/*! 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 47 Three objects with masses \(m_{1... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 47

Three objects with masses \(m_{1}=36.5 \mathrm{~kg}, m_{2} 19.2 \mathrm{~kg},\) and \(m_{3}=12.5 \mathrm{~kg}\) are hanging from ropes that run over pulleys. What is the acceleration of \(m_{1} ?\)

Short Answer

Expert verified
Answer: The acceleration of mass \(m_1\) is approximately \(-2.59 m/s^2\) in the upward direction.

Step by step solution

01

Identify forces acting on the masses

Each mass is acted upon by two forces: 1. The gravitational force (weight) which pulls the mass downward with a force \(F_g = mg\), where \(m\) is the mass and \(g\) is the gravitational acceleration (\(9.81 m/s^2\) in the downward direction). 2. The tension force (\(T\)) in the rope, which acts upward and counteracts the gravitational force. The net force acting on each mass will be a combination of these two forces, and we will use that to find the acceleration of \(m_1\).
02

Apply Newton's second law to each mass

For each mass, we write Newton's second law equation based on the net force acting on the mass. Let's denote the acceleration of \(m_1\) as \(a\). For \(m_1\), the net force is the difference between the tension force \(T_1\) and the gravitational force: \(T_1 - m_1g = m_1a\). For \(m_2\), the net force is the difference between the tension force \(T_2\) and the gravitational force: \(T_2 - m_2g = m_2a\). For \(m_3\), the net force is the difference between the tension force \(T_3\) and the gravitational force: \(T_3 - m_3g = m_3a\).
03

Analyze the tension forces

The tension forces are created by the three masses pulling on the ropes. We have one equation that relates all three tension forces: \(T_3 - T_2 = T_1\). This is because mass \(m_1\) needs to balance out the combined force from masses \(m_2\) and \(m_3\).
04

Solve for acceleration \(a\)

From Step 3, we can write \(T_1 = T_3 - T_2\). Now plug this into the equation for \(m_1\) from Step 2: \(T_3 - T_2 - m_1g = m_1a\) We want to eliminate \(T_2\) and \(T_3\) from the equation and solve for \(a\). We can plug the equations for \(m_2\) and \(m_3\) from Step 2 into the above equation: \((T_3 - m_3g) - (T_2 - m_2g) - m_1g = m_1a\) Solve this equation for \(a\): \(a = \frac{m_3g - m_2g - m_1g}{m_1}\)
05

Calculate the acceleration

Now plug in the given values for masses and gravitational acceleration into the equation and solve for the acceleration \(a\): \(a = \frac{12.5 * 9.81 - 19.2 * 9.81 - 36.5 * 9.81}{36.5} = -2.59~m/s^2\) The acceleration of mass \(m_1\) is approximately \(-2.59 m/s^2\). The negative sign indicates that the acceleration is in the opposite direction of the gravitational force, so mass \(m_1\) is moving upward.

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.

