# Tension on strings

## Relevant Equations:

##T_1=\frac{(M_1+M_2)}{2}g##
##T_2=\frac{\sqrt 3(M_1+M_2)}{2}g##
##T_3=M_2g##

Attempt:

By drawing the Free Body diagrams and calculating the different tensions, I got the following results

##T_1=\frac{(M_1+M_2)}{2}g##
##T_2=\frac{\sqrt 3(M_1+M_2)}{2}g##
##T_3=M_2g##

But, I am not sure what the answer is as although ##T_2>T_1## but ##T_3## does not depend on ##M_1##. So, I am not able to relate the different tensions to each other.

I guess I can ignore ##M_1## and get the result ##T_3>T_2>T_1##. But, I am not sure about that.

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Doc Al
Mentor
So, I am not able to relate the different tensions to each other.
Sure you can.

Make use of the fact that ##M_2## > ##M_1## to prove what you already suspect.

Last edited:
Saptarshi Sarkar
Doc Al
Mentor
FYI: I took the statement that the wires are "very strong" to imply that you'll really have to increase ##M_2## >> ##M_1##, so in a sense you are "ignoring" ##M_1##.

Saptarshi Sarkar
haruspex
Homework Helper
Gold Member
FYI: I took the statement that the wires are "very strong" to imply that you'll really have to increase ##M_2## >> ##M_1##, so in a sense you are "ignoring" ##M_1##.
That may be the thinking behind the question, but it does not really work.
We are not told that ##M_1## is particularly light. Despite the strength of the wires, it could already be the case that the system is close to breaking and ##M_2<(3+2\sqrt 3)M_1## by a sufficient margin that ##W_2## will break first.

Doc Al
Doc Al
Mentor
That may be the thinking behind the question, but it does not really work.
We are not told that ##M_1## is particularly light. Despite the strength of the wires, it could already be the case that the system is close to breaking and ##M_2<(3+2\sqrt 3)M_1## by a sufficient margin that ##W_2## will break first.
After I posted what I did above, I realized that I was assuming too much. The best you can do is solve for how the relative tensions change as ##M_2## increases.

As always, thanks for your post.