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

Question

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 (A) F1/m (B) F2F3/mF1 (C) (F2-F3)/m (D) F2/m

Tardigrade Admin · Accepted Answer

For equilibrium of system, F1=√F22+F32 As θ=90° In the absence of force F1, Acceleration =( textNet force/ textMass) =(√F22+F32/m) =(F1/m)

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

$F_{1} / m$

$F_{2} F_{3} / m F_{1}$

$(F_{2} - F_{3}) / m$

$F_{2} / m$

For equilibrium of system, $F_{1} = F_{2}^{2} + F_{3}^{2}$ As $θ = 9 0^{\circ}$

In the absence of force $F_{1}$ ,

Acceleration $= \frac{Net force}{Mass}$

$= \frac{F _{2}^{2} + F _{3}^{2}}{m}$

$= \frac{F _{1}}{m}$

Q. 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

F1​/m

F2​F3​/mF1​

(F2​−F3​)/m

F2​/m