friction friction problem situations

FrictionFriction Problem Situations

PhysicsMontwood High School

R. Casao

Friction

• Friction Ff is a force that resists motion•Friction involves objects in contact

with each other.•Friction must be overcome before

motion occurs.• Friction is caused by the uneven

surfaces of the touching objects. As surfaces are pressed together, they tend to interlock and offer resistance to being moved over each other.

Microscopic Friction

Magnified section of a polished steel surface showing surface bumps about 5 x 10-7 m (500 nm) high, which corresponds to several thousand atomic diameters.

Computer graphic from a simulation showing gold atoms (below) adhering to the point of a sharp nickel probe (above) that has been in contact with the gold surface.

Surface Roughness

Adhesion

Friction

• Frictional forces are always in the direction that is opposite to the direction of motion or to the net force that produces the motion.

• Friction acts parallel to the surfaces in contact.

Types of Friction• Static friction: maximum frictional force

between stationary objects.• Until some maximum value is reached and

motion occurs, the frictional force is whatever force is necessary to prevent motion.

• Static friction will oppose a force until such time as the object “breaks away” from the surface with which it is in contact.

• The force that is opposed is that component of an applied force that is parallel to the surface of contact.

Types of Friction• The magnitude of the static friction force Ffs

has a maximum value which is given by:

• where μs is the coefficient of static friction and FN is the magnitude of the normal force on the body from the surface.

fs s NF F

Types of Friction• Sliding or kinetic friction: frictional force

between objects that are sliding with respect to one another.• Once enough force has been applied to the

object to overcome static friction and get the object to move, the friction changes to sliding (or kinetic) friction.

• Sliding (kinetic) friction is less than static friction.• If the component of the applied force on the

object (parallel to the surface) exceeds Ffs then the magnitude of the opposing force decreases rapidly to a value Fk given by:

where μk is the coefficient of kinetic friction.

k k NF F

The static frictional force keeps an object from starting to move when a force is applied. The static frictional force has a maximum value, but may take on any value from zero to the maximum, depending on

Static FrictionStatic Friction

what is needed to keep the sum of forces zero.

Types of Friction

• From 0 to the maximum value of the static frictional force Fs in the figure, the applied force is resisted by the static frictional force until “breakaway”.

• Then the sliding (kinetic) frictional force Fk is approximately constant.

Types of Friction

• Static and sliding friction are dependent on:• The nature of the surfaces in contact.

Rough surfaces tend to produce more friction.

• The normal force (Fn) pressing the surfaces together; the greater Fn is, the more friction there is.

Friction vs. Area

Question: Why doesn’t friction depend on contact area?

The microscopic area of contact between a box and the floor is only a small fraction of the macroscopic area of the box’s bottom surface.

If the box is turned on its side, the macroscopic area is increased, but the microscopic area of contact remains the same (because the contact is more distributed). Therefore the frictional force f is independent of contact area.

Types of Friction

• Rolling friction: involves one object rolling over a surface or another object.

• Fluid friction: involves the movement of a fluid over an object (air resistance or drag in water) or the addition of a lubricant (oil, grease, etc.) to change sliding or rolling friction to fluid friction.

Coefficient of Friction

• Coefficient of friction (): ratio of the frictional force to the normal force pressing the surfaces together. has no units.

• Static:

• Sliding (kinetic):

n

fss F

Fμ

n

fkk F

Fμ

•The maximum frictional force is 50 N. As the applied force increases from 0 N to 50 N, the frictional force also increases from 0 N to 50 N and will be equal to the applied force as it increases.

•Once the static frictional force of 50 N has been overcome, only a 40 N force is needed to overcome the 40 N kinetic frictional force and produce constant velocity (a = 0 m/s2).

•As the applied force increases beyond 40 N, the kinetic frictional force remains at 40 N and the 100 N block will accelerate.

A Model of Friction

Friction

Static Friction

Kinetic Friction

push k k NF f F

The kinetic frictional force is also independent of the relative speed of the

surfaces, and of their area of contact.

