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Above are two scenarios, the horizontal string is just cut in the second one. I am confused in the approach to find the value of tension in the string. I was under the impression that when a string is cut, the force components contributed by it disappear in the moment just after the cut, every other force remains the same. Though here, the two values of tension are different.
Your working for both the cases are correct. The tension changes instantaneously when the string is cut.
I will try and explain it best I can.
The case you are referring to is when there is a spring attached instead of a string. When we have a spring and the right most string is cut, the force applied by spring does not change instantly, and thus remains $\frac{mg}{\cos\left(\theta\right)}$.
If you think about it, if the tension (in string) did not change, then the ball would be accelerated towards the string and go upwards for a bit before coming back down, thus contradicting intuition and observations. This makes perfect sense in case of a spring, spring will pull the ball towards it after the string is cut.
I think this might help. Always make draw a picture of the situation and go from there.
Before the string is cut the mass is in equilibrium, so the tension in the string is based on the sum of the vertical and horizontal forces being zero.
After the string is cut the mass is no longer in equilibrium and will experience both a horizontal and vertical acceleration. That means there has to be a net force on the mass. For that to occur the tension in the string has to be less than what it was when the mass was in equilibrium, which is borne out by the two equations.
Hope this helps.
