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Chapter 13: Problem 8

Sketch the following systems on a number line and find the location of the center of mass. $$\begin{aligned} &m_{1}=8 \mathrm{kg} \text { located at } x=2 \mathrm{m} ; m_{2}=4 \mathrm{kg} \text { located at } x=-4 \mathrm{m}\\\ &m_{3}=1 \mathrm{kg} \text { located at } x=0 \mathrm{m} \end{aligned}$$

Short Answer

Expert verified
Answer: The center of mass for this system is at x=0m on the number line.

Step by step solution

01

Sketch the system

First, let's plot each mass (m1, m2, and m3) on a number line. Place m1 at x=2m, m2 at x=-4m, and m3 at x=0m. Draw a dot or a small circle at each of these positions and label them with their corresponding masses.
02

Calculate total mass

To find the center of mass, we need to compute the total mass of the system, which is the sum of the individual masses: Total Mass = \(m_1 + m_2 + m_3\) Total Mass = \(8 kg + 4 kg + 1 kg = 13 kg\)
03

Calculate the center of mass

Using the equation for the center of mass, we can now calculate the weighted average of the positions: $$\bar{x} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}$$ Replace with given values: $$\bar{x} = \frac{(8 kg)(2 m) + (4 kg)(-4 m) + (1 kg)(0 m)}{13 kg}$$ Calculate the numerator: $$= \frac{16 kgm - 16 kgm + 0 kgm}{13 kg}$$ $$= \frac{0 kgm}{13 kg}$$ and then the quotient: $$\bar{x} = 0 m$$
04

Plot the center of mass

The center of mass is at x=0m. On the number line, draw a vertical dashed line or arrow pointing towards the number line, label it with "center of mass" and indicate that its position is at x=0m. In conclusion, the center of mass for this system is located at x=0m on the number line.

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Most popular questions from this chapter

An important integral in statistics associated with the normal distribution is \(I=\int_{-\infty}^{\infty} e^{-x^{2}} d x .\) It is evaluated in the following steps. a. Assume that $$\begin{aligned} I^{2} &=\left(\int_{-\infty}^{\infty} e^{-x^{2}} d x\right)\left(\int_{-\infty}^{\infty} e^{-y^{2}} d y\right) \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}-y^{2}} d x d y \end{aligned}$$ where we have chosen the variables of integration to be \(x\) and \(y\) and then written the product as an iterated integral. Evaluate this integral in polar coordinates and show that \(I=\sqrt{\pi} .\) Why is the solution \(I=-\sqrt{\pi}\) rejected? b. Evaluate \(\int_{0}^{\infty} e^{-x^{2}} d x, \int_{0}^{\infty} x e^{-x^{2}} d x,\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\) (using part (a) if needed).

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