Gravitational Force
Gravitational force is a fundamental concept that plays a pivotal role in understanding motion and interaction of bodies under gravity. It is the attractive force exerted by any mass towards the center of the Earth. This force acts on all objects with mass, pulling them downward. For any object, you can calculate gravitational force using the formula: \[ F_g = mg \]where:
  • \( F_g \) represents the gravitational force
  • \( m \) signifies the mass of the object
  • \( g \) is the acceleration due to gravity, approximately \( 9.81 \ m/s^2 \) on Earth
In the given exercise, each mass, whether it is \( m_1, m_2, \) or \( m_3 \), experiences a gravitational pull downward with its own distinct amount of force based on its mass. This downward force must be counteracted by the tension in the pulleys for the system to be in equilibrium.
Tension Force
Tension force is crucial when dealing with objects connected by strings or ropes over pulleys. It refers to the force that is transmitted through a string, cable, or rope when it is pulled tight by forces acting from opposite ends. This force acts parallel to the string in the opposite direction to gravitational pull, aiming to balance the system.In the exercise, the tension force \( T \) is the counteracting force for each of the hanging masses. For each mass, the rope provides a tension that opposes the weight, creating a balance necessary for calculating acceleration:
  • For \( m_1 \), tension \( T_1 \) must balance against \( m_1g \) and provide the net force leading to the calculated acceleration.
  • Similarly, for \( m_2 \) and \( m_3 \), tensions \( T_2 \) and \( T_3 \) relate similarly counteracting their respective weights.
The relationship and difference between these tensions play a key role in resolving the net force and hence allowing the computation of acceleration.
Acceleration
Acceleration is the rate of change of velocity of an object. Newton's second law defines it as directly proportional to the net force acting on an object and inversely proportional to its mass. Mathematically, this relationship is expressed as:\[ F_{net} = ma \]where:
  • \( F_{net} \) is the net force acting on the object
  • \( m \) is the mass
  • \( a \) stands for acceleration
In the scenario given, knowing the masses and gravitational force, applying Newton's second law helps us pinpoint the acceleration of \( m_1 \). The exercise involves calculating the net forces using the relationship between the gravitational and tension forces. The net force equation:\[T_3 - T_2 - m_1g = m_1a\]is rearranged to solve for \( a \), revealing that the calculated acceleration is negative, meaning that \( m_1 \) is actually accelerating in the upward direction away from gravity鈥攏ot a very intuitive result, but absolutely crucial in understanding the system's behavior.

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

Two blocks of masses \(m_{1}\) and \(m_{2}\) are suspended by a massless string over a frictionless pulley with negligible mass, as in an Atwood machine. The blocks are held motionless and then released. If \(m_{1}=3.50 \mathrm{~kg}\). what value does \(m_{2}\) have to have in order for the system to experience an acceleration \(a=0.400 g\) ? (Hint: There are two solutions to this problem.

A spring of negligible mass is attached to the ceiling of an elevator. When the elevator is stopped at the first floor, a mass \(M\) is attached to the spring, stretching the spring a distance \(D\) until the mass is in equilibrium. As the elevator starts upward toward the second floor, the spring stretches an additional distance \(D / 4\). What is the magnitude of the acceleration of the elevator? Assume the force provided by the spring is linearly proportional to the distance stretched by the spring.

\(\bullet 4.60\) A block of mass \(m_{1}=21.9 \mathrm{~kg}\) is at rest on a plane inclined at \(\theta=30.0^{\circ}\) above the horizontal. The block is connected via a rope and massless pulley system to another block of mass \(m_{2}=25.1 \mathrm{~kg}\), as shown in the figure. The coefficients of static and kinetic friction between block 1 and the inclined plane are \(\mu_{s}=0.109\) and \(\mu_{k}=0.086\) respectively. If the blocks are released from rest, what is the displacement of block 2 in the vertical direction after 1.51 s? Use positive numbers for the upward direction and negative numbers for the downward direction.

A block of mass \(m_{1}=3.00 \mathrm{~kg}\) and a block of mass \(m_{2}=4.00 \mathrm{~kg}\) are suspended by a massless string over a friction less pulley with negligible mass, as in an Atwood machine. The blocks are held motionless and then released. What is the acceleration of the two blocks?

Two blocks are stacked on a frictionless table, and a horizontal force \(F\) is applied to the top block (block 1). Their masses are \(m_{1}=2.50 \mathrm{~kg}\) and \(m_{2}=3.75 \mathrm{~kg}\). The coefficients of static and kinetic friction between the blocks are 0.456 and 0.380 , respectively a) What is the maximum applied force \(F\) for which \(m_{1}\) will not slide off \(m_{2} ?\) b) What are the accelerations of \(m_{1}\) and \(m_{2}\) when \(F=24.5 \mathrm{~N}\) is applied to \(m_{1}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Engineering Physics

Read Explanation

Electrostatics

Read Explanation

Fundamentals of Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Geometrical and Physical Optics

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.