Tension on strings

[Saptarshi Sarkar]

Homework Statement:

The question is added below

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.

 

[Doc Al]

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.

 

[Doc Al]

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" $M_1$.

 

[haruspex]

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]

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.