3 equal masses of m kg …
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Preeti Dabral 1 year, 5 months ago
{tex}\left| \overrightarrow { F _ { A } } \right| = \left| \overrightarrow { F _ { B } } \right| = \left| \overrightarrow { F _ { C } } \right| = \frac { G M \cdot 2 M } { ( 1 ) ^ { 2 } } = 2 G M ^ { 2 }{/tex}
If we consider direction parallel to BC as x-axis and perpendicular direction as y-axis, then as shown in figure, we have
{tex}\vec { F } _ { A } = 2 G M ^ { 2 } \hat { j }{/tex}
{tex}\vec { F } _ { B } = \left( - 2 G M ^ { 2 } \cos 30 ^ { \circ } \hat { i } - 2 G M ^ { 2 } \sin 30 ^ { \circ } \hat { j } \right){/tex}
and {tex}\overrightarrow { F _ { C } } = \left( 2 G M ^ { 2 } \cos 30 ^ { \circ } \hat { i } - 2 G M ^ { 2 } \sin 30 ^ { \circ } \hat { j } \right){/tex}
Therefore, the net force on mass 2M placed at the centroid O is given by,
{tex}\vec { F } = \vec { F } _ { A } + \vec { F } _ { B } + \vec { F } _ { C } {/tex}
{tex}= 2 G M ^ { 2 } \left[ j + \left( - \frac { \sqrt { 3 } } { 2 } \hat { i } - \frac { 1 } { 2 } \hat { i } \right) + \left( \frac { \sqrt { 3 } } { 2 } \hat { i } - \frac { 1 } { 2 } \hat { j } \right) \right]{/tex}
{tex}=0{/tex}
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