Three forces F1- F2 and F3 are simultaneously acting on a particle of mass -m- and keep it in equilibrium- If F1 force were reversed in directiononly- the acceleration of the particle will be

The correct option is D 2F_1/m Now, #160; [ ↓ F_ 1 + F_ 1 #160;; ∴ a ce= 2 F_ 1 #160; / m ...

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Standard X

Physics

Translational and Rotational Equilibrium

Three forces ...

Question

Three forces $_{1}$ , $_{2}$ and $_{3}$ are simultaneously acting on a particle of mass 'm' and keep it in equilibrium. If $_{1}$ force were reversed in directiononly, the acceleration of the particle will be

_{1} / m

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2_{1} / m

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-_{1} / m

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- 2_{1} / m

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Solution

The correct option is D $- 2_{1} / m$
Now,
$\begin{matrix} ↓ - \to F_{1} + - \to F_{1} ∴ \to a c e = \frac{- 2 - \to F_{1}}{m} \end{matrix}$
$∴$ Option $D$ is correct.

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Three forces ¯¯¯¯¯¯F1, ¯¯¯¯¯¯F2 and ¯¯¯¯¯¯F3 are simultaneously acting on a particle of mass 'm' and keep it in equilibrium. If ¯¯¯¯¯¯F1 force were reversed in directiononly, the acceleration of the particle will be

The correct option is D −2¯¯¯¯¯¯F1/mNow, ↓−→F1+−→F1∴→ace=−2−→F1m∴ Option D is correct.

Three forces $_{1}$ , $_{2}$ and $_{3}$ are simultaneously acting on a particle of mass 'm' and keep it in equilibrium. If $_{1}$ force were reversed in directiononly, the acceleration of the particle will be

The correct option is D $- 2_{1} / m$
Now,
$\begin{matrix} ↓ - \to F_{1} + - \to F_{1} ∴ \to a c e = \frac{- 2 - \to F_{1}}{m} \end{matrix}$
$∴$ Option $D$ is correct.