1 7/1/04 Midterm 1: July 9 Will cover material from Chapters 1-6 Go to the room where you usually have recitation July 6 Recitation will be a review session.

17/1/04

Midterm 1: July 9

Will cover material from Chapters 1-6 Go to the room where you usually have

recitation July 6 Recitation will be a review session Practice exam available Friday

Note: HW 3 is still due on Wed, July 7

27/1/04

Review

Newton’s 1st Law: Inertia Newton’s 2nd Law: F=ma Newton’s 3rd Law: “Action and Reaction”

Forces are vectors Units: Newtons [kg m/s2]

37/1/04

Review: Free Body DiagramsThe first step in solving any force problem:

1. Sketch the object in question2. Draw an arrow for each force acting on the

object3. Label each force4. Indicate the direction of acceleration off to the

side (acceleration is NOT a force)

aF1

F3

F2

47/1/04

Review: How to Solve a Force Problem

1. Draw a free body diagram

Guess at forces of unknown magnitude or direction

2. Break forces into components

3. Sum of all the forces in one direction = mass * acceleration in that direction

i.e. Fnet,x=m ax

4. Repeat step 3 for each direction

5. Solve for unknown quantities

57/1/04

Mass on a string

If we pull steadily on the bottom string, which will break first?

A) Top

B) BottomC) It’s a matter of luck

67/1/04

Mass on a string

F

Tension in a string transmits a force along the direction of the string.

77/1/04

Mass on a string

Free body diagram for bottom string (at rest):

F

Tbottom

F

Tbottom = F

87/1/04

Mass on a stringFree body diagram for mass m (at rest):

F

Ttop

mg

F

Ttop = F+mg

97/1/04

What happens if we pull fast?

Which breaks first?

A) TopB) BottomC) It’s a matter of luck

107/1/04

Why did the bottom string break?

If we attempt to accelerate too fast tension on bottom string becomes too large and string snaps.

Newton’s First Law: mass is at rest and cannot accelerate instantaneously to speed of the hand yanking down…

7/1/04 11

Chapter 6

Forces (Part the Second)

127/1/04

Dissipative Forces

So far - Forces don’t depend on history

Direction of motion

Velocity of particle

Dissipative Forces DO

Friction

Viscosity (Air Resistance)

137/1/04

Frictional Forces

Frictional forces are from object's surface interacting with another material

Frictional forces always oppose (intended) motion

FpushFfriction

147/1/04

Kinetic Friction

For an object that is moving, the magnitude of the frictional force is proportional to the magnitude of the normal force on an object:

NfF kkfrictionkinetic

k is the “coefficient of kinetic friction”

157/1/04

Static FrictionThe maximum magnitude of the static frictional force is proportional to the magnitude of the normal force on an object:

NfF ssfrictionstatic

Once an object begins to move, kinetic friction takes over.

k is the “coefficient of kinetic friction”

167/1/04

Kinetic vs. Static Friction

Material Kinetic Static

Glass on Glass 0.40 0.94

Copper on Glass 0.53 0.68

Rubber on Concrete (dry) 0.8 1.0

Rubber on Concrete (wet) 0.25 0.30

Teflon on Teflon 0.04 0.04

Objects are harder to start moving than to keep moving

s > k

Some typical values:

No need to memorize, this will be given when needed

177/1/04

Static and Kinetic Friction

If you push with steadily increasing force:

Ffriction

Fpush

μsN

μkN

Fpush

Ffriction

187/1/04

A Classic Demo…

Pull the “table cloth” slowly, then quickly

197/1/04

Pig on a Frictional Plane

What angle should we tilt the plane so that the pig slides?

μs=0.5

N

Fg

Ffr

y

x

207/1/04

Pig on a Frictional Plane

Fg,x = mg sin θ

N

Fg

Ffy

x

Breaking Fg into components:

Fg,y = mg cos θ

x direction:

NfmgF

fF

maF

ssxg

sxg

xxnet

max,,

max,,

,

sin

0

0

Summing forces in the y direction:

cos

0

0

,

,

,

mgFN

FN

maF

yg

yg

yynet

217/1/04

Pig on a Frictional Plane

N

Fg

Ffy

x

We have:

Nmg s sin

cosmgN

)cos(sin mgmg s

)(cossin s

s tan

So for the pig to slip:

6.26)5.0(tan 1

227/1/04

Pig on a Frictional Plane

N

Fg

Ffy

x

x direction:

xkxg

xkxg

xxnet

maNF

mafF

maF

,

max,,

,

Summing forces in the y direction:

cos

0

0

,

,

,

mgFN

FN

maF

yg

yg

yynet

Say we tilt the ramp just pass = 26.6° anda = 2m/s2. What is μk?

237/1/04

Pig on a Frictional Plane

N

Fg

Ffy

x

xkxg maNF ,

27.0

)6.26cos()/8.9(

)6.26sin()/8.9()/2(2

22

sm

smsmk

cos

sin

g

ag xk

xk mamgmg cossin

247/1/04

Heavy Box

Say you can bench 500 N (about 110 lbs), can you push a 1000 N box across the floor?

