I'm having trouble determining the direction of tension confidently for harder examples. For example,
A light, smooth ring, $R$, is threaded on a light, inextensible string. One end of the string is attached to the fixed point $A$. The other end off the string is threaded through a fixed smooth ring, S directly below $A$, and attached to particle of mass 2kg. Angle SAR equals $α$.
I understand that tension will counteract the 2kg on the particle. But what about the other tensions, in which direction would I put them?
Also, I've heard that sometimes tension is 'directed into itself' like this : —>–<——. What is that, and when do I use it?
Best Answer
See, basically the thing is tension in classical mechanical systems are used for balancing applied forces in a network of applied forces. Having said that it is also important to keep in mind that tension is a bi-directional force which means whatever you have heard like ------>---<----- is correct but it depends upon the situation.
As I said, a tension or a tensional force is always away from an object or a point in the system. To understand this, let us analyse the given system that you have mentioned....
Consider point A, tension T1 (say) is acting through the string towards point R.
Consider point R, tension T1 (bi-direction in nature) is acting towards point A. Tension T2 (say) is acting towards point S.
Notice that T1 and T2 are perpendicular to eachother.
Consider point S, Tension T3 say is acting towards 2g wt. Tension T2 is acting towards the point R.
Now we will frame equations as per requirements and we will try to eliminate the tension forces from the equations, this is our objective for solving any mechanical problem.
In this case,
Look at the diagram and try to understand the following:
$T_3 = 0.002 kg wt.$
Again at point S,
$T_2 sin (\alpha) = T_3$
At point R,
$T_2 sin (\alpha) = F sin (\theta) + T_1 cos (\alpha)$
$T_1 sin (\alpha) = F cos (\theta)$
Now we can eliminate all of T1, T2 and T3 in terms of given propositions and finally ....
From 2nd equation
$T_2 sin (\alpha) = 0.002 kg wt.$
From 3rd equation,
$T_1 cos (\alpha) = 0.002 kg wt. - F sin (\theta)$
From 4th equation,
$T_1 sin (\alpha) = F cos (\theta)$
Hence, $$tan (\alpha) = \frac {T_1 sin (\alpha)} {T_1 cos (\alpha)}$$
$$ =\frac {F cos (\theta)} {0.002 kg wt. - F sin (\theta)}$$
Have a wonderful day!