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, N
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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
• f µs N

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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 unknown
• m 2 kg, applied force 2 N, v 0,
• µs ?

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• Appropriate equation f µ N.
• What is N?
• On a level surface the normal force upward is
  equal to the weight of the object downward, i.e.
  N W mg. So,
• f µ mg or µ f/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 0.85. How far
  does the car slide?
• Knowns vo 25 m/s, v 0 m/s, µs 0.4.
• Unknown x ?
• Equations v2 vo2 2 ax, F ma, N mg, f µN.

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• Solve for x
• x -vo2/2a
• -vo2/2(-F/m)
• -vo2/2(-µN/m)
• -vo2/2(-µmg/m)
• vo2/2µg
• (25 m/s)2/(20.859.8 m/s2)
• 54 m

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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!

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Frictional force and normal force.

• Frictional force is proportional to the normal
  force, f a N. On a level surface N W, but what
  if someone is lifting up on the object?
• Wont that reduce the normal force?

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Free Body Diagram

• N T W, so
• N W T.
• If T pulls at some angle, then just decompose
  into components.

Lift from person,T
Normal,N
Weight, W
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• Example
• A person pulls on a 300 N crate with a rope that
  makes a 370 angle to the ground. If the
  coefficient of static friction is 0.6, how much
  tension must the person exert to get the crate
  moving?
• Knowns W 300 N, ? 37o, µ 0.6.
• Unknown T ?
• Equations f µN. All forces balance when at
  rest.

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• Horizontal Tension T cos ?
• Vertical Tension T sin ?
• Net vertical force N Tsin ? W 0, so
• N W - Tsin ?.
• Net horizontal force Tcos ? f 0
• Tcos ? f µN µ(W - Tsin ?) , so

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• Tcos ? µ(W - Tsin ?)
• Tcos ? µ Tsin ? µW
• T(cos ? µ sin ?) µW
• T µW/(cos ? µ sin ?)
• T 0.6300N/(cos300 .6sin300)
• T 180N/(0.866.3)
• T 180 N/(1.166)
• T 154 N