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Chapter 14: Problem 7

Sketch the following systems on a number line and find the location of the center of mass. $$m_{1}=10 \mathrm{kg} \text { located at } x=3 \mathrm{m} ; m_{2}=3 \mathrm{kg} \text { located at } x=-1 \mathrm{m}$$

Short Answer

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Question: Determine the center of mass for a system of two masses (m1 = 10 kg at x1 = 3 m and m2 = 3 kg at x2 = -1 m) on a number line. Answer: The center of mass for the given system of two masses is approximately 2.08 m on the number line.

Step by step solution

01

Label the given parameters

(m1 = 10 kg, x1 = 3 m, m2 = 3 kg, x2 = -1 m) The first mass (m1) has a mass of 10 kg and is located at position x1 = 3 m. The second mass (m2) has a mass of 3 kg and is located at position x2 = -1 m.
02

Calculate the center of mass using the formula

Calculate x_cm using the formula: $$x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$$ Plugging the values into the formula, we get: $$x_{cm} = \frac{10 \times 3 + 3 \times (-1)}{10 + 3}$$ Solve for x_cm: $$x_{cm} = \frac{30 - 3}{13} = \frac{27}{13} \approx 2.08 \ \mathrm{m}$$
03

Sketch the number line and locate the masses and center of mass

Draw a number line from -2 m to 4 m and mark the positions of the masses and the center of mass: 1. m1 = 10 kg at x1 = 3 m 2. m2 = 3 kg at x2 = -1 m 3. Center of mass at x_cm = 2.08 m, approximately. It is important to visualize the problem. This allows us to see that, since m1 is heavier than m2, the center of mass is closer to m1, as confirmed by the calculation.

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Key Concepts

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

Mass Distribution
In physics, understanding how mass is distributed is crucial to determining the center of mass. Mass distribution refers to how masses are positioned in a given system.
The way mass is apportioned or scattered affects the balance and motion of the system. In the example provided, we have two masses: 10 kg and 3 kg. These are not only different in magnitude, but they are also positioned at different points in space.
  • The larger mass (10 kg) is situated at 3 meters
  • The smaller mass (3 kg) is located at -1 meter
This uneven mass distribution results in a center of mass that is influenced by the distances and sizes of the masses. Generally, a larger mass closer to the origin will pull the center of mass toward itself, as seen in our example.
Number Line
A number line is a visual tool used to represent positions of objects relative to a given starting point, or origin. In this scenario, the number line extends from -2 meters to 4 meters. It helps to illustrate where each mass is located and to provide a reference for calculations.
When sketching a number line:
  • Mark all relevant positions. For example, place the first mass (10 kg) at 3 meters and the second mass (3 kg) at -1 meter.
  • Highlight the calculated center of mass, in this case, approximately 2.08 meters.
This visual layout makes it easier to see how distances affect the center of mass. It shows that the center leans toward the larger mass, which is heavier and farther right on the number line.
Calculation Formula
Finding the center of mass involves using a specific formula that takes into account both the mass and position of each object. The formula for the center of mass, denoted as \(x_{cm}\), is:\[x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}\]In this formula:
  • \(m_1\) and \(m_2\) represent the masses of the objects.
  • \(x_1\) and \(x_2\) denote the positions of these masses along the number line.
By inserting the given values (10 kg at 3 m and 3 kg at -1 m) into the formula, we find that:\[x_{cm} = \frac{10 \times 3 + 3 \times (-1)}{10 + 3} = \frac{27}{13} \approx 2.08 \, \text{meters}\]This computation reveals that while the 10 kg mass has a significantly stronger influence due to its greater mass and positive distance, the overall center of mass lies at 2.08 meters.

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Most popular questions from this chapter

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