Ch. 05 - Using Newton's Laws: Friction, Circular Motion, Drag Forces

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 05 - Using Newton's Laws: Friction, Circular Motion, Drag Forces

Problem 9

Chapter 5, Problem 9

Show that if a skier moves at constant speed straight down a slope of angle θ (Example 5–6), then the coefficient of kinetic friction between skis and snow is μₖ = tanθ. Ignore air resistance.

Verified step by step guidance

Start by analyzing the forces acting on the skier. The skier is moving at a constant speed, which means the net force along the slope is zero. The forces involved are the gravitational force, the normal force, and the kinetic friction force.

Break the gravitational force into components. The component parallel to the slope is given by

F_{g, \(\text{parallel}\)} = mg \, \(\sin\[\theta\)

, and the component perpendicular to the slope is

F_{g, \(\text{perpendicular}\)} = mg \, \(\cos\]\theta\)

, where

m

is the mass of the skier,

g

is the acceleration due to gravity, and

\(\theta\)

is the angle of the slope.

Write the expression for the kinetic friction force. The kinetic friction force is given by

F_k = \(\mu\)_k F_N

, where

\(\mu\)_k

is the coefficient of kinetic friction and

F_N

is the normal force. Since the normal force balances the perpendicular component of gravity,

F_N = mg \, \(\cos\)\(\theta\)

Set up the condition for constant speed. Since the skier is moving at constant speed, the downhill gravitational force component is exactly balanced by the kinetic friction force. This gives the equation

mg \, \(\sin\[\theta\) = \(\mu\)_k \(\cdot\) mg \, \(\cos\]\theta\)

Simplify the equation to solve for

\(\mu\)_k

. Cancel out

mg

from both sides (since it is nonzero), leaving

\(\sin\[\theta\) = \(\mu\)_k \(\cdot\) \(\cos\]\theta\)

. Divide both sides by

\(\cos\[\theta\)

to get

\(\mu\)_k = \(\tan\]\theta\)

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In the context of the skier, this law helps us understand how the forces acting on the skier (gravity, friction, and normal force) balance out when the skier moves at constant speed, leading to the conclusion about the coefficient of kinetic friction.

Friction and Coefficient of Kinetic Friction

Friction is the force that opposes the relative motion of two surfaces in contact. The coefficient of kinetic friction (μₖ) quantifies this frictional force relative to the normal force. In this scenario, the relationship between the slope angle and the frictional force is crucial, as it allows us to derive that μₖ equals the tangent of the slope angle (tanθ) when the skier moves at constant speed.

Forces on an Inclined Plane

When an object is on an inclined plane, the gravitational force can be resolved into two components: one parallel to the slope, which causes the object to accelerate down the slope, and one perpendicular to the slope, which is balanced by the normal force. Understanding these components is essential for analyzing the skier's motion and determining how the angle of the slope affects the forces at play, leading to the relationship between the slope angle and the coefficient of kinetic friction.

Recommended video:

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06:59

Intro to Inclined Planes

Related Practice

Textbook Question

Suppose you are standing on a train accelerating at 0.20 g. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide?

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Textbook Question

Piles of snow on slippery roofs can become dangerous projectiles as they melt. Consider a chunk of snow at the ridge of a roof with a slope of 24°. As the snow begins to melt, the coefficient of static friction decreases and the snow finally slips. Assuming that the distance from a chunk of snow to the edge of the roof is 6.0 m and the coefficient of kinetic friction is 0.20, calculate the speed of the snow chunk when it slides off the roof.

Textbook Question

A flatbed truck is carrying a heavy crate. The coefficient of static friction between the crate and the bed of the truck is 0.75. What is the maximum rate at which the driver can decelerate and still avoid having the crate slide against the cab of the truck?

Textbook Question

Police investigators, examining the scene of an accident involving a car and an old truck, measure 72-m-long skid marks for the truck, which nearly came to a stop before colliding with the car at rest. The coefficient of kinetic friction between rubber and the pavement is about 0.80. Estimate the initial speed of the truck assuming a level road.

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Textbook Question

(II) How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius 85 m at a speed of 95 km/h?

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