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 17b

Chapter 5, Problem 17b

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.

Verified step by step guidance

Step 1: Identify the forces acting on the snow chunk. The forces include the gravitational force (mg), the normal force (N), and the kinetic frictional force (f_k). The gravitational force can be broken into two components: one parallel to the slope (mg sin(θ)) and one perpendicular to the slope (mg cos(θ)).

Step 2: Write the equation for the net force along the slope. The net force (F_net) is the difference between the downslope gravitational force (mg sin(θ)) and the kinetic frictional force (f_k). The kinetic frictional force is given by f_k = μ_k * N, where N = mg cos(θ). Thus, F_net = mg sin(θ) - μ_k * mg cos(θ).

Step 3: Use Newton's second law to find the acceleration of the snow chunk. According to Newton's second law, F_net = ma, where a is the acceleration. Substituting the expression for F_net, we get a = g(sin(θ) - μ_k * cos(θ)).

Step 4: Use kinematic equations to calculate the final speed of the snow chunk. The initial velocity (v_0) is 0 since the snow starts from rest. The distance traveled along the slope is 6.0 m. Using the kinematic equation v^2 = v_0^2 + 2a * d, substitute v_0 = 0, a = g(sin(θ) - μ_k * cos(θ)), and d = 6.0 m to find the final velocity v.

Step 5: Simplify the expression for the final velocity. After substituting the known values for g (9.8 m/s^2), θ (24°), μ_k (0.20), and d (6.0 m), calculate the final velocity v. Ensure that the trigonometric functions sin(θ) and cos(θ) are evaluated correctly.

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Key Concepts

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

Friction

Friction is the force that opposes the relative motion of two surfaces in contact. It is categorized into static friction, which prevents motion, and kinetic friction, which acts when objects are sliding past each other. The coefficient of friction quantifies this force, with values depending on the materials involved. In this scenario, the transition from static to kinetic friction is crucial as the snow chunk begins to slide off the roof.

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, expressed as F = ma. This principle is essential for analyzing the forces acting on the snow chunk as it slides down the roof. The gravitational force and frictional force will determine the net force and, consequently, the acceleration of the snow chunk.

Kinematics

Kinematics is the branch of mechanics that describes the motion of objects without considering the forces that cause the motion. It involves concepts such as displacement, velocity, and acceleration. In this problem, kinematic equations will be used to calculate the final speed of the snow chunk as it slides off the roof, taking into account the distance traveled and the acceleration due to gravity and friction.

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Related Practice

Textbook Question

Two blocks made of different materials connected together by a thin cord slide down a ramp inclined at an angle θ to the horizontal, Fig. 5–40 (block B is above block A). The masses of the blocks are m_A and m_B, and the coefficients of friction are μ_A and μ_B. If m_A = m_B=4.0kg, and μ_A = 0.20 and μ_B = 0.30, determine the acceleration of the blocks.

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Textbook Question

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.

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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

Two blocks made of different materials connected together by a thin cord slide down a ramp inclined at an angle θ to the horizontal, Fig. 5–40 (block B is above block A). The masses of the blocks are m_A and m_B, and the coefficients of friction are μ_A and μ_B. If m_A = m_B= 4.0kg, and μ_A = 0.20 and μ_B = 0.30, determine the tension in the cord, for an angle θ = 32°.

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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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