Question

Four particles each of mass 'm' are placed at the four vertices of a square of side 'a'. Find the net force on any one of the particle.

Answer 1

According to pythagors theorem

In $△ACD$

$(AC{)}^2=(AD{)}^2+(DC{)}^2$

$\Rightarrow (AC{)}^2=a^2+a^2$

$\Rightarrow Ac=a\sqrt{2}$

Now,

we know that

$F=\frac{G{m}_1{m}_2}{r^2}$ [gravitational force]

$\Rightarrow F_{AB}=\frac{Gm\times m}{a^2}=\frac{G{m}^2}{a^2}$

$F_{AD}=\frac{G{m}^2}{a^2}$

$F_{AC}=\frac{G{m}^2}{(AC{)}^2}=\frac{G{m}^2}{2a^2}$

$F_{net}=\sqrt{(F_{AB}^2+(F_{AD}^2\left)\right)}$

$=\sqrt{{\left(\frac{G{m}^2}{a^2}\right)}^2+{\left(\frac{G{m}^2}{a^2}\right)}^2}$

$\Rightarrow F_{net}=\sqrt{2}\frac{G{m}^2}{a^2}$

$\tan \theta =\frac{F_{AB}}{F_{AD}}$

$\Rightarrow \tan \theta =1$

$\Rightarrow \theta ={45}^o$

hence

$F_{net}$ and $F_{CD}$ is in same

direction.

$\Rightarrow$ Net force on particle $A=F_{net}+F_{CD}$

$\sqrt{2}\frac{G{m}^2}{a^2}+\frac{G{m}^2}{2a^2}$

$=\frac{G{m}^2}{a^2}[\sqrt{2}+\frac{1}{2}]$

$=\frac{G{m}^2}{2a^2}(\sqrt{2}+1)$