chapter 5 using newton’s laws: friction, circular …...5-1 applications of newton’s laws...

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Lecture PowerPoints

Chapter 5

Physics for Scientists &Engineers, with Modern

Physics, 4th edition

Giancoli


Chapter 5

Using Newton’s Laws: Friction,Circular Motion, Drag Forces

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Units of Chapter 5

• Applications of Newton’s Laws Involving Friction

• Uniform Circular Motion—Kinematics

• Dynamics of Uniform Circular Motion

• Highway Curves: Banked and Unbanked


Units of Chapter 5

Which path willthe ball take ifyou let go of thestring?

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5-1 Applications of Newton’s LawsInvolving Friction

Friction is always present when two solidsurfaces slide along each other.

The microscopic detailsare not yet fullyunderstood.



Sliding friction is called kinetic friction.

Approximation of the frictional force:

Ffr = μkFN .

Here, FN is the normal force, and μk is thecoefficient of kinetic friction, which isdifferent for each pair of surfaces.

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Static friction applies when two surfacesare at rest with respect to each other(such as a book sitting on a table).

The static frictional force is as big as itneeds to be to prevent slipping, up to amaximum value.

Ffr ≤ μsFN .

Usually it is easier to keep an objectsliding than it is to get it started.

Generally S > K



Note that, in general, μs > μk.

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Example 5-1: Friction: static and kinetic.Our 10.0-kg mystery box rests on a horizontal floor. The coefficient ofstatic friction is 0.40 and the coefficient of kinetic friction is 0.30.Determine the force of friction acting on the box if a horizontal externalapplied force is exerted on it of magnitude:(a) 0, (b) 10 N, (c) 20 N, (d) 38 N, and (e) 40 N.If the box moves what is its acceleration?

As the applied force is increased inmagnitude, the force of static frictionincreases linearly to just match it, untilthe applied force equals μSFN.

If the applied force increases further,the object will begin to move, and thefriction force drops to a roughlyconstant value characteristic of kineticfriction.



Since there is no vertical motionFN = mg in magnitude.

FN = 10.0G9.80 = 98.0 N.

The maximum static frictional forceFfr = SFN = 0.4G98.0 = 39.2 N

If the applied force is 39.2 N thereis no motion and Ffr = FA inmagnitude

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For cases (a), (b), (c) & (d) theapplied force is 39.2 N there is nomotion and Ffr = FA in magnitude



2

frA

NKfr

frA

m/s1.0610.0

10.6

m

ΣFamaΣF

N 10.629.440.0FFΣF

N 29.498.00.3FμF

motiondaccelerateF(static)F(e)

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5-1 Applications of Newton’s Laws Involving Friction

Conceptual Example 5-2: A box against a wall.

You can hold a box against a roughwall and prevent it from slipping downby pressing hard horizontally. Howdoes the application of a horizontalforce keep an object from movingvertically?

Horizontal forces cancel: F = FN

Provided Ffr mg (Ffr SFN)the box will not slide downwards



Example 5-3: Pulling against friction.

A 10.0-kg box is pulled along a horizontalsurface by a force of 40.0 N applied at a 30.0°angle above horizontal. The coefficient ofkinetic friction is 0.30. Calculate theacceleration.

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A 10.0-kg box is pulled along a horizontalsurface by a force of 40.0 N applied at a 30.0°angle above horizontal. The coefficient ofkinetic friction is 0.30. Calculate theacceleration.

Vertically: FPsin +FN = mg

40.0sin30 + FN = 10.0G9.80

20.0 + FN = 98.0

FN = 98.0 - 20.0 = 78.0 N



A 10.0-kg box is pulled along ahorizontal surface by a force of 40.0N applied at a 30.0° angle abovehorizontal. The coefficient of kineticfriction is 0.30. Calculate theacceleration.

Horizontally: FPcos = 40.0cos30 = 34.6 N

Fx = 34.6 - Ffr = 34.6 - KFN = 34.6 - 0.3G78.0 = 11.2 N

Fx = ma a = Fx/m = 11.2/10.0 = 1.1 m/s2

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Conceptual Example 5-4: To push or to pull a sled?

Your little sister wants a ride onher sled. If you are on flatground, will you exert lessforce if you push her or pullher? Assume the same angle θin each case.

Pulling decreases thenormal force, whilepushing increases it.Better pull.



