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When forces $\bar{F_{1}}, \bar{F_{2}}, \bar{F_{3}}$ are acting on a particle of mass m such that $\bar{F_{2}}$ , and $\bar{F_{3}}$ are mutually perpendicular then the particle remains stationary, if the force $\bar{F_{1}}$ is now removed then the magnitude of acceleration of the particle is

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F₁ / m

F₁ F₃ /m F₁

(F₂– F₃) / m

F₂ / m

answer is A.

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Detailed Solution

When ${\bar{F}}_{1} = {\bar{F}}_{2} + {\bar{F}}_{3}$

After removal of ${\bar{F}}_{1}$ resultant force on the particle is equal to ${\bar{F}}_{1}$

Since , ${\bar{F}}_{1} = {\bar{F}}_{2} + {\bar{F}}_{3}$

Acceleration $= \frac{|{\bar{F}}_{1}|}{m}$

$= \frac{F_{1}}{m}$

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