Question

1 mark for review
object mass (kg) x position (m) y position (m)
a 1 0 12
b 2 0 0
c 1 24 0
three small objects a, b, and c each have the masses and positions given in the table. what is the position (x, y) of the center of mass of the three - object system?
a (6 m, 3 m)
b (8 m, 4 m)
c (12 m, 6 m)
d (24 m, 12 m)

Explanation:

Step1: Calculate x - coordinate of center - of - mass

The formula for the x - coordinate of the center of mass $x_{cm}=\frac{\sum_{i = 1}^{n}m_ix_i}{\sum_{i = 1}^{n}m_i}$. Here, $m_A = 1$ kg, $x_A=0$ m, $m_B = 2$ kg, $x_B = 0$ m, $m_C=1$ kg, $x_C = 24$ m. $\sum_{i = 1}^{3}m_i=m_A + m_B+m_C=1 + 2+1=4$ kg, $\sum_{i = 1}^{3}m_ix_i=m_Ax_A + m_Bx_B+m_Cx_C=1\times0+2\times0 + 1\times24=24$ kg·m. So, $x_{cm}=\frac{24}{4}=6$ m.

Step2: Calculate y - coordinate of center - of - mass

The formula for the y - coordinate of the center of mass $y_{cm}=\frac{\sum_{i = 1}^{n}m_iy_i}{\sum_{i = 1}^{n}m_i}$. Here, $m_A = 1$ kg, $y_A = 12$ m, $m_B = 2$ kg, $y_B = 0$ m, $m_C=1$ kg, $y_C = 0$ m. $\sum_{i = 1}^{3}m_i = 4$ kg, $\sum_{i = 1}^{3}m_iy_i=m_Ay_A + m_By_B+m_Cy_C=1\times12+2\times0+1\times0 = 12$ kg·m. So, $y_{cm}=\frac{12}{4}=3$ m.

Answer:

A. $(6\ m,3\ m)$