Two particles A and B, each O of mass m are kept stationary by applying a horizontal force F = mg on particle B as shown in figure. Then,

Resultant of mg and mg is $\sqrt{2}$ mg.

Therefore, T₂ should be equal and opposite of this. Or

$T_2=\sqrt{2}mg ....\left(i\right)$

Further,

$T_2\cos \beta =mg ....\left(ii\right)$

and      $T_2\sin \beta =mg ....\left(iii\right)$

or         $\sin \beta =\cos \beta \Rightarrow \beta =45°$

$$\begin{gathered} T_1\cos \alpha =mg+T_2\cos \beta \\ =mg+\sqrt{2}mg\left(\frac{1}{\sqrt{2}}\right) \end{gathered}$$

or         $T_1\cos \alpha =2mg ....\left(iv\right)$

$$\begin{gathered} T_1\sin \alpha =T_2\sin \beta =\sqrt{2}mg\left(\frac{1}{\sqrt{2}}\right) \\ \therefore T_1\sin \alpha =mg ....\left(v\right) \end{gathered}$$

From Eqs. (iv) and (v), we get

$\tan \alpha =\frac{1}{2}$ and $T_1=\sqrt{5}mg$

$\tan \beta = \tan 45°=1$ and $T_2=\sqrt{2}mg$

$\therefore \tan \beta =2\tan \alpha$ and _T$\sqrt{2}{T}_1=\sqrt{5}{\mathrm{T}}_2$