Finding tension in a freely sliding ring on a wire

Question

A heavy small ring of weight $W$ is free to slide on a smooth surface wire of radius $a$, fixed in a vertical plane. It is attached by a string of length $l$ where

$$2a > l > a\sqrt{2}$$

to a point on the wire in a horizontal line with the centre. Find tension in the string.

Approach :

1.

Here, If A be the point where string is attached to wire, P be the equilibrium position of string, I get Tension as

$$ \dfrac{W(l^2-2a^2)}{a\sqrt{4a^2-l^2}}$$ 2.

Here, If A be the point where string is attached to wire, P be the equilibrium position of string, I get Tension as

$$ \dfrac{- W(l^2-2a^2)}{a\sqrt{4a^2-l^2}}$$

Clearly, 2nd Approach is wrong as magnitude of tension can't be negative. But why is it wrong ? Why isn't this diagram possible ?

I have verified that with given restriction on $l$, the 2nd diagram should very well be possible. Can anyone point out where am I going wrong ? Thanks!

I had it under Maths topic. So posted here. Can you direct to the link ? Thanks! — user1611542, Jul 20 '18 at 18:21
Did you use the same local coordinate system for $P$ in both cases? If so, your answer may be correct because the two tensions are in the opposite directions. — John Douma, Jul 22 '18 at 18:55
@JohnDouma Since I have already considered tension with directions, it shouldn't come out to be -ive. — user1611542, Jul 23 '18 at 17:59
In the first case, $T$ and $W$ are pointing downward while in the second case $W$ points downward while $T$ points upward. — John Douma, Jul 23 '18 at 18:02
Yeah, That's my question. Why can't the second case be possible ? Given the restrictions on l, it is very much possible. But you can say after solving, that it isn't. But can we do without solving, just by testing some conditions. — user1611542, Jul 23 '18 at 18:08

Cesareo · Accepted Answer · 2018-07-22T17:48:44.283

Considering first the first position

By geometric considerations the angle $\angle PAB = \alpha $ is such that

$$ 2 a \cos\alpha = l $$

Now calling

$$ \vec R = r(\cos(2\alpha),\sin(2\alpha))\\ \vec W = w(0,-1)\\ \vec T =- t(\cos\alpha,\sin\alpha) $$

in equilibrium we have

$$ \vec R + \vec W + \vec T = 0 $$

or

$$ \left\{ \begin{array}{rcl} r \cos (2 \alpha )-t \cos (\alpha )& = & 0 \\ -w-t \sin (\alpha )+r \sin (2 \alpha )& = & 0 \\ 2 a \cos (\alpha )& = & l \\ \end{array} \right. $$

and solving for $r,t,\alpha$ we obtain

$$ \left[ \begin{array}{ccc} t & r & \alpha \\ \frac{\left(2 a^2-l^2\right) w}{a \sqrt{4 a^2-l^2}} & -\frac{l w}{\sqrt{4 a^2-l^2}} & \tan ^{-1}\left(\frac{l}{a},-\frac{\sqrt{4 a^2-l^2}}{a}\right) \\ \frac{\left(l^2-2 a^2\right) w}{a \sqrt{4 a^2-l^2}} & \frac{l w}{\sqrt{4 a^2-l^2}} & \tan ^{-1}\left(\frac{l}{a},\frac{\sqrt{4 a^2-l^2}}{a}\right) \\ \end{array} \right] $$

one of them is discarded.

in the second position, the string can not remain taut.

Finding tension in a freely sliding ring on a wire

1 Answers1