Question

Three masses each of mass m are placed at the vertices of an equilateral triangle ABC of side as shown in figure. The force acting on a mass 2m placed at the centroid O of the triangle is

• A. Zero
• B. $\frac { 2 \mathrm { Gm } ^ { 2 } } { \mathrm { l } ^ { 2 } }$
• C. $\frac { 4 \mathrm { Gm } ^ { 2 } } { 1 ^ { 2 } }$
• D. $\frac { 6 \mathrm { Gm } ^ { 2 } } { 1 ^ { 2 } }$

Answer 1

Answer: (A)

Zero

Answer 2

Draw a perpendicular AD to the side BC.

$\therefore \mathrm { AD } = \mathrm { AB } \sin 60 ^ { \circ } = \frac { \sqrt { 3 } } { 2 } 1$

Distance AO of the centroid O from A is $\frac { 2 } { 3 } \mathrm { AD }$

$\therefore \mathrm { AO } = \frac { 2 } { 3 } \left( \frac { \sqrt { 3 } } { 2 } \mathrm { l } \right) = \frac { 1 } { \sqrt { 3 } }$

By Symmetry, AO = BO = CO $= \frac { 1 } { \sqrt { 3 } }$

Force on mass 2m at O due to mass m at A is

$\mathrm { F } _ { \mathrm { OA } } = \frac { \mathrm { Gm } ( 2 \mathrm {~m} ) } { ( 1 / \sqrt { 3 } ) ^ { 2 } } = \frac { 6 \mathrm { Gm } ^ { 2 } } { 1 ^ { 2 } }$ along OA

Force on mass 2m at O due to mass m at B is

Force on mass 2m at O due to mass m at C is

Draw a line PQ parallel to BC passing through O. Then

Resolving $\overrightarrow { \mathrm { F } } _ { \mathrm { OC } }$into two components.

Resolving acting along OP and OQ are equal in magnitude and opposite in directions. So, they will cancel out while the components acting along OD will add up.

$\therefore$The resultant force on te mass 2m at O is

$\mathrm { F } _ { \mathrm { R } } = \mathrm { F } _ { \mathrm { OA } } - \left( \mathrm { F } _ { \mathrm { OR } } \sin 30 ^ { \circ } + \mathrm { F } _ { \mathrm { OC } } \sin 30 ^ { \circ } \right)$