Solve This Physics Problem

Question

Three point masses (0.07 kg, 0.16 kg, and 0.115 kg) are placed at the corners of a right triangle of sides 6 cm and 7 cm. Find the center of mass of the three-mass system (in cm). (Assume that the +x-axis is to the right and the +y-axis is up along the page. Take the origin to be at the position of the 160 g mass.)

I managed to get ycm right, but not xcm.

Answer 1

Hi Elle

To find the center of mass of the three-mass system, you can use the following equations:

\[x_{\text{cm}} = \frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i}\]

\[y_{\text{cm}} = \frac{\sum_{i=1}^{n} m_i y_i}{\sum_{i=1}^{n} m_i}\]

Where:

- \(x_{\text{cm}}\) and \(y_{\text{cm}}\) are the x and y coordinates of the center of mass.

- \(m_i\) is the mass of each particle.

- \(x_i\) and \(y_i\) are the x and y coordinates of each particle.

- \(n\) is the number of particles.

For this problem:

- \(m_1 = 0.07 \, \text{kg}\), \(x_1 = 0 \, \text{cm}\), and \(y_1 = 6 \, \text{cm}\) (the 0.07 kg mass is at the origin).

- \(m_2 = 0.16 \, \text{kg}\), \(x_2 = 7 \, \text{cm}\), and \(y_2 = 0 \, \text{cm}\).

- \(m_3 = 0.115 \, \text{kg}\), \(x_3 = 0 \, \text{cm}\), and \(y_3 = 0 \, \text{cm}\).

Now, plug these values into the center of mass equations:

\[x_{\text{cm}} = \frac{(0.07 \cdot 0) + (0.16 \cdot 7) + (0.115 \cdot 0)}{0.07 + 0.16 + 0.115}\]

Simplify the numerator:

\[x_{\text{cm}} = \frac{1.12}{0.345}\]

\[x_{\text{cm}} \approx 3.25 \, \text{cm}\]

So, the x-coordinate of the center of mass is approximately 3.25 cm. The y-coordinate, as you mentioned, can be found separately using a similar method.

To find the y-coordinate of the center of mass (\(y_{\text{cm}}\)), you can use the same center of mass equation:

\[y_{\text{cm}} = \frac{\sum_{i=1}^{n} m_i y_i}{\sum_{i=1}^{n} m_i}\]

Plug in the values for this problem:

- \(m_1 = 0.07 \, \text{kg}\), \(x_1 = 0 \, \text{cm}\), and \(y_1 = 6 \, \text{cm}\).

- \(m_2 = 0.16 \, \text{kg}\), \(x_2 = 7 \, \text{cm}\), and \(y_2 = 0 \, \text{cm}\.

- \(m_3 = 0.115 \, \text{kg}\), \(x_3 = 0 \, \text{cm}\), and \(y_3 = 0 \, \text{cm}\).

Now, calculate the \(y_{\text{cm}}\):

\[y_{\text{cm}} = \frac{(0.07 \cdot 6) + (0.16 \cdot 0) + (0.115 \cdot 0)}{0.07 + 0.16 + 0.115}\]

Simplify the numerator:

\[y_{\text{cm}} = \frac{0.42}{0.345}\]

\[y_{\text{cm}} \approx 1.22 \, \text{cm}\]

So, the y-coordinate of the center of mass is approximately 1.22 cm.

I hope this helps out