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  4. When forces F1, F2, F3 are acting on a particle of mass m such that F2 and F3 are mutually perpendicular, then the particle remains stationary. If the force F1 is now removed then the acceleration of the particle is :

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

AIEEEAIEEE 2002Laws of Motion

Solution:

The particle is remains stationary under the action of three forces $\vec{ F }_{ 1 }, \vec{ F }_{2}$, and $\vec{ F }_{3}$, it means resultant force is zero.
$\vec{ F }_{1}=-\left(\vec{ F }_{2}+\vec{ F }_{3}\right)$
Since, in second case $F_{1}$ is removed (in terms of magnitude we are talking now), the forces acting are $F_{2}$ and $F_{3}$ the resultant of which has the magnitude as $F_{1}$, so acceleration of particle is $\frac{F_{1}}{m}$ in the direction opposite to that of $\vec{ F }_{1}$.