1.

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 isA. (a) `F//m`B. (b) `F_2F_3//mF_1`C. (c) `(F_2-F_3)//m`D. (d) `F_2//m`

Answer» Correct Answer - A
When `F_1`, `F_2` and `F_3` are acting on a particle then the particle remains stationary. This means that the resultant of `F_1`, `F_2` and `F_3` is zero. When `F_1` is removed, `F_2` and `F_3` will remain. But the resultant of `F_2` and `F_3` should be equal and opposite to `F_1`, .i.e. `|vecF_2+vecF_3|=|vecF_1|`
`:.` `a=(|vecF_2+vecF_3|)/(m)impliesa=(F_1)/(m)`


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