Max static friction = μsN = (0.75)(1000 N) = 750 N

Ffriction

μs = 0.75 and μk= 0.45

Kinetic friction = μkN = (0.45)(1000 N) = 450 N

Not a chance…

No problem!

257/1/04

Pulling a Box

μk=0.25

Frope

θ

x: Fropecosθ - Ffr = ma Fropecosθ - μk(mg - Fropesinθ) = ma

y: Fropesinθ + N - Fg = 0 N = mg - Fropesinθ

Frope

Fg

N

Ffr

a

Summing forces:

267/1/04

Pulling a Box Frope

μk=0.25

θ

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0 5 10 15 20 25

Acce

lera

tion

Angle

m

mgFa kkrope

)sin(cos

0)cossin(

)sin(cos

krope

kkrope

m

F

m

mgF

d

d

d

da

25.0for 0.14)(tan 1 kk

To find the maximum:

sincos k

277/1/04

Accelerating a Car

The coefficients of friction for rubber on concrete are:

Dry: s=1.0 k=0.80

Wet: s=0.30 k=0.25

A car of mass 1000 kg tries to accelerate from a stop sign. What is the minimum time to accelerate to 30 m/s on dry pavement? On wet pavement?

287/1/04

Accelerating a Car

Dry: s=1.0 t = (30 m/s)/(1.0 × 9.8 m/s2) = 3.06 s

Wet: s=0.30 t = (30 m/s)/(0.30 × 9.8 m/s2) = 10.2 s

If a tire does not slip, friction is static

Ff=μsN

ga s

atvv f 0 g

v

a

vt

s

ff

mamgf ss max,

297/1/04

Stopping a Car

What are the minimum stopping distances for a 1000 kg car at 30 m/s in the following circumstances?

Dry pavement, wheels rolling? (ABS)

Dry pavement, wheels locked?

Wet pavement, wheels rolling? (ABS)

Wet pavement, wheels locked?

307/1/04

Stopping a Car

Wheels locked:

Ff=μkN

v0=30 m/s

Dry: μk = 0.80 xf = 57.4 mWet: μk = 0.25 xf = 183.7 m

)(2 02

02 xxavv ff

a

vx f 2

20

ga

mamgf

k

ks

max,

g

vx

kf 2

20so…

317/1/04

Stopping a car

Wheels rolling:

Ff=μsN

vi=30 m/s

Dry: μs = 1.0 xf = 45.9 mWet: μs = 0.30 xf = 153.1 m

Same as before, but we replace k with s.

g

vx

sf 2

20

Distance is shorter with ABS

327/1/04

Drag Force

Opposes Motion

Depends on velocity

A correct treatment is complicated,but an approximation of the “drag force” in the case of high velocity is given by:

is the fluid density, C is the “drag coefficient”

221 AvCFD

337/1/04

Terminal Velocity

An object stops accelerating when Fg=FD

gFAvC 221

AC

Fv g

22

Fg

FD

This is the “terminal velocity”

347/1/04

Force and Uniform Circular Motion

Object travels around a circle at constant speed

a

v

R

Recall: centripetal acceleration

center thetowards2

R

va

center thetowards2

R

vmF

amF

c

c

Centripetal Force

357/1/04

Demo: Water in a bucket

367/1/04

Why Didn’t I Get Wet (hopefully...)?

a

v

Accelerates faster than g downward, water cannot fall out and is pushed in a circle.

377/1/04

Ferris Wheel

Radius: R = 9mPeriod: T = 20 sec (fast)

What is the force on an 80 kg rider from the seat when he is at the top and at the bottom?

387/1/04

Ferris Wheel

The centripetal force is given by:

center thetowards2

R

vmamF cc

T

Rv

2Recall:

Ns

mkg

T

RmFc 71

)20(

)9)(80(442

2

2

2

Then,

397/1/04

Ferris Wheel

At the top:

NNsmkgN

FFN

FFN

FmaF

cg

cg

ccynet

71371)/8.9)(80( 2

,

At the bottom:

NNsmkgN

FFN

FFN

FmaF

cg

cg

ccynet

85571)/8.9)(80( 2

,

N

Fg

ac

N

Fg

ac

407/1/04

Rotation with FrictionHow fast can a car round an unbanked curve (dry pavement) with a radius of 50 m without slipping sideways?

R=50 m

vFf

vmax = μsgR

Dry: μs = 1.0 vmax = 1.0×9.8×50 = 22.1 m/s

R

vac

2

Nf ss max,

fs ac

RmvN

mafF

s

csnet

/2

417/1/04

Rotation with FrictionHow much does the curve need to be banked to provide the centripetal acceleration without using sideways frictional force from the tires?

θ

N

mg

y

x

y: N cosθ = mg N = mg / cosθ

x: N sinθ = mac but ac = v2/R N sinθ = mv2/R

Thus:R

mvmg 2

cos

sin

45)/8.9)(50(

)/1.22(tantan

2

21

2

smm

sm

Rg

v

427/1/04

Example: Problem 6.41

A puck of mass m slides on a frictionless table while attached to a hanging cylinder of mass M by a cord through a hole in the table. What speed must the puck travel at to keep the cylinder at rest?