Question

Consider a system of $2$ boxes, each with mass $M$, which are stacked as shown in the diagram. The top box has height $a$ and the bottom height $b$.
At what $y$-coordinate is the system's center of mass located, if the bottom of the lower box is at $y=0$? Assume both boxes have uniform density (the center of mass of each box is located at its center).

• A. $\frac{1}{4}(a+4b)$
• B. $\frac{1}{2}a$
• C. $\frac{1}{2}(a+3b)$
• D. $\frac{1}{2}(a+b)$
• E. $\frac{1}{4}(a+3b)$

Answer 1

Answer: (E)

$\frac{1}{4}(a+3b)$

Given : $m_1=m_2=M$

Centre of mass of an individual block lies at its geometric centre.

From figure, $y_1=\frac{b}{2}$ $y_2=b+\frac{a}{2}=\frac{2b+a}{2}$

Using $y_{cm}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}=\frac{M\left(\frac{b}{2}\right)+M\left(\frac{2b+a}{2}\right)}{M+M}=\frac{a+3b}{4}$