Kinetic Friction and SpeedKinetic Friction and Speed

Rolling Friction

Horizontal Surface – Constant Speed

•Constant speed: a = O m/s2.•The normal force pressing the surfaces together is the weight; Fn = Fw

fx

fx

fx

x

FF

N0FF

amFF

amFΣ

wkfx

wkf

w

f

n

fk

FμFF

FμF

FF

FF

μ

Horizontal Surface: a > O m/s2

wkf

w

f

n

fk

wn

fx

x

fx

FμF

FF

FF

μ

FF

amFF

amFΣ

FF

Horizontal Surface: a > O m/s2

• If solving for:• Fx:

• Ff:

• a:

gmμamF

FμamF

FamF

kx

wkx

fx

amFF xf

mFF

a fx

Horizontal Surface: Skidding to a Stop or Slowing Down (a < O m/s2)

• The frictional force is responsible for the negative acceleration.

• Generally, there is no Fx.

wkf

w

f

n

fk

wn

f

FμF

FF

FF

μ

FF

amF

)ta(vv

)ta5.0()tv(xΔ

)xΔa2(vv

if

2i

2i

2f


• Most common use involves finding acceleration with a velocity equation and finding k:

• Acceleration will be negative because the speed is decreasing.


• The negative sign for acceleration a is dropped because k is a ratio of forces that does not depend on direction.

• Maximum stopping distance occurs when the tire is rotating. When this happens, a = -s·g.

• Otherwise, use a = -k·g to find the acceleration, then use a velocity equation to find distance, time, or speed.

ga

gmam

FF

FF

μw

f

n

fk

Friction, Cars, & Antilock Brakes The diagram shows forces acting on a car with front-wheel drive. Typically, Fn > Fn’ because the engine is over the front wheels. The largest frictional force fs the tire can exert on the road is µs·Fn. Attempts to make the tire exert a force larger than this causes the tire to “burn rubber” and actually reduces the force, since µk<µs.

Note that while all points on the rolling tire have the same speed v in the reference frame of the car, in the reference frame of the road the bottom of the tire is at rest, while top is moving forward with a speed of 2·v.

Antilock brakes sense the wheel rotation and “ease off” if it close to stopping, maintaining static friction with the road and allowing better control of steering than if the wheels were locked.

Antilock Brakes

Example:The Effect of Antilock Brakes

22 2 0

0 2 and 0, so 2

v

v v a x v xa

A car is traveling at 30 m/s along a horizontal road. The coefficients of friction are ms=0.50 and mk=0.40.(a) What is the braking distance xa with antilock brakes?(b) What is the braking distance xb if the brakes lock?

and a s b ka g a g2 20

2

(30 m/s)91.7 m

2 2 (0.5) (9.81 m/s )

as

vx

g2 20

2

(30 m/s)114.7 m

2 2 (0.4) (9.81 m/s )

bk

vx

g

Example: A Game of Shuffleboard

k k nf F

A cruise-ship passenger uses a shuffleboard cue to push a shuffleboard disk of mass 0.40 kg horizontally along the deck, so that the disk leaves the cue at a speed of 8.5 m/s. The disk then slides a distance of 8.0 m.

What is the coefficient of kinetic friction between the disk and deck?

0 y yF m a

0 n nF m g F m g

x xF m a

k k xf m g m a x ka g

2 2 20 02 0 2 x x x x kv v a x v g x

2 20

2

(8.5 m/s)0.46

2 2 (9.81 m/s ) (8.0 m)

x

k

v

g x

Down an Inclined Plane


• Resolve Fw into Fx and Fy.

• The angle of the incline is always equal to the angle between Fw and Fy.

• Fw is always the hypotenuse of the right triangle formed by Fw, Fx, and Fy.

θsinFF

θcosFF

FF

θsinF

Fθcos

wx

wy

w

x

w

y


• The force pressing the surfaces together is NOT Fw, but Fy; Fn = Fy.

or

ykf

y

f

n

fk

fx

fx

FμF

FF

FF

μ

mFF

a

amFF

amΣF

sin cos

mass m cancels out

( sin ) ( cos )

( sin )cos

m g m g m a

g g a

g ag

Down an Inclined Plane• If we place an object on an inclined plane and

increase the tilt angle to the point at which the object just begins to slide.

• What is the relation between and the static coefficient of friction µs?

