Problem 7 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The center of mass is located at approximately \(x_{cm} \approx 2.0769\, \mathrm{m}\) on the number line.

Step 1: Sketch the number line

Draw a horizontal number line and mark the positions of the two masses. Place \(m_1 = 10\, \text{kg}\) at \(x = 3\, \text{m}\) and \(m_2 = 3\, \text{kg}\) at \(x = -1\, \text{m}\).

Step 2: Apply the center of mass formula

Now, we will use the center of mass formula to find the location of the center of mass: $$x_{cm} = \frac{\sum x_i m_i}{\sum m_i}$$ First, calculate the sum of the product of each mass and its location on the number line: $$\sum x_i m_i = (3\,\text{m})(10\,\text{kg}) + (-1\,\text{m})(3\,\text{kg}) = 30\,\text{kg}\cdot\text{m} - 3\,\text{kg}\cdot\text{m} = 27\, \text{kg}\cdot\text{m}$$ Next, find the total mass: $$\sum m_i = m_1 + m_2 = 10\,\text{kg} + 3\,\text{kg} = 13\,\text{kg}$$ Now, divide the sum of the product of each mass and its location on the number line by the total mass: $$x_{cm} = \frac{27\, \text{kg}\cdot\text{m}}{13\,\text{kg}} = 2.0769\, \mathrm{m}$$ Thus, the center of mass is located at \(x_{cm} \approx 2.0769\, \mathrm{m}\) on the number line.

Step 3: Add the center of mass to the sketch

Mark the location of the center of mass, \(x_{cm} \approx 2.0769\, \mathrm{m}\), on the number line. You will find it closer to mass \(m_1\) since it has a larger mass and thus contributes more weight to the center of mass.