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 95b

Chapter 5, Problem 95b

A 28.0-kg block is connected to an empty 2.00-kg bucket by a cord running over a frictionless pulley (Fig. 5–56). The coefficient of static friction between the table and the block is 0.42 and the coefficient of kinetic friction between the table and the block is 0.34. Sand is gradually added to the bucket until the system just begins to move. Calculate the acceleration of the system.

Verified step by step guidance

Identify the forces acting on the system. For the block on the table, the forces include the tension in the cord, the frictional force, and the normal force. For the bucket, the forces include the tension in the cord and the gravitational force due to its weight.

Determine the maximum static friction force for the block using the formula:

f

_max=μ_sN, where

μ

_s is the coefficient of static friction and

N

is the normal force. The normal force is equal to the weight of the block:

N = m

_blockg, where

g

is the acceleration due to gravity.

Find the weight of the bucket and the sand when the system just begins to move. At this point, the tension in the cord equals the maximum static friction force. Use the equation:

T = m

_bucketg, where

T

is the tension and

m

_bucket is the total mass of the bucket and sand.

Once the system begins to move, calculate the net force acting on the system. The kinetic friction force opposes the motion of the block and is given by:

f

_k=μ_kN, where

μ

_k is the coefficient of kinetic friction.

Use Newton's second law to calculate the acceleration of the system. The net force is the difference between the gravitational force on the bucket and the kinetic friction force on the block. Apply the equation:

F

_net=m_totala, where

m

_total is the combined mass of the block and bucket, and

a

is the acceleration.

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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. This principle is fundamental in analyzing the motion of the block and bucket system, as it allows us to relate the forces acting on both objects to their resulting acceleration.

Friction

Friction is the force that opposes the relative motion of two surfaces in contact. In this scenario, both static and kinetic friction are relevant; static friction prevents the block from moving until a certain threshold is reached, while kinetic friction acts once the block is in motion. The coefficients of friction provided are essential for calculating the forces involved.

Pulley Systems

A pulley system allows for the transfer of force and motion between different objects. In this case, the frictionless pulley changes the direction of the tension force in the cord connecting the block and the bucket. Understanding how tension is distributed in the system is crucial for applying Newton's laws and calculating the acceleration.

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06:23

Systems Connected By Pulleys

Related Practice

Textbook Question

A train traveling at a constant speed rounds a curve of radius 215 m. A lamp suspended from the ceiling swings out to an angle of 18.5° throughout the curve. What is the speed of the train? [Hint: See Example 4–15.]

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Textbook Question

A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic friction between the skier's skis and the water surface is μₖ = 0.25 (Fig. 5–59). What is the skier's acceleration if the rope pulling the skier behind the boat applies a horizontal tension force of magnitude F_T = 240N to the skier (θ = 0°)?

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Textbook Question

A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic friction between the skier's skis and the water surface is μₖ = 0.25 (Fig. 5–59). What is the skier's horizontal acceleration if the rope pulling the skier exerts a force of F_T = 240N on the skier at an upward angle θ = 12°?

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Textbook Question

The 70.0-kg climber in Fig. 5–53 is supported in the 'chimney' by the friction forces exerted on his shoes and back. The static coefficients of friction between his shoes and the wall, and between his back and the wall, are 0.80 and 0.60, respectively. What is the minimum normal force he must exert? Assume the walls are vertical and that the static friction forces are both at their maximum. Ignore his grip on the rope.

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Textbook Question

A car starts rolling down a 1-in-4 hill (1-in-4 means that for each 4 m traveled along the road, the elevation change is 1 m). How fast is it going when it reaches the bottom after traveling 55 m? Assume an effective coefficient of rolling friction equal to 0.10.

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