# System of 3-pulley in lagrangian approach

I have the following problem that is in Symon Mechanics cap. 9 exercise 8:

Masses $m$ and $2m$ are suspended from a string of lenght $l_1$ wich passes over a pulley. Masses $3m$ and $4m$ are similarly suspended by a string of lenght $l_2$ over another pulley. There two pulleys hang from the ends of a string of lenght $l_3$ over a third pulley fixed. Set up Lagrange eq's and find the accelerations and tensions in the strings.

Then this is my goal: See the picture below, it represents the system.

In the figure 'Polia' $n =$ 'Pulley' $n$. Note that we have marked an origin at the center of the pulley that hang the other two. With that I've constructed the coordinates of each mass:

$$1m : \left \{ \begin{matrix} \tilde{x}_{1} = -R_3-R_1 \\ \tilde{y}_1 = -(x_5+x_1) \end{matrix}\right.$$

$$2m : \left \{ \begin{matrix} \tilde{x}_{2} = -R_3+R_1 \\ \tilde{y}_2 = -(x_5+x_2) = -x_5 - (l_1-x_1) \end{matrix}\right.$$

$$3m : \left \{ \begin{matrix} \tilde{x}_{3} = R_3-R_2 \\ \tilde{y}_3 = -(x_6+x_3) = -(l_3-x_5)-x_3 \end{matrix}\right.$$

$$4m : \left \{ \begin{matrix} \tilde{x}_{4} = R_3+R_2 \\ \tilde{y}_4 = -(x_6+x_4) = -(l_3-x_5) - (l_2-x_3) \end{matrix}\right.$$

with the constrains

$$\left \{ \begin{matrix} x_1+x_2 = l_1 \\ x_3+x_4 = l_2 \\ x_5+x_6= l_3 \end{matrix}\right.$$

This will imply in the below kinetic and potential energy

$$T = \frac{m}{2}(\dot{x}_5+\dot{x}_1)^2 + \frac{2m}{2}(\dot{x}_5 - \dot{x}_1)^2 + \frac{3m}{2}(-\dot{x}_5+\dot{x}_3)^2 +\frac{4m}{2}(\dot{x}_5+\dot{x}_3)^2$$

and

$$U = -mg(x_1+x_5) - 2mg(x_5 + l_1 -x_1) - 3mg(l_3-x_5+x_3)-4mg(l_3+l_2-x_5-x_3)$$

Question: Have I missed something? The accelerations I'll get from the equations of motion for $x_1,x_3,x_5$ will be the accelerations of the string, so I just need to construct $\mathcal{L} = T - U$ and set the Euler-Lagrange Equations? Is this a wrong way to write the lagrangian? If I construct the lagrangian as I stated, it will be the correct lagrangian for the system?

• a very minor detail which will not affect your EOMs because it changes the Lagrangian by a constant, is that $x_1 + x_2 = l_1 - \pi R_1$ and similarly with the other two; a more important one, shouldn't $U = -mgh$ (emphasis on the minus) Nov 24 '17 at 19:59
• Yes, the minus is important, I've just copied from my notes that has the lagrangian and the lagrangian is $T - U$. I'll fix that. The other part on the constrains is that it is usual to not count the $\pi R$ part isn't? This is way I did not considered (but I saw this on it). Mostly because a very great book on the field wrote in my language does not considers.
– R.W
Nov 24 '17 at 22:33
• well, the length of the rope is what it is, if we are going to include it as an actual parameter in the Lagrangian, we can't just cut off a piece of it. Since it is a constant though (and is only multiplied by other constants), you might as well drop it altogether and just write $x_1+x_2 = 0$, ultimately it is irrelevant. If you really want, you could redefine your coordinates so that $x_1+x_2 = l_1$. Nov 24 '17 at 23:40

$$T=\frac{1}{2} m (\dot x_1+\dot x_3)^2+m (\dot x_1-\dot x_5)^2+\frac{3}{2} m (\dot x_3-\dot x_5)^2+2 m (\dot x_3+\dot x_5)^2$$
$$V=-2 g m (l_1-x_1+x_5)-4 g m (l_2+l_3-x_3-x_5)-3 g m (l_3+x_3-x_5)-g m (x_1+x_3)$$