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Information about Calculating friction

Slides for teaching the factors affecting the magnitude of friction force, and how to calculate this for static and kinetic friction on horizontal surfaces and inclined planes.

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fs max & fk are affected by

fs max & fk are affected by Smoothness

fs max & fk are affected by Smoothness affected by materials

fs max & fk are affected by Smoothness Normal force

fs max & fk are affected by Smoothness Normal force

fs max & fk are affected by µs or µk Smoothness Normal force

Materials in contact µs µk Steel on Teflon 0,04 0,04 Wood on snow 0,08 0,06 Metal on metal (lubricated) 0,10 0,07 Wood on wood 0,7 0,4 Steel on steel 0,75 0,57 Glass on glass 0,9 0,4 Rubber on concrete 0,9 0,7

fs max & fk are affected by µs or µk Smoothness N Normal force

fs max & fk are affected by µs or µk Smoothness N Normal force

fs max = µs N

fk = µ k N

Maximum static friction Fstatic friction max = µs N normal force surface exerts on object; depends on object’s weight, downward forces on object and surface’s slope; if no slope and no extra forces, = weight coefficent of static friction; depends on the materials of the rubbing surfaces

Kinetic friction Fkinetic friction = µk N normal force surface exerts on object; depends on object’s weight, downward forces on object and surface’s slope; if no slope and no extra forces, = weight coefficent of kinetic friction; depends on the materials of the rubbing surfaces

Question 1 A 4kg concrete slab is pulled horizontally along a piece of rubber. Calculate the frictional force: a) just before the slab starts to move (maximum static friction) b) as the slab moves along (kinetic friction) Rubber-concrete: µs = 0,9 µd = 0,7

mass = 4 kg

mass = 4 kg weight = 40 N

Normal Weight

40 N 40 N

a) fs max = µs N fs max = 0,9 • 40 N fs max = 36 N

b) fk = µ k N fk = 0,7 • 40 N fk = 28 N

Question 2 A 4 kg wooden sled is pulled up a snow-covered slope which is at 20° to the horizontal.

Question 2 A 4 kg wooden sled is pulled up a snow-covered slope which is at 20° to the horizontal. Calculate the frictional force: a) just before the sled starts to move (maximum static friction) b) as the sled moves along (kinetic friction) Wood-snow: µs = 0,08 µd = 0,06

mass = 4 kg 20°

mass = 4 kg weight = 40 N 20°

Weight

Wy W

N Wy W

N Wy 20° W 20°

N Wy Wy = W • cos 20° 20° W 20°

N Wy Wy = W • cos 20° Wy = 40 N • cos 20° 20° W 20°

N Wy Wy = W • cos 20° Wy = 40 N • cos 20° Wy = 37,59 N 20° W 20°

37,59 N 37,59 N W

Materials in contact µs µk Steel on Teflon 0,04 0,04 Wood on snow 0,08 0,06 Metal on metal (lubricated) 0,10 0,07 Wood on wood 0,7 0,4 Steel on steel 0,75 0,57 Glass on glass 0,9 0,4 Rubber on concrete 0,9 0,7

a) fs max = µs N fs max = 0,08 • 37,59 N fs max = 3 N

b) fk = µ k N fk = 0,06 • 37,59 N fk = 2,26 N

fs max & fk are affected by µs or µk Smoothness N Normal force

Question 2c A 4 kg wooden sled is pulled up a snow-covered slope which is at 20° to the horizontal. c) How hard must a person pull parallel to the slope upward for the sled to move at a constant velocity?

Question 2c A 4 kg wooden sled is pulled up a snow-covered slope which is at 20° to the horizontal. c) How hard must a person pull parallel to the slope upward for the sled to move at a constant velocity? (to balance the forces of: - kinetic friction and - the component of the sled’s weight down the slope)

For CONSTANT VELOCITY: Fdown slope Fup slope

For CONSTANT VELOCITY: Fdown slope Fup slope

N Wy W

W Component of W down slope

W

Wx W

Wx W Wx

Wy Wx Wx W

Wx 20° Wx = W • sin 20° W Wx

Wx 20° Wx = W • sin 20° Wx = 40 • sin 20° 40 N Wx

Wx 20° 40 N Wx Wx = W • sin 20° Wx = 40 • sin 20° Wx = 13,68 N down the slope

13,68 N 20° 40 N

Person pulls 13,68 N

Person pulls 13,68 N f

13,68 N f

b) fk = µ k N fk = 0,06 • 37,59 N fk = 2,26 N

13,68 N fk = 2,26 N

Fdown slope = 15,94 N 13,68 N fk = 2,26 N

For CONSTANT Fdown slope = 15,94 N VELOCITY: 13,68 N Fup slope = 15,94 N fk = 2,26 N

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