M1 Jan 2004 Q4: Resolving forces?

Two small rings, A and B, each of mass 2m, are threaded on a rough horizontal pole. The coefficient of friction between each ring and the pole is miu. The rings are attached to the ends of a light inextensible string. A smooth ring C, of mass 3m, is threaded, on the string and hangs in equilibrium under the pole. The rings A and B are in limiting equilibrium on the pole, with angle BAC = angle ABC = theda, where tan(theda)= 3/4.

(a) Show that the tension in the string is 5/2mg.

Done this.

(b) Find the value of miu.

According to the mark scheme, the reaction at A or B is 2mg + Tsin(theda). Why is not 2mg - Tsin(theda)? In which way is the reaction acting?

Thanks a million

The normal reaction is acting upwards, and there is a vertical component of tension and the weight acting downwards. So in limiting equilibrium, $N = T sin\theta + 2mg$

A diagram here will help you.

Also
$\mu$=mu
$\theta$=theta

I've read the diagram in the mark scheme, but the tension in the string goes both upwards and downwards!?

Yea, I looked at the mark scheme myself, she's right, "T" goes upwards and downwards (so will they not cancel out according to Newton III). I've attached the diagram below:

Any tension in any string acts both upwards and downwards. They cancel out when you consider the whole system, not just one body within it.

Heh, I just figured that out one second as well! Thanks for the help!

I've read the diagram in the mark scheme, but the tension in the string goes both upwards and downwards!?

The vertical component of tension is acting downwards? Why is that? Shouldn't it be acting upwards?