Conservation of the string length in pulleys

[Bunny-chan]

Homework Statement

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 $l_1$ length is equivalent to $l_2$. Can somebody explain?

Homework Equations

The Attempt at a Solution

 

[haruspex]

doesn't seem to me that two times the $l_1$ length is equivalent to $l_2$

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?

 

[Bunny-chan]

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?

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

 

[Chestermiller]

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.

 

[Bunny-chan]

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.

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

 

[Chestermiller]

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

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.

 

[Bunny-chan]

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.

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

 

[Chestermiller]

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

What part don't you understand?

 

[Bunny-chan]

What part don't you understand?

Where did the $y$ term go to?

 

[Chestermiller]

Where did the $y$ term go to?

What is the derivative of y with respect to time?

 

[Bunny-chan]

What is the derivative of y with respect to time?

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

 

[Chestermiller]

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

Sure

 

[Bunny-chan]

Sure

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?

 

[haruspex]

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?

No, why?

 

[Chestermiller]

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?

No

 

[Bunny-chan]

No

Nevermind. I can see that now. Thank you!