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Question

A particle of mass $m$ is at the centre of a regular hexagon and is acted upon by six forces each of magnitude $F$ directed towards the vertices of the hexagon. If one of the forces is reversed in direction, the acceleration of the particle is

A

0

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B

\frac{F}{m}

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C

\frac{\sqrt{3} F}{m}

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D

\frac{2 F}{m}

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Solution

The correct option is D $\frac{2 F}{m}$
$L e t t h e s i x f o r c e s b e_{1},_{2},_{3},_{4},_{5} a n d_{1}$
$I n i t i a l l y,$
$_{1} +_{2} +_{3} +_{4} +_{5} +_{6} = 0$
$∴_{2} +_{3} +_{4} +_{5} +_{6} = -_{1} . . . e q (i)$
$B u t w h e n_{1} i s r e v e r s e d,$
$_{n e t} = -_{1} +_{2} +_{3} +_{4} +_{5} +_{6}$
From eq(i),
$_{n e t} = -_{1} -_{1}$
$∴ A c c e l e r a t i o n = \frac{2 F_{1}}{m}$

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