Frictional Forces Two types:

- static – applies to stationary objects - kinetic – applies to sliding (moving) objects

Like FN, the Frictional Force is a contact force, but acts parallel to the interface of two objects

Chapter 6. Applications of Newton’s Laws

y

x

FN

mg

FA f

Apply Newton’s 2nd Law

xAx

N

yNy

mafFF

mgF

mamgFF 0

If applied force is small, book does not move (static), ax=0, then f=fs

Increase applied force, book still does not

move

Increase FA more, now book moves, ax0

SA fF

SxSAxSA fmafFmafF

There is some maximum static frictional force, fs

max. Once the applied force exceeds it, the book moves

NSS Ff max

s is the coefficient of static friction, it is a dimensionless number, different for each surface-object pair (wood-wood, wood-metal); also depends on surface preparation

s does not depend on the mass or surface area of the object

Has value: 0 < s < 1.5

If no applied vertical force

Magnitudes not vectors

mgf SS max

Push down on book

Apply Newton’s 2nd Law

y

x

FN

mg

FA f

)(

0

maxPSNSS

PN

yPNy

FmgFf

FmgF

maFmgFF

FP

What is FA needed just to start book moving? )(0 max

PSSASAx FmgfFfFF

When an Object is Moving? fs

max is exceeded so the object can move, but friction force is still being applied.

However, less force is needed to keep an object moving (against friction) than to get it started

We define kinetic friction

k is the coefficient of kinetic friction, similar to S but always less than S

Now, let’s consider incline plane problem, but with friction

Nkk Ff

y yFBD

mg

FN

x

FN

mg

x

fsfs

Book is at rest

m = 1.00 kg, = 10.0°, s = 0.200

N 9.65

)0.10)(cos80.9)(kg0.1(

cos

0cos

2sm

o

N

Ny

mgF

mgFF

N 70.1

)0.10)(sin80.9)(kg00.1(

sin

0sin

2sm

o

S

Sx

mgf

fmgF

Book can move (slide) if fs>fsmax

What is fsmax?

S

NSS

f

Ff

N 1.93

N) 65.9)(200.0(max Book does not move.

What angle is needed to cause book to slide?

S

S

NS

SS

mgmg

Fmg

ff

tan

cossin

sin

max

o

1

11.3

)(tan

S

As is increases, FN decreases, therefore fs

max decreases

Once book is moving, we need to use the kinetic coefficient of friction

Lets take, = 15.0° and k = 0.150 < s

2

2

sm

sm

12.1

)0.15cos150.00.15)(sin80.9(

)cos(sinor

cossin

sin

oo

kx

xk

xkx

ga

mamgmg

mafmgF

The Tension Force

Crate, mrope T

frictionless

Assume rope is massless and taut

T-T

0TTFx

FBD of rope

FBD of crate

T

mg

Tm

Frictionless pulley

Crate at rest

mgT

mgTFy

0

T

mg

Like the normal force, the friction and tension forces are all manifestations of the electromagnetic force

They all are the result of attractive (and repulsive) forces of atoms and molecules within an object (normal and tension) or at the interface of two objects

Applications of Newton’s 2nd Law

Equilibrium – an object which has zero acceleration, can be at rest or moving with constant velocity

0,00 yx FFF

Example: book at rest on an incline with friction

Non-equilibrium – the acceleration of the object(s) is non-zero yyxx maFmaFamF ,

Example Problem

Three objects are connected by strings that pass over massless and frictionless pulleys. The objects move and the coefficient of kinetic friction between the middle object and the surface of the table is 0.100 (the other two being suspended by strings). (a) What is the acceleration of the three objects? (b) What is the tension in each of the two strings?

m2

m1 m3

Given: m1 = 10.0 kg, m2=80.0 kg, m3=25.0 kg, k=0.100

Find: a1, a2, a3, T1, and T2Solution: 1) Draw free-body diagrams

m1

T1

m1g

T2

m3

m3g m2g

T1

FN

T2

fk

y

x

2) Apply Newton’s 2nd Law to each object

)( 111

1111

111

gamT

amgmT

amF

y

y

yy

)( 332

3332

333

gamT

amgmT

amF

y

y

yy

gmF

gmF

F

N

N

y

2

2

2

0

0

2212

222

xk

xx

amfTT

amF

gmFf kNkk 2 Also, ax1=0, ax3=0, ay2=0

Three equations, but five unknowns: ay1,ay3,ax2,T1,and T2

But, ay1 = ax2 = -ay3 = a

Substitute 2nd and 3rd equations into the 1st equation

0

)()(

221

133

2213

amgmgmamgmam

amgmgamgam

k

k

)(

)(

332

111

22212

gamT

gamT

amgmTT

y

y

xk

321sm

321

213

213321

213

213

260.0 0.250.800.10

)100.0(0.800.100.2580.9

)(

)()()(

0)( )(

yxy

k

k

k

aaa

mmm

mmmga

mmmgmmmammmg

mmma

N 230

)60.080.9(0.25)(

N 104

)80.960.0(0.10)(

32

11

agmT

gamT

Example Problem

A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 25.0° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier’s mass is 55.0 kg , and the coefficient of kinetic friction between the skis and the snow is 0.120. Find the magnitude of the force that the tow bar exerts on the skier.

Given: m = 55.0 kg, k = 0.120, = 25.0°

Infer: since velocity is constant, ax=0; also ay=0 since skier remains on slope

equilibrium 0,00 yx FFF

Draw FBD, apply Newton’s 2nd Law

mg

FN

x

FN

mg

x

FpFp

fk fk

cos

0cos

0

mgF

mgF

F

N

N

y

N 286

)sincos(

sincos

sin

sin

0sin

0

k

k

Nk

kP

Pk

x

mg

mgmg

mgF

mgfF

mgFf

F