Center of Mass — Question 6

Question 6

Masses (2,3,5) kg sit at $x=0,4,10$. Where must a (2)-kg mass be placed so the center is $x=5$?

Question 6 – Solution

Step 1: Let $a$ be the location of the new mass. The four masses are $2$, $3$, $5$, and $2$ kg, so the total mass is $$m=2+3+5+2=12\text{ kg}.$$

Step 2: Write the center-of-mass equation. For point masses on a line, $$\bar x=\frac{\sum m_i x_i}{\sum m_i}.$$ Requiring $\bar x=5$ gives $$5=\frac{2(0)+3(4)+5(10)+2a}{12}.$$

Step 3: Solve for $a$. $$\begin{align*} 60&=0+12+50+2a,\\ 60&=62+2a,\\ 2a&=-2,\\ a&=-1. \end{align*}$$

Step 4: Check. The total moment is $62+2(-1)=60$, and $60/12=5$, as required. $$\boxed{\text{Place the $2$-kg mass at }x=-1.}$$