Conservation of the string length in pulleys

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1. May 27, 2017

Bunny-chan

1. The problem statement, all variables and given/known data

This is a common massless string and pulleys problem. I'd just like to understand why, according to the solution, $l_2 + 2l_1 =$ constant. It doesn't seem to me that two times the $l_1$ lenght is equivalent to $l_2$. Can somebody explain?

2. Relevant equations

3. The attempt at a solution

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2. May 27, 2017

haruspex

Indeed not, but that would lead to $2l_1-l_2$ is constant, which is not what is claimed.
Besides, you need to think about changes in the lengths, not the absolute lengths. If l1 increases by x, what happens to l2?

3. May 27, 2017

Bunny-chan

Hm. Will it decrease? Because if $m_1$ goes down, it means $m_2$ has to go up, right?

4. May 27, 2017

Staff: Mentor

Let x be the distance in elevation between the center of pulley 1 and the center of pulley 2, y be the distance between the elevation of the center of pulley 2 and the ceiling, z be the distance in elevation between mass 2 and the center of pulley 2, and w be the distance in elevation between mass 1 and the center of pulley 1. So the total length of string is L=2x+y+z. The distance l2 is given by $l_2=z+y$ and the distance l1 is $l_1=x+y+w$. So, $$2l_1+l_2=(2x+z)+3y+2w=L+3y+2w$$ But, since L, y, and w are constant, $2l_1+l_2$ must be constant.

5. May 27, 2017

Bunny-chan

I was able to understand everything before this point. Why do you take $2l_1$?

6. May 27, 2017

Staff: Mentor

I do $2l_1$ because I need 2x.

Incidentally, I would never have done this problem in terms of l1 and l2. I would have immediately written down the total length as $L=2x+y+z$, and then differentiated to get $$\frac{dL}{dt}=0=2\frac{dx}{dt}+\frac{dz}{dt}$$ If m1 moves down at the rate of dx/dt, m2 moves up at the rate of -dz/dt.

7. May 27, 2017

Bunny-chan

OK. I get it now, but I don't quite understand the differentiation.

8. May 27, 2017

Staff: Mentor

What part don't you understand?

9. May 27, 2017

Bunny-chan

Where did the $y$ term go to?

10. May 27, 2017

Staff: Mentor

What is the derivative of y with respect to time?

11. May 27, 2017

Bunny-chan

$\frac{dy}{dt}$? But since it is constant, it equals 0? Is that so?

12. May 27, 2017

Staff: Mentor

Sure

13. May 27, 2017

Bunny-chan

But isn't $x$ also constant? I mean, the distance in elevation between the center of pulley 1 and the center of pulley 2 will always be the same, won't it?

14. May 27, 2017

haruspex

No, why?

15. May 27, 2017

Staff: Mentor

No

16. May 27, 2017

Bunny-chan

Nevermind. I can see that now. Thank you!