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Frictio n Friction Problem Situations Chapter 5.2 and 5.3

Friction Friction Problem Situations Chapter 5.2 and 5.3

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Page 1: Friction Friction Problem Situations Chapter 5.2 and 5.3

FrictionFriction Problem Situations

Chapter 5.2 and 5.3

Page 2: Friction Friction Problem Situations Chapter 5.2 and 5.3

Review of Forces• Remember to draw your free body

diagram

• F=ma

• Weight=mg

• Practice problem: If you lift a stuffed suitcase with a force of 105N, with an acceleration of 0.705 m/s.• What is the mass of the suitcase?• What is the weight?

Page 3: Friction Friction Problem Situations Chapter 5.2 and 5.3

Your Weight seems different when you accelerate

• When you are in an elevator how do you feel when you are going up?

• How about when it is going down?

• Newton’s Law explains this…• You feel the normal force that is pushed

up from the floor. This is what weight feels like

• When the floor exerts a force greater than your weight, you feel heavy.

• This is called apparent weight.

Page 4: Friction Friction Problem Situations Chapter 5.2 and 5.3

Forces in Springs• A compressed or stretched spring

exerts a force when it tries to return to its starting place.

• The amount of stretch in a spring depends on the force you apply.

• This changes the length of the spring (either compressed or stretched)

• Use Hooke’s Law to solve these problems• Force=spring constant X change in

length

Page 5: Friction Friction Problem Situations Chapter 5.2 and 5.3

Spring Constant• F=kx• K=spring constant• N/m• The larger the spring constant, the larger

the force exerted by the spring.• The larger the change in length, the larger

the force exerted by the spring.• Problem: what is the force required to

cause 3.4cm of stretch of a spring with a spring constant of 21 N/m

Page 6: Friction Friction Problem Situations Chapter 5.2 and 5.3

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.

Page 7: Friction Friction Problem Situations Chapter 5.2 and 5.3
Page 8: Friction Friction Problem Situations Chapter 5.2 and 5.3

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

Page 9: Friction Friction Problem Situations Chapter 5.2 and 5.3

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.

Page 10: Friction Friction Problem Situations Chapter 5.2 and 5.3

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.

Page 11: Friction Friction Problem Situations Chapter 5.2 and 5.3

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

Page 12: Friction Friction Problem Situations Chapter 5.2 and 5.3

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

Page 13: Friction Friction Problem Situations Chapter 5.2 and 5.3

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.

Page 14: Friction Friction Problem Situations Chapter 5.2 and 5.3

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.

Page 15: Friction Friction Problem Situations Chapter 5.2 and 5.3

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.

Page 16: Friction Friction Problem Situations Chapter 5.2 and 5.3

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.

Page 17: Friction Friction Problem Situations Chapter 5.2 and 5.3

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

n

fkk F

Page 18: Friction Friction Problem Situations Chapter 5.2 and 5.3

A Model of Friction

Friction

Page 19: Friction Friction Problem Situations Chapter 5.2 and 5.3

Static Friction

Page 20: Friction Friction Problem Situations Chapter 5.2 and 5.3

Kinetic Friction

push k k NF f F

Page 21: Friction Friction Problem Situations Chapter 5.2 and 5.3

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

Page 22: Friction Friction Problem Situations Chapter 5.2 and 5.3

Rolling Friction

Page 23: Friction Friction Problem Situations Chapter 5.2 and 5.3

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

Page 24: Friction Friction Problem Situations Chapter 5.2 and 5.3

Horizontal Surface: a > O m/s2

wkf

w

f

n

fk

wn

fx

x

fx

FμF

FF

FF

μ

FF

amFF

amFΣ

FF

Page 25: Friction Friction Problem Situations Chapter 5.2 and 5.3

Horizontal Surface: a > O m/s2

• If solving for:• Fx:

• Ff:

• a:

gmμamF

FμamF

FamF

kx

wkx

fx

amFF xf

mFF

a fx

Page 26: Friction Friction Problem Situations Chapter 5.2 and 5.3

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

Page 27: Friction Friction Problem Situations Chapter 5.2 and 5.3

)ta(vv

)ta5.0()tv(xΔ

)xΔa2(vv

if

2i

2i

2f

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

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

• Acceleration will be negative because the speed is decreasing.

Page 28: Friction Friction Problem Situations Chapter 5.2 and 5.3

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

• 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