Question

Centre of mass of $3$ particles $10kg,20kg$ and $30kg$ is at $(0,0,0).$ Where should a particle of mass $40kg$ be placed so that the combined centre of mass will be at (3,3,3)$?$

• A. $(0,0,0)$
• B. $(7.5,7.5,7.5)$
• C. $(1,2,3)$
• D. $(4,4,4)$

Answer 1

Answer: (B)

Let the position coordinates of $40kg$ be $(x,y,z)$.
$\therefore X_{CM}=\frac{\sum m_i{x}_i}{\sum m_i}$ or $3=\frac{10\times 0+20\times 0+30\times 0+40\times x}{10+20+30+40}$
$\therefore x=\frac{300}{40}=7.5$
$Y_{CM}=\frac{\sum m_i{y}_i}{\sum m_i}$
$3=\frac{10\times 0+20\times 0+30\times 0+40\times y}{10+20+30+40}\Rightarrow y=\frac{300}{40}=7.5$
$Z_{CM}=\frac{\sum m_i{z}_i}{\sum m_i}$
$3=\frac{10\times 0+20\times 0+30\times 0+40\times z}{10+20+30+40}\Rightarrow z=\frac{300}{40}=7.5$