Answer to Question #113842 in Classical Mechanics for Ettu Oloruntoba

Question

A light inextensible string of lenght l has one end attached to the end A of a smooth heavy rod AB whose lenght is 2a.The other end of the string is attached to a small smooth ring C (whose weight may be neglected) which slides over the rod.The rod is suspended with the string over a smooth peg P. Show that, in the position of equilibrium
(i) The portion PC of the string is at right angles to the rod.
(ii)The length x of the portion PA is given by the equation
l((x^2)+(a^2))=2x^3 .

Answer 1

As per the question,

As it will release, then due to self weight of the string string will start to slide, and it will come in equilibrium, if sum of all force the force will get zero.

there is force gravity in downward direction, tension in the two strings are T1 and T2.

So, for the equilibrium condition,$\dfrac{T_1}{\sin\theta_1}=\dfrac{T_2}{\sin(\theta_2)}=\dfrac{w}{\sin\theta}$

But here the string is uniform, so here the left side of the string and right side of the string will be at the same length, $\sin \theta_1=\sin\theta_2$

The tension in the string connected in both side will have the equal tension, so, T1=T2

it is possible if length of the both side of the string is same, hence the angle between the string must be 90.

Now, from the above $x^2+a^2=2x^3$