Example 5-5: Two boxes and a pulley.Two boxes are connected by a cordrunning over a pulley. The coefficient ofkinetic friction between box A and thetable is 0.20. We ignore the mass of thecord and pulley and any friction in thepulley, which means we can assume that aforce applied to one end of the cord willhave the same magnitude at the other end.We wish to find the acceleration, a, of thesystem, which will have the samemagnitude for both boxes assuming thecord doesn’t stretch. As box B movesdown, box A moves to the right.

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Example 5-5: Two boxes and a pulley.

NB: Both masses have the same magnitude ofacceleration.

A moves to the right and B moves downwards.

Taking the +ve horizontal (x) direction toRIGHT we must take +ve vertical (y) directionDOWNWARDS.

Kinetic Frictional Force: Ffr = KFN = KmAg




Box A: No vertical motion as FN = mAg

Horizontally:

Fx = FT - Ffr = FT - KmAg = mAa FT = KmAg + mAa - (1)

Box B: Vertically: Fy = mBg - FT = mBa FT = mBg - mBa - (2)

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Equating (1) & (2)

KmAg + mAa = mBg - mBa mAa + mBa = mBg - KmAg a(mA + mB) = g(mB - KmA)

2

BA

AKB

m/s1.42.05.0

5.0)0.29.8(2.0

mm

mμmga



Example 5-6: The skier.

This skier is descending a30° slope, at constantspeed. What can you sayabout the coefficient ofkinetic friction?

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0.58tantancos

sin

cossin

0sin

cos&

speed)(constant0cos

cos

sin

30θθ

θμ

0θmgμθmg

maFμθmgΣF

θmgF

maθmgFΣF

θ-mgF

θmgF

mgF

k

k

xNkx

N

yNy

Gy

Gx

G



Example 5-7: A ramp, a pulley, and two boxes.Box A, of mass 10.0 kg, rests on a surface inclined at 37° to thehorizontal. It is connected by a lightweight cord, which passesover a massless and frictionless pulley, to a second box B, whichhangs freely as shown. (a) If the coefficient of static friction is0.40, determine what range of values for mass B will keep thesystem at rest. (b) If the coefficient of kinetic friction is 0.30, andmB = 10.0 kg, determine the acceleration of the system.

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Example 5-7: A ramp, a pulley, and two boxes.

(a) If the coefficient of static friction is 0.40, determine what range ofvalues for mass B will keep the system at rest.

If there is no motion then the tension in the cord FT = mBgand friction between box A and the surface is sufficient toprevent motion.

The frictional force always opposes any possible motionso two cases need to be considered:




Case i: mB is small Ffr must act up slope to preventmotion

Case ii: mB is large Ffr must act down slope to preventmotion

Ffr SFN = SmAgcos

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Case i: Max Ffr corresponds to min mB

Fx = mAgsin - FT – Ffr = 0= mAgsin - mBg – SmAgcos = 0

mB = mA(sin - S cos)

= 10.0(sin37 - 0.40cos37) = 2.8 kg

If mB < 2.8 kg box will slide down slope




Case ii: Max Ffr corresponds to max mB

Fx = mAgsin + Ffr - FT = 0= mAgsin + SmAgcos - mBg = 0

mB = mA(sin + S cos)

= 10.0(sin37 + 0.40cos37) = 9.2 kg

If mB > 9.2 kg box will slide up slope

Answer: For system at rest: 2.8 kg < mB < 9.2 kg

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Example 5-7: A ramp, a pulley, and two boxes.(b) If the coefficient of kinetic friction is 0.30, and mB = 10.0 kg,determine the acceleration of the system.

With mB = 10 kg box A will slide up slope so weneed to use free body diagram for case ii.

Both boxes have same acceleration.

NB: Once boxes accelerate FT mBg



Consider Box A:FA = mAa = FT - mAgsin - KFN

mAa = FT - mAgsin - K mAgcos

Consider Box B:FB = mBa = mBg - FT

FT = mBg – mBa

Substitute for FT into box A equation:

mAa = mBg – mBa - mAgsin - K mAgcos

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mAa = mBg – mBa - mAgsin - K mAgcos

Solve for a:

2

BA

AKAB

AKABBA

m/s0.78

10.010.0

3710.00.3371010.09.80

mm

θmμθmmga

θmμθmmgmma

cossin

cossin

cossin


5-2 Uniform Circular Motion—Kinematics

Uniform circular motion: motion in a circle ofconstant radius at constant speed

Instantaneous velocity is always tangent to thecircle.

