Find the center of mass of the point masses lying on the x-axis. $\begin{array} { l }
m _ { 1 } = 10 , m _ { 2 } = 1 , m _ { 3 } = 7 \
x _ { 1 } = 3 , x _ { 2 } = 10 , x _ { 3 } = 4
\end{array}$
A) $\bar { x } = \frac { 35 } { 9 }$
B) $\bar { x } = \frac { 67 } { 18 }$
C) $\bar { x } = \frac { 34 } { 9 }$
D) $\bar { x } = \frac { 71 } { 21 }$
E) $\bar { x } = \frac { 71 } { 18 }$

Question

Find the center of mass of the point masses lying on the x-axis. $\begin{array} { l } 
m _ { 1 } = 10 , m _ { 2 } = 1 , m _ { 3 } = 7 \
x _ { 1 } = 3 , x _ { 2 } = 10 , x _ { 3 } = 4
\end{array}$
A) $\bar { x } = \frac { 35 } { 9 }$
B) $\bar { x } = \frac { 67 } { 18 }$
C) $\bar { x } = \frac { 34 } { 9 }$
D) $\bar { x } = \frac { 71 } { 21 }$
E) $\bar { x } = \frac { 71 } { 18 }$

Quizplus · Accepted Answer

We can use the formula for the center of mass of a system of particles:
$$\bar{x} = \frac{\sum\limits_{i=1}^n m_ix_i}{\sum\limits_{i=1}^nm_i}$$
Plugging in the values from the problem:
$$\bar{x}=\frac{10\cdot3+1\cdot10+7\cdot4}{10+1+7}=\frac{34}{9}$$
So the center of mass is at $\frac{34}{9}$ on the x-axis. Therefore, the answer is C.

Find the Center of Mass of the Point Masses Lying m1=10,m2=1,m3=7x1=3,x2=10,x3=4\begin{array} { l } m _ { 1 } = 10 , m _ { 2 } = 1 , m _ { 3 } = 7 \\x _ { 1 } = 3 , x _ { 2 } = 10 , x _ { 3 } = 4\end{array}m1​=10,m2​=1,m3​=7x1​=3,x2​=10,x3​=4​

Find the Center of Mass of the Point Masses Lying $\begin{array} { l } m _ { 1 } = 10 , m _ { 2 } = 1 , m _ { 3 } = 7 \\x _ { 1 } = 3 , x _ { 2 } = 10 , x _ { 3 } = 4\end{array}$