/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 A light string passing over a sm... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 42

A light string passing over a smooth light pulley connects two blocks of masses \(m_{1}\) and \(m_{2}\) (vertically). If the acceleration of the system is \((g / 8)\), then the ratio of masses is a. \(5: 3\) b. \(4: 3\) c. \(9: 7\) d. \(8: 1\)

Short Answer

Expert verified
The ratio of masses is \(9:7\).

Step by step solution

01

Understand Forces

Identify that the system consists of two blocks connected by a string over a pulley. The tension in the string is the same on both sides; however, the gravitational force differs due to different masses.
02

Write Equations of Motion

For mass \(m_1\) moving up, the equation is \(T - m_1g = m_1a\). For mass \(m_2\) moving down, the equation is \(m_2g - T = m_2a\), where \(T\) is the tension and \(a = \frac{g}{8}\) is the acceleration.
03

Solve for Tension \(T\)

From the two equations: \(T = m_1(g + a)\) and \(T = m_2(g - a)\). Set them equal to solve for \(T\):\[m_1(g + a) = m_2(g - a)\]
04

Substitute and Simplify

Substitute \(a = \frac{g}{8}\) into the equation:\[m_1\left(g + \frac{g}{8}\right) = m_2\left(g - \frac{g}{8}\right)\] resulting in \[m_1\left(\frac{9g}{8}\right) = m_2\left(\frac{7g}{8}\right)\].
05

Find the Ratio

Cancel \(g\) from both sides and solve for the ratio of masses\( \frac{m_1}{m_2} \):\[m_1 \cdot 9 = m_2 \cdot 7 \implies \frac{m_1}{m_2} = \frac{7}{9}\].
06

Final Answer

The masses ratio is \(\frac{m_1}{m_2} = \frac{7}{9}\), which corresponds to option (c) \(9:7\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equations of Motion in Atwood's Machine
When analyzing the movement of blocks in an Atwood's Machine, understanding the equations of motion is essential. The system consists of two blocks connected by a string that passes over a pulley. This setup reflects a classic physics problem where we want to determine how these blocks move under the influence of gravity.
The motion of each block can be described by Newton's second law: the net force on an object is equal to the mass of the object times its acceleration. In this case, we have two separate equations, one for each mass.
  • For the block with mass \(m_1\) moving upwards, the equation is \(T - m_1g = m_1a\).
  • For the block with mass \(m_2\) moving downwards, we have \(m_2g - T = m_2a\).
Here, \(T\) is the tension in the string, \(g\) is the acceleration due to gravity, and \(a\) is the acceleration of the system. Setting these equations correctly allows us to solve for unknowns like tension or the mass ratio.
Understanding Tension in the String
Tension in the string is a crucial concept in analyzing Atwood's Machine. It refers to the force exerted by the string on the blocks. In an ideal system, the string is massless and the tension throughout remains constant. This simplifies our calculations considerably.
The tension is affected by the forces acting on each block. Despite the fact that both blocks experience the same tension, they are moving in opposite directions with a common acceleration \(a\). Using the equations derived from Newton's second law:
  • Tension for \(m_1\): \(T = m_1(g + a)\)
  • Tension for \(m_2\): \(T = m_2(g - a)\)
Equating these two expressions allows us to delve deeper into the dynamics of the system. Analyzing tension helps us connect the equations of motion and derive other values like mass ratios.
Role of Gravitational Force
Gravitational force plays a fundamental role in the behavior of Atwood's Machine. It is the force that acts downwards on each of the masses due to gravity. This force depends on the mass of the object and can be expressed as \(m_1g\) for the smaller block and \(m_2g\) for the larger one.
In Atwood's Machine, gravity causes the heavier block to accelerate downward, while the lighter block is pulled upward. The disparity in gravitational force between the two masses leads to the movement of the system. The difference in these gravitational pulls, combined with the tension in the string, allows us to solve for system properties, such as acceleration or mass ratios.
A vital aspect to remember is that gravitational force does not operate in isolation but in tandem with tension and acceleration; thus, comprehending its role is essential for a full understanding of the system's dynamics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A block \(A\) of mass \(2 \mathrm{~kg}\) is placed over another block \(B\) of mass \(4 \mathrm{~kg}\) which is placed over a smooth horizontal ffoor. The coefficient of friction between \(A\) and \(B\) is \(0.4\). When a horizontal force of magnitude \(10 \mathrm{~N}\) is applied on \(A\), the acceleration of blocks \(A\) and \(B\) are a. \(1 \mathrm{~ms}^{-2}\) and \(2 \mathrm{~ms}^{-2}\), respectively. b. \(5 \mathrm{~ms}^{-2}\) and \(2.5 \mathrm{~ms}^{-2}\), respectively. c. Both the blocks will moves together with acceleration \(1 / 3 \mathrm{~ms}^{-2}\) d. Both the blocks will move together with acceleration \(5 / 3 \mathrm{~ms}^{-2}\)

A heavy uniform chain lies on a horizontal table top. If the coefficient of friction between the chain and the table surface is \(0.25\), then the maximum fraction of the length of the chain, that can hang over one edge of the table is a. \(20 \%\) b. \(25 \%\) c. \(35 \%\) d. \(15 \%\)

Friction force can be reduced to a great extent by a. Lubricating the two moving parts. b. Using ball bearings between two moving parts. c. Introducing a thin cushion of air maintained between two relatively moving surfaces. d. All of the above.

A lift is moving down with an acceleration \(a .\) A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift, and a man standing stationary on the ground are, respectively a. \(a, g\) b. \((g-a) ; g\) c. \(a_{1} a\) d. \(g, g\)

An object moving with a constant acceleration in a noninertial frame a. must have non-zero net force acting on it. b. may have zero net force acting on it. c. may have no force acting on it. d. this situation is practically impossible. (The pseudo force acting on the object has also to be considered)
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Linear Momentum

Read Explanation

Oscillations

Read Explanation

Classical Mechanics

Read Explanation

Force

Read Explanation

Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.