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 22

Chapter 5, Problem 22

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?

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Identify the forces acting on the crate: The forces include the gravitational force (weight of the crate), the normal force exerted by the truck bed, and the static frictional force that prevents the crate from sliding.

Write the equation for the maximum static frictional force: The maximum static frictional force is given by \( f_{\text{max}} = \mu_s \cdot F_N \), where \( \mu_s \) is the coefficient of static friction (0.75) and \( F_N \) is the normal force. Since the crate is on a flat surface, \( F_N = m \cdot g \), where \( m \) is the mass of the crate and \( g \) is the acceleration due to gravity.

Relate the static frictional force to the deceleration: The static frictional force is what prevents the crate from sliding. Using Newton's second law, \( f_{\text{max}} = m \cdot a \), where \( a \) is the maximum deceleration of the truck. Substitute \( f_{\text{max}} \) from the previous step into this equation.

Simplify the equation to solve for the maximum deceleration: Substitute \( f_{\text{max}} = \mu_s \cdot m \cdot g \) into \( m \cdot a = f_{\text{max}} \). The mass \( m \) cancels out, leaving \( a = \mu_s \cdot g \).

Substitute the known values: Use \( \mu_s = 0.75 \) and \( g = 9.8 \ \text{m/s}^2 \) to calculate the maximum deceleration \( a \). This will give the maximum rate at which the truck can decelerate without the crate sliding.

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Key Concepts

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

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces in contact. It is characterized by a coefficient, which in this case is 0.75, indicating the maximum ratio of the force of static friction to the normal force. This concept is crucial for determining the maximum force that can act on the crate without causing it to slide.

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 (F = ma). In this scenario, understanding how the deceleration of the truck affects the forces acting on the crate is essential for calculating the maximum deceleration rate without sliding.

Deceleration

Deceleration refers to a decrease in the speed of an object, which is a form of acceleration in the opposite direction. In the context of the truck, the rate of deceleration must be calculated to ensure that the force exerted on the crate does not exceed the maximum static friction force, preventing it from sliding forward.

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Textbook Question

(III) A 4.0-kg block is stacked on top of a 12.0-kg block, which is accelerating along a horizontal table at a = 5.2m/s² (Fig. 5–43). Let μ_k = μ_s = μ. What is the force that must be applied to the 12.0-kg block in (a) and in (b), assuming that the table is frictionless?

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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

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

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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