Three force `F_(1), F_(2), F_(3)` -of which `F_(2)` and `F_(3)` are mutually perpendicular__ act on a particle of mass m so that the particle is stationary. Find the acceleration of the particle when `F_(1)` is withdrawn.

Explanation

As the particle is stationary the resultant force,
`vecF_(1)+vecF_(2)+vecF_(3)=vec0`
Hence `vecF_(2)+vecF_(3)=-vecF_(1)`
`:. vecF_(2)+vecF_(3)` is equal and opposite to `vecF_(1)`.
`vecF_(2) "and" vecF_(3)` are mutually perpendicular, so the magnitude of `(vecF_(2)+vecF_(3)) "is" sqrt(F_(2)^(2)+F_(3)^(2))`.
When `vecF_(1)` is withdrawn, the resultant force on the particle = `vecF_(2)+vecF_(3)` , in that case the acceleration of the particle will be in the opposite direction of `vecF_(1)`.
`:.` Magnitude of the acceleration = `(sqrt(F_(2)^(2)+F_(2)^(3)))/(m)`.