Question

Four cubes of side \(a=1m\)and of mass \(40g, 20g,10g ~\text{and}~20g\) are arranged in \(xy\) plane as shown in the figure. The coordination of the center of mass of the combination with respect to \(0~\text{are}~x_{cm}~\text{and}~y_{cm}\centerdot\)
Find \(\left | x_{cm}+y_{cm} \right |\centerdot\)

Answer 1

Given: \(a=1m,m_{1}=20g,m_{2}=10g,m_{3}=20g,m_{4}=40g\)
Since all plate have uniform distribution of mass, hence they will have center of mass exactly at center of square
\(x_{cm}=\dfrac{20\times \dfrac{a}{2}+30\times \dfrac{3a}{2}+20\times \dfrac{5a}{2}+40\times \dfrac{a}{2}}{90}\)

\(x_{cm}=\dfrac{19a}{18}\)

\(y_{cm}=\dfrac{20\times \dfrac{a}{2}+10\times \dfrac{a}{2}+20\times \dfrac{a}{2}+40\times \dfrac{3a}{2}}{90}\)

\(y_{cm}=\dfrac{17a}{18}\)

\(\left | x_{cm}+y_{cm} \right |\) \(=\dfrac{19a}{18}\) +\(\dfrac{17a}{18}\)

\(=\dfrac{36a}{18}\)

= 2

Final Answer: 2