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Ch 07: Newton's Third Law
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 07: Newton's Third LawProblem 47
Chapter 7, Problem 47

FIGURE P7.47 shows a 200 g hamster sitting on an 800 g wedge-shaped block. The block, in turn, rests on a spring scale. An extra-fine lubricating oil having μs = μk = 0 is sprayed on the top surface of the block, causing the hamster to slide down. Friction between the block and the scale is large enough that the block does not slip on the scale. What does the scale read, in grams, as the hamster slides down?

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Understand the problem: The hamster slides down a frictionless wedge-shaped block, which rests on a spring scale. The goal is to determine the reading of the scale in grams as the hamster slides down. Note that the block does not slip on the scale, and the lubricating oil ensures no friction between the hamster and the block.
Analyze the forces: The spring scale measures the total normal force exerted by the block and the hamster on it. This force is equal to the combined gravitational forces of the block and the hamster, as there is no vertical acceleration of the system.
Express the gravitational forces: The gravitational force acting on the hamster is given by \( F_{\text{hamster}} = m_{\text{hamster}} g \), where \( m_{\text{hamster}} = 0.2 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). Similarly, the gravitational force acting on the block is \( F_{\text{block}} = m_{\text{block}} g \), where \( m_{\text{block}} = 0.8 \, \text{kg} \).
Combine the forces: Since the block does not slip on the scale and the hamster's motion is along the inclined surface, the vertical forces add directly. The total force measured by the scale is \( F_{\text{total}} = F_{\text{hamster}} + F_{\text{block}} = (m_{\text{hamster}} + m_{\text{block}}) g \).
Convert to grams: The scale reading is typically given in grams. To convert the total force to grams, use the relationship \( 1 \; \text{kg} = 1000 \; \text{g} \). Thus, the scale reading is \( (m_{\text{hamster}} + m_{\text{block}}) \times 1000 \; \text{g} \).

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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. In this scenario, the hamster is sliding down the wedge-shaped block due to the absence of friction (μₛ = μₖ = 0) between them. Understanding friction is crucial as it determines whether an object will remain stationary or move when a force is applied.
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Newton's Second Law

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). This principle is essential for analyzing the forces acting on the hamster and the wedge, allowing us to calculate the resultant forces and the reading on the spring scale as the hamster slides down.
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Weight and Normal Force

Weight is the force exerted by gravity on an object, calculated as the product of mass and gravitational acceleration (W = mg). The normal force is the support force exerted by a surface perpendicular to the object. In this problem, the weight of the hamster and the wedge affects the reading on the scale, as the normal force changes when the hamster slides down the block.
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Related Practice
Textbook Question

What is the acceleration of the 3.0 kg block in FIGURE CP7.55 across the frictionless table? Hint: Think carefully about the acceleration constraint.

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Textbook Question

The 1.0 kg physics book in FIGURE P7.40 is connected by a string to a 500 g coffee cup. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are μs = 0.50 and μk = 0.20. At the highest point, does the book stick to the slope, or does it slide back down?

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Textbook Question

In FIGURE CP7.54, find an expression for the acceleration of m1. The pulleys are massless and frictionless. Hint: Think carefully about the acceleration constraint.

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Textbook Question

A house painter uses the chair-and-pulley arrangement of FIGURE P7.45 to lift himself up the side of a house. The painter's mass is 70 kg and the chair's mass is 10 kg. With what force must he pull down on the rope in order to accelerate upward at 0.20 m/s².

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Textbook Question

A 2.0 kg block on a horizontal, frictionless surface is connected by a massless spring and a massless, frictionless pulley to a hanging mass. For what value of the hanging mass does the block accelerate at 1.5 m/s²?

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