Question

A block of mass $3m$ and base ${}^{\prime }{a}^{\prime }$, height ${}^{\prime }{b}^{\prime }$, is resting on the smooth horizontal surface. A man of mass $m$ goes from P to Q on the wedge. Find the displacement of the centre of mass of (man + block) system.

Answer 1

As shown in fig. $\left(a\right)$

$X_{com}=\frac{ma+3m\left(\frac{a}{3}\right)}{4m}=\frac{2ma}{4m}=\frac{a}{2}{Y}_{com}=\frac{0\times m+3m\times \frac{a}{3}}{4m}=\frac{1}{4}a$

When man is at position then,

${X^{\prime }}_{com}=\frac{1}{4}a{Y^{\prime }}_{com}=\frac{1}{2}a$

Displacement of centure of mass$=\sqrt{{(X_{com}-{X^{\prime }}_{com})}^2+{(Y_{com}-{Y^{\prime }}_{com})}^2}=\sqrt{{(\frac{a}{2}-\frac{2a}{4})}^2+{(\frac{a}{4}-\frac{a}{2})}^2}=\sqrt{\frac{a^2}{4^2}+\frac{a^2}{4^2}}=\frac{a}{2\sqrt{2}}$