/*! 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 28 Three masses are placed on the \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 28

Three masses are placed on the \(y\) -axis: \(2 \mathrm{~kg}\) at \(y=300 \mathrm{~cm}, 6 \mathrm{~kg}\) at \(y=150 \mathrm{~cm}\), and \(4 \mathrm{~kg}\) at \(y=-75 \mathrm{~cm}\). Find their center of mass.

Short Answer

Expert verified
The center of mass is located at \( y = 100 \, \text{cm} \) on the y-axis.

Step by step solution

01

Identify the Masses and Positions

First, we need to identify the different masses and their coordinates on the y-axis. The three masses are: - Mass 1: 2 kg located at y = 300 cm - Mass 2: 6 kg located at y = 150 cm - Mass 3: 4 kg located at y = -75 cm. These values will be used to calculate the center of mass.
02

Calculate Total Mass

Calculate the total mass of the system by summing the individual masses:\[ M = m_1 + m_2 + m_3 = 2 \, \text{kg} + 6 \, \text{kg} + 4 \, \text{kg} = 12 \, \text{kg} \].
03

Calculate Weighted Positions

Take the product of each mass's value and its position on the y-axis:- For Mass 1: \( 2 \, \text{kg} \times 300 \, \text{cm} = 600 \, \text{kg}\cdot\text{cm} \)- For Mass 2: \( 6 \, \text{kg} \times 150 \, \text{cm} = 900 \, \text{kg}\cdot\text{cm} \)- For Mass 3: \( 4 \, \text{kg} \times -75 \, \text{cm} = -300 \, \text{kg}\cdot\text{cm} \).
04

Sum of Weighted Positions

Add up the results from Step 3 to get the sum of the weighted positions:\[ \text{Sum} = 600 + 900 - 300 = 1200 \, \text{kg}\cdot\text{cm} \].
05

Calculate Center of Mass

Use the formula for the center of mass along the y-axis:\[ y_{\text{cm}} = \frac{\text{Sum of weighted positions}}{\text{Total mass}} = \frac{1200 \, \text{kg}\cdot\text{cm}}{12 \, \text{kg}} = 100 \, \text{cm} \].

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
The concept of mass distribution is foundational in understanding the center of mass. In any set of objects, knowing how mass is spread out helps determine where the center of mass lies. In the given problem, the mass distribution includes three points:
  • A 2 kg mass at 300 cm on the y-axis.
  • A 6 kg mass at 150 cm on the y-axis.
  • A 4 kg mass at -75 cm on the y-axis.
Each mass's position is pivotal, as it affects the overall center of mass. Think of mass distribution as a collection of unique weights scattered along a line, each influencing the 'center' based on how far it is and how heavy it is.
Weighted Positions
Weighted positions represent the influence of each mass based on its position relative to a point of reference. Here, the weight is not just the mass itself but also includes its distance from the origin along the y-axis. For instance, the weighted positions for our three masses are:
  • For 2 kg at 300 cm: the weight is calculated as \(2 \times 300 = 600\, \text{kg} \cdot \text{cm}\).
  • For 6 kg at 150 cm: \(6 \times 150 = 900\, \text{kg} \cdot \text{cm}\).
  • For 4 kg at -75 cm: \(4 \times (-75) = -300\, \text{kg} \cdot \text{cm}\).
Combining these weighted positions, we get a sum that informs us about the net effect of all masses acting together.
Center of Mass Calculation
Calculating the center of mass involves finding a point that considers these weighted contributions equally. After determining the weighted positions, the next step is to find their sum: \(600 + 900 - 300 = 1200\, \text{kg} \cdot \text{cm}\). Now, to find the center of mass (\(y_{\text{cm}}\)), we divide this sum by the total mass of all objects. With a total mass of 12 kg: \[ y_{\text{cm}} = \frac{1200\, \text{kg} \cdot \text{cm}}{12\, \text{kg}} = 100\, \text{cm} \] This result indicates that the center of mass lies 100 cm along the y-axis, balancing out the distributed mass values accurately.
Mass and Position Relationship
The interplay between mass and position is essential in understanding balance points like the center of mass. Each mass, depending on its weight and position, contributes differently to the final balance point. Heavier masses have a more significant impact when they are further from the reference point. Similarly, smaller masses can have an outsized influence if placed at a strategic distance. This relationship highlights why simply averaging the positions of the objects wouldn't suffice. Instead, by using weighted positions, we respect both mass and position in determining an accurate center of mass. Understanding this relationship is key to grasping why the center of mass is a reflection of both how much mass there is and where it acts.

