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 acceleration of the particle is:

• A. $F_2/m$
• B. $F_1/m$
• C. $\frac{\sqrt{F_2^2+F_3^2}}{m}$
• D. Both 2 and 3

Answer 1

The correct option is B $F_1/m$

When forces $F_1$, $F_2$ and $F_3$ are acting on the particle, it remains in equilibrium. Now, it is given that $F_2$ and $F_3$ are perpendicular to each other. So, we can infer that $F_1$ has to balance the resultant of the other two forces.

$F_1=F_2\perp F_3$

$\therefore F_1=\sqrt{F_2^2+F_3^2}$

The force $F_1$ is now removed. So, the resultant of $F_2$ and $F_3$ will now make the particle move with force equal to $F_1$.

Hence, the acceleration of the particle is;

$a=\frac{F_1}{m}$