Question

three particles a b and c each of mass m are lying at the corners of an equilateral triangle of side l with the particle is released keeping the particles p and c fixed the magnitude of instantaneous acceleration of a is

Answer 1

If the particle A is released, it move by the gravitational forces of attraction due to the particles B and C each, which are at angle 30° with the vertical.

The force of attraction due to the particle B is,

• $\bf{F_{AB}}=\sf{\dfrac{Gm^2}{L^2}\,\left(-\hat i\,\sin30^o-\hat j\,\cos30^o\right)}$

The force of attraction due to the particle B is,

• $\bf{F_{AC}}=\sf{\dfrac{Gm^2}{L^2}\,\left(\hat i\,\sin30^o-\hat j\,\cos30^o\right)}$

Hence net force of attraction is,

$\longrightarrow\bf{F=F_{AB}+F_{AC}}$

$\longrightarrow\bf{F}=\sf{\dfrac{Gm^2}{L^2}\,\left(-\hat i\,\sin30^o-\hat j\,\cos30^o\right)+\dfrac{Gm^2}{L^2}\,\left(\hat i\,\sin30^o-\hat j\,\cos30^o\right)}$

$\longrightarrow\bf{F}=\sf{-\dfrac{2Gm^2}{L^2}\,\cos30^o\,\hat j}$

$\longrightarrow\bf{F}=\sf{-\dfrac{2Gm^2}{L^2}\cdot\dfrac{\sqrt3}{2}\,\hat j}$

$\longrightarrow\bf{F}=-\sf{\sqrt3\,\sf{\dfrac{Gm^2}{L^2}\,\hat j}$

By Second Law, the net acceleration of the particle is,

$\longrightarrow\bf{a}=\dfrac{\bf{F}}{\sf{m}}$

$\longrightarrow\bf{a}=-\sf{\sqrt3\,\sf{\dfrac{Gm}{L^2}\,\hat j}$

So the magnitude of the acceleration is,

$\longrightarrow\underline{\underline{\sf{a=\sqrt3\,\dfrac{Gm}{L^2}}}}$

Hence (4) is the answer.