Question

Three particles $A,B$ and $C$ each of mass $m$ are lying at the corners of an equilateral triangle of side $L$. If the particle $A$ is released keeping the particles $B$ and $C$ fixed, the magnitude of instantaneous acceleration of $A$ is

• A. $\sqrt{3}\frac{G{m}^2}{L^2}$
• B. $\sqrt{2}\frac{G{m}^2}{L^2}$
• C. $\sqrt{4}\frac{Gm}{L^2}$
• D. $\sqrt{3}\frac{Gm}{L^2}$

Answer 1

The correct option is A $\sqrt{3}\frac{G{m}^2}{L^2}$

The forces acting on particle at point $A$ are :

$\left(i\right)F_{AB}=\frac{GMM}{L^2}$ along the line joining $B$ form $A$

$\left(ii\right)F_{AC}=\frac{Gmm}{L^2}$ along the line joining $C$ from $A$

The resultant force on A is the vector sum of $F_{AB}$ and $F_{AC}$

$F_{net}=\sqrt{F_{A{B}^2}+F_{A{C}^2}+2F_{AB}{F}_{AC}\cos \theta }=\frac{Gmm}{L^2}\left(\sqrt{2(1+cos\Theta )}\right)$

As $\theta ={60}^{\circ }$ ( equilateral triangle )

So $F_{net}=\frac{G{m}^2}{L^2}\left(\sqrt{2(1+\frac{1}{2})}\right)=\sqrt{3}\frac{G{m}^2}{L^2}$

If instantaneous accelaration of particle is a then :

$F_{net}=ma$

$\Rightarrow a=\frac{F_{net}}{m}=\frac{Gm}{L^2}$