1.

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

Answer»

`F_(1)//m`
`F_(2)F_(3)//mF_(1)`
`(F_(2)-F_(3))//m`
`F_(2)//m`

SOLUTION :For the equilibrium of the body `F_(1)` must act opposite to `F_(2)` and `F_(3)` and must be EQUAL and opposite to the resultant of `F_(2)` and `F_(3)` as shown

Thus `F_(1)=F =ma`or `a=(F)/(m)=(F_(1))/(m)`
HENCE choice is (a).


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