- : cos N wy axis F F

- : sin s s N wx axis f F F

sincos

sintan

cos

s ws

n w

s

f FF F

Down an Inclined Plane• If the object slides down the incline at constant

speed (a = 0 m/s2), the relation between and the kinetic coefficient of friction µk:

θtanμ

θtanθcosθsin

θcosFθsinF

FF

FF

μ

FF

N0FFs

m0mFF

k

w

w

y

x

n

fk

fx

fx

2fx


• To determine the angle of the incline:• If moving:

• If at rest:

k1 μtanθ

s1 μtanθ

Example: A Sliding Coin A hardcover book is resting on a tabletop with its front cover facing upward. You place a coin on the cover and very slowly open the book until the coin starts to slide. The angle is the angle of the cover just before the coin begins to slide.

Find the coefficient of static friction µs between the coin and book.

, cos s s N s sf F at so f m g

cos 0 cos y y n NF m a F m g or F m g

sin 0 or sin x x s sF m a m g f f m g

, cos sin tan s sTherefore or

Example:Dumping a file cabinet

A 50.0 kg steel file cabinet is in the back of a dump truck. The truck’s bed, also made of steel, is slowly tilted. What is the size of the static friction force when the truck’s bed is tilted by 20°? At what angle will the file cabinet begin to slide?

Steel on dry steel

Free-body diagram

Example:Dumping a file cabinet

sin sin 0;

cos cos 0;

x s s

y

F w f m g f

F n w n m g

2sin (50.0kg) (9.80m/s ) sin 20 168 N; sf m g

max cos ; s s s sf f n m g

sin sin cos 0; s sm g f m g m g

sintan ; arctan arctan(0.80) 38.7

coss s

File cabinet will begin to slide when:

Non-Parallel Applied Force on Ramp

m·gm·g ·cos

m·g ·sin

fk

N

If an applied force acts on the box at an angle above the horizontal, resolve FA into parallel and perpendicular components using the angle + :

FA ·cos ( + θ) and FA ·sin ( + θ)

FA serves to increase acceleration directly and indirectly: directly by FA ·cos ( + θ) pulling the box down the ramp, and indirectly by FA ·sin ( + θ) lightening the normal support force with the ramp (thereby reducing friction).

FA

FA ·cos( + )

FA ·sin( + )


m·gm·g

·cos

m·g ·sin

fk

N

FA

FA ·cos( + )

FA ·sin ( + )

If FA ·sin( + ) is not big enough

to lift the box off the ramp, there is no acceleration in the perpendicular direction. So, FA ·sin( + ) + FN = m·g·cos.

Remember, FN is what a scale would read if placed under the box, and a scale reads less if a force lifts up on the box. So, FN = m·g ·cos - FA ·sin( + ), which means fk = k ·FN = k ·[m·g ·cos - FA ·sin( + )].


m·g

m·g ·cos

m·g ·sin

fk

N

FA

FA ·cos( + )

FA ·sin( + )

If the combined force of FA ·cos( + ) + m·g ·sin is is enough to move the box: FA ·cos( + ) + m·g·sin

- k ·[m·g·cos - FA ·sin( + )] = m·a

Up an Inclined Plane

Up an Inclined Plane• Resolve Fw into Fx and Fy.

• The angle of the incline is always equal to the angle between Fw and Fy.

• Fw is always the hypotenuse of the right triangle formed by Fw, Fx, and Fy.

θsinFF

θcosFF

FF

θsinF

Fθcos

wx

wy

w

x

w

y


• Fa is the force that must be applied in the direction of motion.

• Fa must overcome both friction and the x-component of the weight.

• The force pressing the surfaces together is Fy.


ykf

y

f

n

fk

xfa

xfa

x

yn

FμF

FF

FF

μ

mFFF

a

amFFF

amFΣ

FF

•For constant speed, a = 0 m/s2.

Fa = Fx + Ff

•For a > 0 m/s2.Fa = Fx + Ff +

(m·a)

Pulling an Object on a Flat Surface


•The pulling force F is resolved into Fx and Fy.

θsinFF

θcosFFF

Fθsin

FF

θcos

y

x

y

x


•Fn is the force that the ground exerts upward on the mass. Fn equals the downward weight Fw minus the upward force Fy from the pulling force.•For constant speed, a = 0 m/s2.

mFF

a

amFF

amFΣ

)FF(μF

FFF

FF

μ

FFF

N0FFF

N0FΣ

fx

fx

x

ywkf

yw

f

n

fk

ywn

wyn

y

Example: Pulling A Sled

2

sin 0 sin

50 9.8 100 sin 40 425.72

y y N N

N

F m a F T m g m or F m g T

mF kg N Ns

Two children sitting on a sled at rest in thesnow ask you to pull them. You pull on the sled’s rope, which makes an angle of 40° withthe horizontal. The children have a combinedmass of 45 kg, and the sled has a mass of 5.0 kg. The coefficients of static and kineticfriction are µs=0.20 and µk=0.15, and the sled is initially at rest.

