When forces `F_(1), F_(2)` and `F_(3)` are acting on a particle of mass m such that `F_(2)` and `F_(3)` are m utually perpendicular, then the particle remains stationary. If the force F x is now removed, then the acceleration of the particle is .........

Options

A. `(R_(3))/(m)`
B. `(R_(1)+R_(2))/(m)`
C. `(R_(1)-R_(2))/(m)`
D. `(R_(1))/(m)`

Answer

A

Explanation

In equilibrium resultant of`R_(1) ` and `R_(2)` is `R_(3) ` and will be in opposite direction
`R_(3) = sqrt(R_(1)^(2) + R_(2)^(2))`
Now acceleration `=("resultant force " ) /( " mass ")`
`a= (R_(3))/(m ) = (sqrt(R_(1)^(2) + R_(2)^(2))/( m ))`