Question

If you can change location of the point A on the wall and hence the orientation of the string OA without altering the orientation of the string OB as shown in figure-II. What angle should the string OA make with the wall so that a minimum tension is developed in it?

Answer 1

For equilibrium horizontal force must be balanced.

$T_1\sin \theta =T_2\cos 60°\to \left(1\right)$

$T_1\cos \theta =T_2\sin 60=10\sqrt{3}\to \left(1\right)$

$T_1\cos \theta +\frac{2T_1\sin \theta \times \sqrt{3}}{2}=10\sqrt{3}$

$T_1[\cos \theta +\sqrt{3}\sin \theta ]=10\sqrt{3}$

$T_1=\frac{10\sqrt{3}}{\cos \theta +\sqrt{3}\sin \theta }$

$T_2=2T_1\sin 60$

$T_2=\frac{30}{\cos \theta +\sqrt{3}\sin \theta }$

For tension to be minimum in $0A$.

$\cos \theta +\sqrt{3}\sin \theta$ shouold be maximum, which is maximum value of $a\cos\theta+\b\sin\theta=\sqrt{a^2+b^2}$

Here, $T_{max}=\frac{10\sqrt{3}}{2}$ when, $\theta =60°$