Friction, Circular Motion, and More Applications of Newtons Laws - - PDF document

Friction, Circular Motion, and More Applications of Newton’s Laws

• Friction
• Uniform Circular Motion
• More Applications of Newton’s Laws
• Homework

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Friction

• Kinetic friction

fk = µkn where µk is the coefficient of kinetic friction and n is the magnitude of the normal force

• Static friction

fs ≤ µsn where µs is the coefficient of static friction

• µs > µk for a given surface

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Friction Example 1

A woman pulls a loaded sled, with a mass of 75 kg, along a horizontal surface at a constant speed. The woman is pulling with a rope at an angle of 42◦ with the horizontal and the coefficient of friction between the runners and the snow is 0.1. What is the tension in the rope?

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Friction Example 1 Solution

A woman pulls a loaded sled, with a mass of 75 kg, along a horizontal surface at a constant speed. The woman is pulling with a rope at an angle of 42◦ with the horizontal and the coefficient of friction between the runners and the snow is 0.1. What is the tension in the rope?

Fx = T cos φ − fk = 0

(1)

Fy = T sin φ + N − mg = 0

(2) (1) ⇒ T cos φ = µkN N = T cos φ µk (2) ⇒ T sin φ + T cos φ µk = mg T = µkmg µk sin φ + cos φ T = (0.1) (75 kg)

• 9.8 m/s2
• (0.1) sin 42◦ + cos 42◦ = 91 N

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Friction Example 2

You are driving your car at 24 m/s. What is the shortest distance in which you can brake to a halt? The coefficient

• f static friction between the tires and the road is 0.6.

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Friction Example 2 Solution

You are driving your car at 24 m/s. What is the shortest distance in which you can brake to a halt? The coefficient

• f static friction between the tires and the road is 0.6.

n mg fs x y

Fx = −fs = max

−µsn = max ax = µsn m

Fy = n − mg = 0

n = mg ax = −µsg vxi = 24 m/s vxf = 0 xi = 0 xf =? v2

xf = v2 xi + 2ax (xf − xi)

xf = − v2

xi

2ax = v2

xi

2µsg = (24 m/s)2 2 (0.6) (9.8 m/s2) = 49 m

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Uniform Circular Motion

• Centripetal acceleration

ac = v2 r

• Centripetal force

Fc = mac = mv2 r

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The Conical Pendulum

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The Conical Pendulum

Fy = T cos θ − mg = 0

T = mg cos θ

Fr = T sin θ = mv2

r mg tan θ = mv2 r v =

• rg tan θ

v =

• Lg sin θ tan θ

The period of the motion is T = 2πr v = 2πr √rg tan θ = 2π

• L cos θ

g

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Homework 9 - Due Wed. Sept. 29

• Read Sections 3.4 & 5.1-5.2
• Answer questions 5.1, 5.2 & 5.6
• Do problems 3.27, 5.3, 5.5, 5.7, 5.11, 5.13 & 5.16

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