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Looking at the change in velocity in the limit thatthe time interval becomes infinitesimally small,we see that



This acceleration is called the centripetal, orradial (aR), acceleration, and it points towardthe center of the circle.


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Looking at the change in velocity in the limit thatthe time interval becomes infinitesimally small,we see that


r

va

vttr

v

t

va

r

vv

rv

v

2

ttt

R

0Δ0Δ0ΔR

Δ

Δlim

Δ

Δlim

Δ

Δlim

ΔΔΔΔ

ll

ll


This acceleration is called the centripetal, orradial (aR), acceleration, and it points towardthe center of the circle.


r

va

a

2

R

R

21 aa

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Example 5-8: Acceleration of a revolving ball.

A 150 g ball at the end of a string is revolvinguniformly in a horizontal circle of radius 0.600 m.The ball makes 2.00 revolutions in a second. Whatis its centripetal acceleration?

22

2

R

m/s94.76.0

54.7

r

va

m/s 7.540.64πr4πr2π2

nce/scircumfere2.00speedlinear

rev/s2.00speedangular


5-2 Uniform Circular Motion—KinematicsExample 5-9: Moon’s centripetal acceleration.

The Moon’s nearly circular orbit about the Earthhas a radius of about 384,000 km and a period T of27.3 days. Determine the acceleration of the Moontoward the Earth.

23

8

23

2

R

m/s1072.2

1084.3

1002.1

r

va

r = 384 000 km = 3.84108 mc = 2r = 23.84108 = 2.41109 m

T = 27.3 d = 27.3246060 = 2.36106 s

m/s101.02102.36

102.41 3

6

9

T

cv

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A centrifuge works byspinning very fast. Thismeans there must be avery large centripetalforce. The object at Awould go in a straightline but for this force; asit is, it winds up at B.




Example 5-10: Ultracentrifuge.

The rotor of an ultracentrifuge rotates at 50,000rpm (revolutions per minute). A particle at the topof a test tube is 6.00 cm from the rotation axis.Calculate its centripetal acceleration, in “g’s.”

sg'101.6780.9

1064.1

m/s1064.1

1000.6

314

56

26

2

2

2

R

r

va

r = 6.00 cm = 6.0010-2 mc = 2r = 26.0010-2 = 0.377 m50000 rpm = 50000/60 = 833 rev/sT = 1/833 = 1.2010-3 s

m/s3141020.1

377.03-

T

cv

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5-3 Dynamics of Uniform Circular Motion

For an object to be in uniform circular motion,there must be a net force acting on it.

We already know theacceleration, so canimmediately write theforce:



We can see that theforce must be inwardby thinking about aball on a string.Strings only pull;they never push.

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There is no centrifugal force pointing outward;what happens is that the natural tendency of theobject to move in a straight line must beovercome.

If the centripetal force vanishes, the object fliesoff at a tangent to the circle.


Example 5-11: Force on revolving ball (horizontal).

Estimate the force a person must exert on a string attachedto a 0.150 kg ball to make the ball revolve in a horizontalcircle of radius 0.600 m. The ball makes 2.00 revolutions persecond. Ignore the string’s mass.

N14

6.0

54.7150.0

:ForceipetalCentr

2

2

C

RC

r

vmF

maF m = 0.150 kgr = 0.6 mc = 2r = 20.6 = 3.77 m2.00 rev/sT = 1/2.00 = 0.500 s

m/s54.7500.0

77.3

T

cv

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Example 5-12: Revolving ball (vertical circle).

A 0.150 kg ball on the end of a1.10 m long cord (negligiblemass) is swung in a verticalcircle. (a) Determine theminimum speed the ball musthave at the top of its arc sothat the ball continuesmoving in a circle. (b)Calculate the tension in thecord at the bottom of the arc,assuming the ball is movingat twice the speed of part (a).



A 0.150 kg ball on the end of a 1.10 mlong cord (negligible mass) is swungin a vertical circle. (a) Determine theminimum speed the ball must have atthe top of its arc so that the ballcontinues moving in a circle.

At point 1: Centripetal force FC = FT1 + mg = mv2/r ( +ve)For circular motion there must be a tension in the string.If FT1 = 0 there will be insufficient centripetal forcefor circular motion.

m/s3.281.109.80

:0For2

CT1

grv

r

mvmgFF

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A 0.150 kg ball on the end of a 1.10 mlong cord (negligible mass) is swung ina vertical circle. (b) Calculate thetension in the cord at the bottom of thearc, assuming the ball is moving attwice the speed of part (a).

