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 is

Laws of Motion PYQ — JEE Main 2002 · Shift 0 · Q172 with Solution

View written solutionFree

Correct answer: A

Since the particle remains stationary under the action of three forces $F_1, F_2, F_3$ , the net force must be zero:

\vec F_1 + \vec F_2 + \vec F_3 = 0

So,

\vec F_1 = -(\vec F_2 + \vec F_3)

This means that the magnitude of $F_1$ is equal to the resultant of $F_2$ and $F_3$ .

Because $\vec F_2 \perp \vec F_3$ ,

F_1 = |\vec F_2 + \vec F_3| = \sqrt{F_2^2 + F_3^2}

Now only $\vec F_2$ and $\vec F_3$ act on the particle. Their resultant is:

\vec F_{\text{net}} = \vec F_2 + \vec F_3

Its magnitude is

F_{\text{net}} = \sqrt{F_2^2 + F_3^2} = F_1

By Newton's second law,

a = \frac{F_{\text{net}}}{m} = \frac{F_1}{m}

Therefore, the correct answer is Option A.