Three particles of masses 50 g, 100 g and 150 g are placed at the vertices of an equilateral triangle of side 1 m (as shown in the figure ). The (x,y) coordinates of the centre of mass will be:

 

 

 

  • Option 1)

    \left [ \frac{\sqrt{3}}{4}m,\frac{5}{12}m \right ]

  • Option 2)

    \left [ \frac{7}{12}m,\frac{\sqrt{3}}{8}m \right ]

  • Option 3)

    \left [ \frac{7}{12}m, \frac{\sqrt{3}}{4} m\right ]

  • Option 4)

    \left [ \frac{\sqrt{3}}{8}m, \frac{7}{12} m\right ]

 

Answers (1)

X_{cm}=\frac{50\times 0+100\times 1+150\times \frac{1}{2}}{50+100+150}=\frac{175}{300}=\frac{7}{12}cm

Y_{cm}=\frac{50\times 0+100\times 0+150\times \frac{\sqrt{3}}{2}}{50+100+150}=\frac{\sqrt{3}}{4}cm

 


Option 1)

\left [ \frac{\sqrt{3}}{4}m,\frac{5}{12}m \right ]

Option 2)

\left [ \frac{7}{12}m,\frac{\sqrt{3}}{8}m \right ]

Option 3)

\left [ \frac{7}{12}m, \frac{\sqrt{3}}{4} m\right ]

Option 4)

\left [ \frac{\sqrt{3}}{8}m, \frac{7}{12} m\right ]

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