At point 2: v = 23.28 = 6.56 m/s ( +ve)

N7.341.10

6.569.800.150

2

2

T2

2

T2C

r

vgmF

r

mvmgFF



Example 5-13: Conical pendulum.

A small ball of mass m,suspended by a cord oflength l, revolves in a circle ofradius r = l sin θ, where θ isthe angle the string makeswith the vertical. (a) In whatdirection is the accelerationof the ball, and what causesthe acceleration? (b)Calculate the speed andperiod (time required for onerevolution) of the ball in termsof l, θ, g, and m.

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A small ball of mass m,suspended by a cord of length l,revolves in a circle of radius r = lsin θ, where θ is the angle thestring makes with the vertical.

(a) In what direction is theacceleration of the ball, and whatcauses the acceleration?

The acceleration is the radial acceleration requiredto keep the ball moving in a circular path and isprovided by the horizontal component FT sin ofthe tension in the cord.



(b) Calculate the speed and period (timerequired for one revolution) of the ball interms of l, θ, g, and m.

θθlgcosθ

θlgv

cosθ

θlgv

l

mvmg

l

mvF

r

mvθFF

θ

mgF

mgθFF

2

tansinsin

sin

sincos

sin

sinΣ :lyHorizontal

cos

0cosΣ :Vertically

2

2

2

2

2

2

T

2

Tx

T

Ty

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(b) Calculate the speed and period (timerequired for one revolution) of the ball interms of l, θ, g, and m.

g

l

lg

l

v

rT

T

rv

cos2

cos

sin

sin2222


5-4 Highway Curves: Banked and Unbanked

When a car goes around a curve, there must bea net force toward the center of the circle ofwhich the curve is an arc. If the road is flat, thatforce is supplied by friction.

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If the frictional force isinsufficient, the car willtend to move morenearly in a straight line,as the skid marks show.



As long as the tires do not slip, the friction isstatic. If the tires do start to slip, the friction iskinetic, which is bad in two ways:

1. The kinetic frictional force is smaller than thestatic.

2. The static frictional force can point toward thecenter of the circle, but the kinetic frictionalforce opposes the direction of motion, makingit very difficult to regain control of the car andcontinue around the curve.

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Example 5-14: Skidding on a curve.

A 1000 kg car rounds a curve on a flatroad of radius 50 m at a speed of 15 m/s(54 km/h). Will the car follow the curve, orwill it skid? Assume:

(a) the pavement is dry and the coefficientof static friction is μs = 0.60;

(b) the pavement is icy and μs = 0.25.




(a) the pavement is dry andthe coefficient of staticfriction is μs = 0.60;

curvefollowcar willm/s15as

m/s17.19.8500.6

0.6:PavementDry

max

Smax

SC

Sfr(max)NSfr

S

v

rgμv

mgμr

mvF

mgμFFμF

μ

2max

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skidcar willm/s1.11as

m/s1.119.8500.25

0.25:PavementIcy

max

Smax

SC

Sfr(max)NSfr

S

v

rgμv

mgμr

mvF

mgμFFμF

μ

2max

(b) the pavement is icy andμs = 0.25.



Banking the curve can help keepcars from skidding. In fact, forevery banked curve, there is onespeed at which the entirecentripetal force is supplied by the

horizontal component ofthe normal force, and nofriction is required. Thisoccurs when:

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Example 5-15: Banking angle.

(a) For a car traveling with speed v around acurve of radius r, determine a formula for theangle at which a road should be banked sothat no friction is required.

(b) What is this angle for an expressway off-ramp curve of radius 50 m at a design speedof 50 km/h?



(a) For a car traveling with speed v around a curveof radius r, determine a formula for the angle atwhich a road should be banked so that no frictionis required.

rg

vtanθ

θmgθ

θmg

r

mvθF

θ

mgFmgθF

2

2

N

NN

tancos

sinsin:lyHorizontal

coscos:Vertically

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(b) What is this angle for an expressway off-rampcurve of radius 50 m at a design speed of 50 km/h?

m/s13.896060

1000505.21

km/h508.950

89.13tantan

21

21

v

rg

v


Summary of Chapter 5

• Kinetic friction:

• Static friction:

• An object moving in a circle at constantspeed is in uniform circular motion.

• It has a centripetal acceleration of

• There is a centripetal force given by

chapter 5 using newton’s laws: friction, circular …...5-1 applications of newton’s laws...

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