Question

The force $F_1,F_{2}and{F}_3$ are acting on a particle of mass in, such that $F_{2}and{F}_3$ are mutually perpendicular and under their effect, the particle remains stationary. What will be the acceleration a II in particle, if the force $F_1$ is removed?

• A. $a=-\frac{2F}{m}$
• B. $a=\frac{F_1}{m}$
• C. $a=-\frac{F/2}{m}$
• D. $a=-\frac{F/3}{m}$

Answer 1

The correct option is B $a=\frac{F_1}{m}$

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 force $F_1$ has to balance the resultant of the other two forces$.$

$F_3=F_2+F_3$

$\therefore F_1=\sqrt{{\left(F_2\right)}^2+{\left(F_3\right)}^2}$

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

Hence,

all option are incorrect answer is $a=\frac{F_1}{m}$