Question

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 :

• A. $F_1/m$
• B. $F_2{F}_3/m{F}_1$
• C. $(F_2-F_3)/m$
• D. $F_2/m$

Answer 1

Answer: (A)

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 ${\vec{F}}_1$.