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 system consists of the following masses in the \(x y\) -plane: \(4.0 \mathrm{~kg}\) at coordinates \((x=0, y=5.0 \mathrm{~m})\), \(7.0 \mathrm{~kg}\) at \((3.0 \mathrm{~m}, 8.0 \mathrm{~m})\), and \(5.0 \mathrm{~kg}\) at \((-3.0 \mathrm{~m},-6.0 \mathrm{~m})\). Find the position of its center of mass. $$ \begin{array}{l} x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(4.0 \mathrm{~kg})+(3.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-3.0 \mathrm{~m})(5.0 \mathrm{~kg})}{(4.0+7.0+5.0) \mathrm{kg}}=0.38 \mathrm{~m} \\ y_{\mathrm{cm}}=\frac{\sum y_{i} m_{i}}{\sum m_{i}}=\frac{(5.0 \mathrm{~m})(4.0 \mathrm{~kg})+(8.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-6.0 \mathrm{~m})(5.0 \mathrm{~kg})}{16 \mathrm{~kg}}=2.9 \mathrm{~m} \end{array} $$ and \(z_{\mathrm{cm}}=0\). These distances are, of course, measured from the origin \((0,0,0,\), .

A \(2.0\) -kg brick is moving at a speed of \(6.0 \mathrm{~m} / \mathrm{s}\). How large a force \(F\) is needed to stop the brick in a time of \(7.0 \times 10^{-4} \mathrm{~s}\) ? Since we have a force and the time over which it acts, that suggests using the impulse equation (i.e., Newton's Second Law): Impulse on brick \(=\) Change in momentum of brick $$ \begin{aligned} F \Delta t &=m v_{f}-m v_{i} \\ F\left(7.0 \times 10^{-4} \mathrm{~s}\right) &=0-(2.0 \mathrm{~kg})(6.0 \mathrm{~m} / \mathrm{s}) \end{aligned} $$ from which \(F=-1.7 \times 10^{4} \mathrm{~N}\). The minus sign indicates that the force opposes the motion.

A 90 -g ball moving at \(100 \mathrm{~cm} / \mathrm{s}\) collides head-on with a stationary 10 -g ball. Determine the speed of each after impact if \((a)\) they stick together, \((b)\) the collision is perfectly elastic, (c) the coefficient of restitution is \(0.90\).

A 6000 -kg truck traveling north at \(5.0 \mathrm{~m} / \mathrm{s}\) collides with a \(4000-\mathrm{kg}\) truck moving west at \(15 \mathrm{~m} / \mathrm{s}\). If the two trucks remain locked together after impact, with what speed and in what direction do they move immediately after the collision?

As shown in Fig. \(8-1\), a 15 -g bullet is fired horizontally into a \(3.000-\mathrm{kg}\) block of wood suspended by a long cord. The bullet lodges in the block. Compute the speed of the bullet if the impact causes the block (and bullet) to swing \(10 \mathrm{~cm}\) above its initial level. Consider first the collision of block and bullet. During the collision, momentum is conserved, so Momentum just before \(=\) Momentum just after $$ (0.015 \mathrm{~kg}) v+0=(3.015 \mathrm{~kg}) V $$ where \(v\) is the speed of the bullet just prior to impact, and \(V\) is the speed of block and bullet just after impact. We have two unknowns in this equation. To find another equation, we can use the fact that the block swings \(10 \mathrm{~cm}\) high. If we let \(\mathrm{PE}_{\mathrm{G}}=0\) at the initial level of the block, energy conservation tells us that $$ \begin{array}{l} \text { KE just after collision }=\text { Final } \mathrm{PE}_{\mathrm{G}} \\ \qquad \frac{1}{2}(3.015 \mathrm{~kg}) V^{2}=(3.015 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.10 \mathrm{~m}) \end{array} $$ From this \(V=1.40 \mathrm{~m} / \mathrm{s}\). Substituting this combined speed into the previous equation leads to \(v=0.28 \mathrm{~km} / \mathrm{s}\) for the speed of the bullet. Notice that we cannot write the conservation of energy equation \(\frac{1}{2} m v^{2}=(m+M) g h\), where \(m=0.015 \mathrm{~kg}\) and \(M=3.000 \mathrm{~kg}\) because energy is lost (through friction) in the collision process.
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Electrostatics

Read Explanation

Rotational Dynamics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Solid State Physics

Read Explanation

Fundamentals of Physics

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.