Find the acceleration of the sled and children if the rope tension is 100 N.

2

cos

0.15 425.72 63.86

100 cos 40 63.86100 cos 40 63.86 50 0.2549

50

x x k x

kk k k N

N

x x

F ma T f m a

ff F N N

F

N N mN N kg a askg

Simultaneous Pulling and Pushing an Object on a Flat Surface

Simultaneous Pulling and Pushing an Object on a Flat Surface

θsinFF

θcosFFF

Fθsin

FF

θcos

y

x

y

x

m

FFFa

amFFF

amFΣ

)FF(μF

FFF

FF

μ

FFF

N0FFF

N0FΣ

fpushx

fpushx

x

ywkf

yw

f

n

fk

ywn

wyn

y

Pushing an Object on a Flat Surface


•The pushing force F is resolved into Fx and Fy.

sinθFF

cosθFFF

Fsinθ

FF

cosθ

y

x

y

x


•Fn is the force that the ground exerts upward on the mass. Fn equals the downward weight Fw plus the upward force Fy from the pushing force.•For constant speed, a = 0 m/s2.

mFF

a

amFF

amFΣ

)FF(μF

FFF

FF

μ

FFF

N0FFF

N0FΣ

fx

fx

x

ywkf

yw

f

n

fk

ywn

wyn

y

Pulling and Tension

• The acceleration a of both masses is the same.

Pulling and Tension

• For each mass:

• Isolate each mass and examine the forces acting on that mass.

n2kf2

n1kf1

w2n2

w1n1

FμF

FμF

FF

FF

Pulling and Tension

•m1 = mass

•T1 may not be a tension, but could be an applied force (Fa) that causes motion.

amFTT

amΣF

11f21

1

Pulling and Tension

•m2 = mass

amFT

amΣF

22f2

2

Pulling and Tension

• This problem can often be solved as a system of equations:

• See the Solving Simultaneous Equations notes for instructions on how to solve this problem using a TI or Casio calculator.

amFT

amFTT

22f2

11f21

Revisiting Tension and Friction


•For the hanging mass, m2:

•The acceleration a of both masses is the same.

•For the mass on the table, m1:

amTgm

amTF

amΣF

gmF

22

22w

2

22w

amT-F

FμF

FF

amΣF

1f

1nf

1w1n


12

f2

12f2

2f12

mmFgm

a

amamFgm

amFamgm

A block of mass m2 = 5.0 kg has been adjusted so that the block m1 = 7.0 kg is just on the verge of sliding.

(a)What is the coefficient of static friction ms between the table and the block?

Example: A Sliding Block

1 1 10 so y N y NF F m g m a F m g

' 2 2 ' 20 so x xF m g T m a T m g

2 21 2

2 2

5Therefore, ; 0.71

7

s s

m g m kgm g m g

m g m kg

1 10 so x x s N sF T f m a f F m g T

Example: A Sliding Block(b) With a slight push, the blocks move with

acceleration a. Find a if µk = 0.54.

1 1 1

1 1

x x k x

x k

F T f m a so T m g m a

T m a m g

' 2 2 ' 2 2 ' x x xF m g T m a T m g m a

1 1 2 2

1 2 2 1

2 1

1 2

22

T=T, therefore,

( )

( )

9.8 m/s 5.0 kg 0.54 7.0 kg1.0 m/s

7.0 kg 5.0 kg

k

k

k

m a m g m g m a

m a m a m g m g

g m ma

m m

a

' x xa a a

Normal Force Not Associated with Weight.

• A normal force can exist that is totally unrelated to the weight of an object.

applied forcefriction

weightnormal

FN = applied force

Friction is Always Parallel to Surfaces….

•In this case, for the block to remain in position against the wall without moving:

• the upward frictional force Ff has to be equal and opposite to the downward weight Fw.•The rightward applied force F has to be equal ad opposite to the leftward normal force FN.

F

FW

Ff

FN

(0.20)

friction friction problem situations

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