/*! 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 13 Consider the following mass dist... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 13

Consider the following mass distribution, where \(x\) - and \(y\) -coordinates are given in meters: \(5.0 \mathrm{~kg}\) at \((0.0,0.0) \mathrm{m}\) \(3.0 \mathrm{~kg}\) at \((0.0,4.0) \mathrm{m}\), and \(4.0 \mathrm{~kg}\) at \((3.0,0.0) \mathrm{m} .\) Where should a fourth object of \(8.0 \mathrm{~kg}\) be placed so that the center of gravity of the four-object arrangement will be at \((0.0,0.0) \mathrm{m} ?\)

Short Answer

Expert verified
The fourth object of mass \(8.0kg\) should be placed at \((-1.6m,-1.5m)\) such that the center of gravity of the four-object arrangement will be at the origin \((0.0m, 0.0m)\).

Step by step solution

01

Understanding the Problem and Identifying the Given Information

Firstly, we are given three masses and their respective positions: \(5.0kg\) at \((0.0,0.0)\), \(3.0kg\) at \((0.0, 4.0)\) and \(4.0kg\) at \((3.0, 0.0)\). We are also told that a fourth object of mass \(8.0kg\) should be placed in such a way that the center of mass of all four objects lies at the origin, which would be \((0.0, 0.0)\).
02

Calculating the Current Center of Mass

The formula for center of mass in a system with three objects is , \(X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3} \) for the x-coordinate and similar one for the y-coordinate , \(Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3} \). Substituting in the given position and mass values for the three objects: \((X_{cm}, Y_{cm}) = \frac{(5.0kg * 0.0m) + (3.0kg * 0.0m) + (4.0kg * 3.0m)}{5.0kg + 3.0kg + 4.0kg}, \frac{(5.0kg * 0.0m) + (3.0kg * 4.0m) + (4.0kg * 0.0m)}{5.0kg + 3.0kg + 4.0kg} = (1.0m, 1.2m) \). So currently, the center of mass (center of gravity) is at \((1.0m, 1.2m)\).
03

Finding the Desired Position of the Fourth Object

In order for the center of mass to be at the origin, the location of the fourth mass must be such that it shifts the current center of mass from \((1.0m, 1.2m)\) to \((0.0m, 0.0m)\). This can be achieved by setting up two equations, one for each coordinate, \(X_{cm}' = \frac{m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4}{M}\) and \(Y_{cm}' = \frac{m_1y_1 + m_2y_2 + m_3y_3 + m_4y_4}{M}\) where \(X_{cm}'\) and \(Y_{cm}'\) are the desired center of mass (which is \((0.0m, 0.0m)\)), \(M = m_1 + m_2 + m_3 + m_4\) and \((x_4, y_4)\) is the desired position of the fourth object. Solving these equations for \((x_4,y_4)\), we get the location of the fourth mass as \(-1.6m, -1.5m\).

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.

Mass Distribution
Mass distribution refers to how mass is spread out in space, affecting how a system behaves. When objects have masses positioned at various points, their combined center of mass is influenced by their individual locations and masses. The center of mass is like the balancing point of the system, where the mass is equally distributed around all directions.

In the exercise, we consider masses placed at different coordinates, and our goal is to add another mass in such a way that the entire system balances at the origin, (0.0, 0.0) m. This is achieved by strategically placing the additional mass using the concept of mass distribution.

When dealing with mass distribution, it is important to take note of the equations used to find the center of mass. These equations are essentially weighted averages of the coordinates, where the weights are the masses themselves. By understanding how these masses influence the system based on their positions, we can manipulate and control the overall center of mass for specific needs.
Coordinate System
A coordinate system helps in defining positions of points in a structured manner. For this exercise, we rely on a two-dimensional coordinate system, which uses x and y coordinates to specify where each mass is located in space. This system allows us to calculate the center of mass more easily.

In practical terms, the coordinate system allows us to represent the positions of the 5.0 kg, 3.0 kg, and 4.0 kg masses, and eventually place the 8.0 kg mass. By plotting each point as \((x, y)\) on a graph, we can visualize how these points interact and balance with each other.

Using a coordinate system is essential in physics problem solving because it gives a defined structure to analyze problems systematically. It allows one to apply mathematical principles such as calculating averages (like the center of mass) effectively, leading to a deeper understanding of spatial relationships.
Physics Problem Solving
Physics problem solving involves breaking down complex systems into manageable parts and applying scientific principles to find solutions. In our exercise, the key task was to balance the mass system such that its center of mass is at the specified target point, the origin.

To solve such a problem, we follow a systematic approach:
  • Identify the known quantities and their positions.
  • Calculate the current center of mass using given data.
  • Determine how a new mass will influence the overall center of mass.
  • Solve equations to find exact coordinates for the new mass that will shift the center to the desired location.

This kind of structured approach not only helps in solving this specific problem but also strengthens general problem-solving skills in physics. By using formulas and coordinate systems, and by understanding mass distributions, students can master the art of physics problem solving gradually.

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 astronauts (Fig. \(\mathrm{P} 8.72\) ), each having a mass of \(75.0 \mathrm{~kg}\), are connected by a \(10.0-\mathrm{m}\) rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). Treating the astronauts as particles, calculate (a) the magnitude of the angular momentum and (b) the rotational energy of the system. By pulling on the rope, the astronauts shorten the distance between them to \(5.00 \mathrm{~m}\). (c) What is the new angular momentum of the system? (d) What are their new speeds? (e) What is the new rotational energy of the system? (f) How much work is done by the astronauts in shortening the rope?

The top in Figure P8.49 has a moment of inertia of \(4.00\) \(\times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\) and is initially at rest. It is free to rotate about a stationary axis \(A A^{\prime}\). A string wrapped around a peg along the axis of the top is pulled in such a manner as to maintain a constant tension of \(5.57 \mathrm{~N}\) in the string. If the string does not slip while wound around the peg, what is the angular speed of the top after \(80.0 \mathrm{~cm}\) of string has been pulled off the peg? Hint: Consider the work that is done.

Halley's comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being \(0.59 \mathrm{~A} . \mathrm{U}\), and its greatest distance being 35 A.U. (1 A.U. is the EarthSun distance). If the comet's speed at closest approach is \(54 \mathrm{~km} / \mathrm{s}\), what is its speed when it is farthest from the Sun? You may neglect any change in the comet's mass and assume that its angular momentum about the Sun is conserved.

A solid uniform sphere of mass \(m\) and radius \(R\) rolls without slipping down an incline of height \(h\). (a) What forms of mechanical energy are associated with the sphere at any point along the incline when its angular speed is \(\omega\) ? Answer in words and symbolically in terms of the quantities \(m, g, y, I, \omega\), and \(u\) (b) What force acting on the sphere causes it to roll rather than slip down the incline? (c) Determine the ratio of the sphere's rotational kinetic energy to its total kinetic energy at any instant.

A uniform solid cylinder of mass \(M\) and radius \(R\) rotates on a frictionless horizontal axle (Fig. P8.81). Two objects with equal masses \(m\) hang from light cords wrapped around the cylinder. If the system is released from rest, find (a) the tension in each cord and (b) the acceleration of each object after the objects have descended a distance \(h .\)
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Translational Dynamics

Read Explanation

Waves Physics

Read Explanation

Linear Momentum

Read Explanation

Geometrical and Physical Optics

Read Explanation

Electricity and Magnetism

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.