch of sticky snow where it experiences a frictional force of 12N

Ff

frictional force of 12N.

a) What is the acceleration of the box?

mg

the box?

b) What is the coefficient of kinetic friction between the b d h ? =→=

=amaF

maFNet

3012box and the snow?

=

=→=

a

amaFf 3012== kNkf mgFF µµ 0.4 m/s/s

==

k

k

µµ )8.9)(30(12

0.04kµ

Optional!!-The “not so much fun” begins….Now suppose your friend decides to help by pulling the box

across the snow using a rope that is at some angle from the horizontal She begins by experimenting with the angle of pullhorizontal. She begins by experimenting with the angle of pulland decides that 40 degrees is NOT optimal. At what angle, θ, will the minimum force be required to pull the sled with a constant velocity?constant velocity?

Let’s start by making a function for “F” in terms of “theta” using our equations of motion. θθ sinsin NN FmgFmgFF −=→=+

FN F θµθµθ

)sin(coscos

K

NkF

FmgFFFF−=

==

Fsinθ

Ff

θ

µθµθµθµθ

θµµθ

)sin(cossincos

sincos

kk

kk

mgFmgFF

FmgF

+=+

−=

Fcosθ

Fsinθ

mgθµθ

µθ

µθµθ

sincos)(

)sin(cos

k

k

kk

mgF

mgF

+=

=+

θµθ sincos k+

Optional-What does this graph look p g plike?

θθµθ

i)( kmgF =

Theta Force

1 11.7536

10 11.8579θµθ sincos k+ 10 11.8579

20 12.3351

30 13.2728

40 14.8531F Th t 50 17.4629

60 21.9961

70 30.9793

Force vs. Theta

30

35

15

20

25

orce

(N) Does this graph tell

you the angle needed f i i f ?

5

10

15Fo for a minimum force?

00 10 20 30 40 50 60 70 80

Theta (degrees)

Optional-What does this graph look like?theta Force

0.5 11.7563

1 11 7536 θµθµθ

sincos)( kmgF

+=

1 11.7536

1.5 11.7517

2 11.7508

2.5 11.7507

θµθ sincos k+

Force vs. Theta

3 11.7515

3.5 11.7532

4 11.7558

4 5 11 759311.762

11.764

11.766

4.5 11.7593

5 11.7638

11 756

11.758

11.76

Forc

e (N

)

Could this graph tell

11.752

11.754

11.756Could this graph tell you the angle needed for a minimum force?

11.750 1 2 3 4 5 6

Theta (degrees)

What do you notice about the SLOPE at this minimum force?

Optional-Taking the derivativep gHere is the point. If we find the

derivative of the function Force vs. Thetaderivative of the function and set the derivative equal to ZERO, we can find the ANGLE t thi i i

11.764

11.766

ANGLE at this minimum. Remember that the derivative is the SLOPE of 11.758

11.76

11.762

rce

(N)

the tangent line. The tangent line’s slope is zero at the minimum force and 11 752

11.754

11.756For

at the minimum force and thus can be used to determine the angle we

11.75

11.752

0 1 2 3 4 5 6

Theta (degrees)

need.Theta (degrees)

This tells us that our minimum force is somewhere between 2 & 3 degrees.

Optional-Taking the derivative using the Chain Rule

θµθµθ mgF

k

k

sincos)(

+=

θµθµθµθµµ

mgdmgd

dF kkk

k

k

))sin(cos()

sincos( 1−+

=+

=θθθ ddd

==

)16()3(cos)()3sin()(

2

2

xxxxfxxxf

++=′

+=

)()()(f

Derivative ofoutside function

Leave insidefunction alone

Derivative ofinside function

)3cos()16()( 2 xxxxf ++=′

Optional-Taking the derivative using the Chain Rule1))sin(cos(

θθµθµ

θkk

dmgd

ddF +

=−

2 functionoutsideofderivative)sincos1(

constants

θµθ

µθθ

kmgdd

+

=−

f ii idfd i iialonefunction inside theleaving as wellas

functionoutsideof derivative)sincos1(-

θθ

θµθ k =+

)cossin()(

functioninside ofderivativecossinθµθµθ

θµθ

kk

k

mgF +−−=′

=+−

2)sincos()(

θµθθ

k

F+

Now we set the derivative equal to ZERO and solve for theta!Now we set the derivative equal to ZERO and solve for theta!

Optional-Setting the derivative equal to zero

++−−

=′θµθ

θµθµθ)sincos(

)cossin()( 2kkmgF

+−−=

+θµθµθµθ

)cossin(0

)sincos(

kk

k

mg

+−=+

=

θµθθµθ

cossin0)sincos(

0 2

k

k

=+

θµθθµθ

cossincossin0

k

k

→=

=−− µθ

µθ

)04.0(tan)(tan

tan11

k

k

=→

θµθ )04.0(tan)(tan k

2.29°

NEWTON’S FIRST LAW ANDPHYSICAL THERAPYPHYSICAL THERAPY

PINCHED NERVESThe neck is a complex place when it The neck is a complex place when it comes to nerves, and neck pain can stem from all manner of things, including a pinched nerve in the g pshoulder. Nerves can become trapped in the shoulder itself, due to the cramped conditions in an area called the brachial plexus. The brachial plexus has a number of cervical nerves travelling through it to the upper limbs.

Note: Pinched shoulder nerves Note: Pinched shoulder nerves prevent people from lifting their arm.

TREATMENTS?Ice – This will reduce inflammation.

What if ice does not work?

Physical Therapy Routine - TRACTION

CERVICAL TRACTIONO t h i d b h i l th i t One technique used by physical therapist

and chiropractors to provide pain relief and improve motion is cervical traction. Traction gently extends the neck g yopening the spaces between the cervical vertebrae and temporarily alleviating pressure on the affected discs. Neck traction can either be done continuously traction can either be done continuously or intermittently, alternating between short periods of pulling and resting.

It’s also possible to do cervical traction at home such as the device shown here. There are pulley systems that you can There are pulley systems that you can hook up to a doorway, or devices that will enable you to perform cervical traction while lying down.

PHYSICS APPLICATION

Cervical traction works based on the idea of EQUILIBRIUM .

netF =∑ 0T

waterbagNeck mgT =∑

T

mg

mg