/*! 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 10 A system consists of the followi... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 10

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 \((0,0,0,\), .

Short Answer

Expert verified
The center of mass is at (0.38 m, 2.9 m, 0).

Step by step solution

01

Write the Formulas for Center of Mass

The center of mass for a system of masses in the plane is given by the coordinates \((x_{cm}, y_{cm})\). The formulas are:\[x_{cm} = \frac{\sum x_i m_i}{\sum m_i} \y_{cm} = \frac{\sum y_i m_i}{\sum m_i}\]where \(x_i\) and \(y_i\) are the coordinates of each mass \(m_i\).
02

Calculate Total Mass

Calculate the total mass of the system by summing up the individual masses:\[\sum m_i = 4.0 + 7.0 + 5.0 = 16.0\, \text{kg}\]
03

Substitute for x-coordinate

Substitute the values into the formula for \(x_{cm}\):\[x_{cm} = \frac{(0 \times 4.0) + (3.0 \times 7.0) + (-3.0 \times 5.0)}{16.0}\]Simplifying gives:\[x_{cm} = \frac{0 + 21.0 - 15.0}{16.0} = \frac{6.0}{16.0} \] \[x_{cm} = 0.38\, \text{m}\]
04

Substitute for y-coordinate

Substitute the values into the formula for \(y_{cm}\):\[y_{cm} = \frac{(5.0 \times 4.0) + (8.0 \times 7.0) + (-6.0 \times 5.0)}{16.0}\]Simplifying gives:\[y_{cm} = \frac{20.0 + 56.0 - 30.0}{16.0} = \frac{46.0}{16.0} \] \[y_{cm} = 2.9\, \text{m}\]
05

Conclusion

The coordinates of the center of mass of the system are \((x_{cm}, y_{cm}, z_{cm})\). In this problem, since it is in the \(xy\)-plane, \(z_{cm} = 0\). Therefore, the center of mass is at:\((0.38, 2.9, 0)\)

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Key Concepts

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

Mass Distribution
Understanding mass distribution can help in solving physics problems related to center of mass. Mass distribution refers to how mass is spread out in a system or object. In our scenario, we have three different masses located at different points in the coordinate plane.
  • Mass One: 4.0 kg at (0, 5.0 m)
  • Mass Two: 7.0 kg at (3.0 m, 8.0 m)
  • Mass Three: 5.0 kg at (-3.0 m, -6.0 m)
The mass distribution is not equal, meaning these masses differ in size and also in their positions. To understand the mass distribution, we analyze how these masses are spread in terms of their coordinates and weights. This helps us accurately determine the center of mass, which acts like a balance point for the whole system. Considering both the magnitude (how heavy) and position (where located) of each mass is essential to finding the center of mass. This understanding plays a key role in calculating where the center of mass is. Depending on how masses are spread out, the center of mass will shift. In scenarios with more complex or asymmetric mass distributions, this calculation becomes vital for stability and balance analysis.
Coordinate System
A coordinate system like the Cartesian plane is often used to locate points in two dimensions, which is exactly how the example presented operates. Understanding a coordinate system is essential to solve problems related to mass distribution and other physics problems. In a Cartesian coordinate system, each point is defined by a pair of numerical coordinates. For example, the mass at (3.0 m, 8.0 m) has its position defined in space by these coordinates. These coordinates provide a clear, mathematical way to specify locations of points and allow us to execute precise calculations.
The fundamental purpose of knowing how to use a coordinate system is to identify the location of different masses, which we then use to calculate the center of mass point. We can draw imaginary lines or axes (x and y) on this plane to help visualize these positions. This mental or visual mapping is vital in transferring coordinates to formulaic representations effectively.
Physics Problem Solving
Solving physics problems often involves a systematic approach, breaking down complex concepts into manageable steps. Here, we solve the problem of finding the center of mass using a step-by-step method, utilizing key physics concepts effectively.
The steps are as follows:
  • **Identify Known Values:** Start by locating the given masses and their coordinates.
  • **Total Mass Calculation:** Calculate the total mass of the system, which is simply adding all the individual masses. This total mass is crucial for solving the center of mass equations.
  • **Apply Center of Mass Formulas:** Use the provided formulas for the x-coordinate and y-coordinate of the center of mass. Substitute each mass and its coordinates into these formulas.
Using these steps and understanding, we can compute that the center of mass for the system is located at (0.38 m, 2.9 m, 0). This methodical approach assures that each part of the problem is addressed, making complex problems approachable and solvable, especially as physics problems can often appear daunting or unclear at first glance.
With practice, using this framework becomes second nature, helping students not only solve textbook problems but also strengthen their problem-solving acumen in general physics.

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Most popular questions from this chapter

A \(1200-\mathrm{kg}\) gun mounted on wheels shoots an \(8.00-\mathrm{kg}\) projectile with a muzzle velocity of \(600 \mathrm{~m} / \mathrm{s}\) at an angle of \(300^{\circ}\) above the horizontal. Find the horizontal recoil speed of the gun.

A 16 -g mass is moving in the \(+x\) -direction at \(30 \mathrm{~cm} / \mathrm{s}\), while a 4.0-g mass is moving in the \(-x\) -direction at \(50 \mathrm{~cm} / \mathrm{s}\). They collide head on and stick together. Find their velocity after the collision. Assume negligible friction. This is a completely inelastic collision for which \(\mathrm{KE}\) is not conserved, although momentum is. Let the \(16-\mathrm{g}\) mass be \(m_{1}\) and the \(4.0\) -g mass be \(m_{2}\). Take the \(+x\) -direction to be positive. That means that the velocity of the 4.0-g mass has a scalar value of \(v_{2 x}=-50 \mathrm{~cm} / \mathrm{s}\). We apply the law of conservation of momentum to the system consisting of the two masses: Momentum before impact = Momentum after impact $$ \begin{aligned} m_{1} v_{1 x}+m_{2} v_{2 x} &=\left(m_{1}+m_{2}\right) v_{x} \\ (0.016 \mathrm{~kg})(0.30 \mathrm{~m} / \mathrm{s})+(0.0040 \mathrm{~kg})(-0.50 \mathrm{~m} / \mathrm{s}) &=(0.020 \mathrm{~kg}) v_{x} \\ v_{x} &=+0.14 \mathrm{~m} / \mathrm{s} \end{aligned} $$ (Notice that the 4.0-g mass has negative momentum.) Hence, \(\overrightarrow{\mathbf{v}}=0.14 \mathrm{~m} / \mathrm{s}\) - POSITIVE \(X\) -DIRECTION

An empty \(15000-\mathrm{kg}\) coal car is coasting on a level track at \(5.00 \mathrm{~m} / \mathrm{s}\). Suddenly \(5000 \mathrm{~kg}\) of coal is dumped into it from directly above it. The coal initially has zero horizontal velocity with respect to the ground. Find the final speed of the car.

A \(2.0\) -kg block of wood rests on a tabletop. A \(7.0\) -g bullet is shot straight up through a hole in the table beneath the block. The bullet lodges in the block, and the block flies \(25 \mathrm{~cm}\) above the tabletop. How fast was the bullet going initially?

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.
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