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Free Body Diagram

Three forces ...

Question

Three forces are acting on a particle of mass m initially in equilibrium. If the first two forces ( $R_{1}$ and $R_{2}$ ) are perpendicular to each other and suddenly the third force ( $R_{3}$ ) is removed, then the acceleration of the particle is

A

\frac{R_{3}}{m}

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B

\frac{R_{1} + R_{2}}{m}

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C

\frac{R_{1} - R_{2}}{m}

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D

\frac{R_{1}}{m}

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Solution

The correct option is A $\frac{R_{3}}{m}$
Given,

Magnitude of Forces on object are $R_{1}, R_{2} & R_{3}$

Two forces $R_{1} & R_{2}$ are normal to each other.

In vector form ${\to R}_{1}, {\to R}_{2} & {\to R}_{3}$ .

Two forces ${\to R}_{1} & {\to R}_{2}$ are normal to each other.

Initially object is in equilibrium, mean vector sum of forces is zero.

${\to R}_{1} + {\to R}_{2} + {\to R}_{3} = 0$ (at equilibrium)

${\to R}_{1} + {\to R}_{2} = - {\to R}_{3}$ (at equilibrium)

In Term of magnitude

$∣ ∣ {\to R}_{1} + {\to R}_{2} ∣ ∣ = ∣ ∣ {\to R}_{3} ∣ ∣$

If force ${\to R}_{3}$ is removed, the remaining forces are ${\to R}_{1}, {\to R}_{2}$ .

$\to F = {\to R}_{1} + {\to R}_{2}$

Acceleration = Force/ Mass $= \frac{\to F}{M}$

$| a | = ∣ ∣ ∣ \frac{\to F}{M} ∣ ∣ ∣ = \frac{∣ ∣ \to F ∣ ∣}{M} = \frac{∣ ∣ {\to R}_{1} + {\to R}_{2} ∣ ∣}{M} = \frac{∣ ∣ {\to R}_{3} ∣ ∣}{M} = \frac{R_{3}}{M}$

Hence, acceleration on mass m is $\frac{R_{3}}{M}$

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