Three forces bar(F_(1)), bar(F_(2)) and bar(F_(3)) are simultaneously acting on a particle of mass 'm' and keep it in equilibrium. If bar(F_(1)) force is reversed in direction only, the acceleration of the particle will be.

Three forces bar(F_(1)), bar(F_(2)) and bar(F_(3)) are simultaneously

1.	Three forces bar(F_(1)), bar(F_(2)) and bar(F_(3)) are simultaneously acting on a particle of mass 'm' and keep it in equilibrium. If bar(F_(1)) force is reversed in direction only, the acceleration of the particle will be.
Answer» `bar(F_(1))//m` `2bar(F_(1))//m` `-bar(F_(1))//m` `-2bar(F_(1))//m` SOLUTION :Under equilibrium condition `VEC(F_(1))+vec(F_(2))+vec(F_(3))=0` `vec(F_(1))=-(F_(1)+F_(2)), a=(-F_(1)+F_(2)+F_(3))/(m)`