Question

Centre of mass of $ 3$ particles $10\, kg , 20\, kg$ and $30 \,kg$ is at $(0,0,0) .$ Where should a particle of mass $40\, kg$ 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 $40 kg$ be $(x, y, z)$.
$\therefore X_{ CM }=\frac{\sum m_{i} x_{i}}{\sum m_{i}}$ or $3=\frac{10 \times 0+20 \times 0+30 \times 0+40 \times x}{10+20+30+40}$
$\therefore x=\frac{300}{40}=7.5$
$Y_{ CM }=\frac{\sum m_{i} y_{i}}{\sum m_{i}}$
$3=\frac{10 \times 0+20 \times 0+30 \times 0+40 \times y}{10+20+30+40} \Rightarrow y=\frac{300}{40}=7.5$
$Z_{ CM }=\frac{\sum m_{i} z_{i}}{\sum m_{i}}$
$3=\frac{10 \times 0+20 \times 0+30 \times 0+40 \times z}{10+20+30+40} \Rightarrow z=\frac{300}{40}=7.5$