Friction

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Friction
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Friction

• Friction causes objects to slow down.
• Friction creates heat.
• Friction degrades an objects energy

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Static Friction
Normal force
Applied Force
Friction, Resistance

• Objects at rest.
• Applied force is insufficient to move object.

Weight mg
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Friction

• Surface imperfections and micro-welds.

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Static Friction

• Increases linearly
• For a given pair of surfaces, the ratio of
  Frictional force to Applied force is a constant.

Frictional Force
Applied Force
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Static Friction

• The relationship is
• µs force of friction/normal force.
• Where µ is called the coefficient of static
  friction.
• It has no units and varies between 0 and 2 in
  general. We usually rearrange the equation
• Ffs µs FN

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Example of Static Friction

• What is the coefficient of static friction
  between a tabletop and a 2 kg block of wood if a
  2 N force is required to start the block moving?

• Identify knowns and unknowns

• m 2 kg

applied force 2 N,

• v 0

• µs ?

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• Appropriate equation Ff µ FN.
• What is FN?

On a level surface the normal force upward is
equal to the weight of the object downward, i.e.
FN m(-g)
FN Fg mg
Ff µFN
Ff µ mg or
µ f/mg
Fg mg
2 N/(2 kg 9.8m/s2)
µ 0.102
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Kinetic Friction

• The word kinetic stems from the Greek word,
  kinema meaning motion, so kinetic friction deals
  with the friction present when motion is
  occurring.
• The resistance is less because the microscopic
  impediments are being sheared off and no time for
  micro-welds to form.

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Kinetic Friction

• Constant
• Less than static friction.
• This is the key to non-skid brakes.

Frictional Force
Applied Force, N
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Compare skidding with no skidding.

• A car moving at 25 m/s slams on its brakes. The
  coefficient of static friction with the road is
  1.2 and the kinetic coefficient is 0.4 How far
  does the car slide?

Equations v2 v02 2 ad Fnet ma FN mg Ff µ
FN
Knowns vo 25 m/s v 0 m/s µs 0.4. Unknown
d ?
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Free body diagram for skidding
Direction of car ()
FN m(-g)
Ff µk FN
Fg mg
Fnet Ff
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Solve for a from net force equation
Solve for d
Vf2 v02 2ad
Fnet ma a fnet/m
d - v02 /2a
Substitute in for a d -v02 / 2(Fnet/m)
-v02m/2Fnet
Since Fnet Ff
-v02m/2Ff
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AND.
Ff µk FN
FN Fg mg
-v02m/2µ FN
SO
-v02m/2µ mg
(-25m/s)2 -.4(9.8m/s2)
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• Now, solve the same problem using non-skid brakes
  so the wheels keep turning while the car slows
  down. In this case we use the state coefficient
  of friction
• x vo2/2µg
• (25 m/s)2/(20.859.8 m/s2)
• 37.5 m
• Imagine if the pavement wear wet and the
  coefficient of kinetic friction was 0.47. The
  stopping distance would be nearly 75 m!