A) \[\left( \frac{7}{12}m,\frac{\sqrt{3}}{4}m \right)\]
B) \[\left( \frac{5}{12}m,\frac{\sqrt{3}}{5}m \right)\]
C) \[\left( \frac{5}{13}m,\frac{\sqrt{3}}{5}m \right)\]
D) \[\left( \frac{8}{12}m,\frac{\sqrt{3}}{5}m \right)\]
Correct Answer: A
Solution :
\[{{X}_{cm}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] \[=\frac{\begin{align} & (1.0\,kg)(0m)+(2.0kg)(1.0m)+(3.0kg) \\ & (1.0m.\cos 60{}^\circ ) \\ \end{align}}{(1.0kg+2.0kg+3.0kg)}\] \[=\frac{7}{12}m\] \[{{y}_{cm}}=\frac{\Sigma {{m}_{1}}{{y}_{1}}}{\Sigma {{m}_{1}}}=\frac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}+{{m}_{3}}{{y}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] \[=\frac{\begin{align} & (1.0kg)(0m)+(2.0kg)(1.0m)+(3.0kg) \\ & (1.0m.\sin 60{}^\circ ) \\ \end{align}}{(1.0kg+2.0kg+3.0kg)}\] \[=\frac{\sqrt{3}}{4}m\]
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