Investigating Lagrangians and Constraints for Tension Calculation

[mcconnellmelany]

Homework Statement

image of the system attached below, there's no friction.

Relevant Equations

$L=E_k-U+T \alpha$

I had used the same constraint as the solution manual says.

So my two Lagrangian would be
$L_1=\frac{1}{2}m_A\dot{x_A}^2+\frac{1}{2}m_B\dot{x_B}^2+\frac{1}{2}m_C\dot{x_C}^2+m_Cgx_C+T(x_A+x_B+2x_C-c)$
whereas c is just a constant.

Of course, I have to write my Lagrangian using constraints. But I can write it three different way.

1. Put values of x_C's
2. Put values of x_B's
3. Put values of x_A's

I don't know why, if I put values for x_B's or x_A's, I get wrong answer for Tension if I put values for x_C's I get the correct answer. I earlier faced this kind of problem but thought that I made mistake somewhere so it happened but I have faced the issue again. This time I am sure I didn't make any mistake.

Let me start with the third step (second and third step is almost same).

$L_2=\frac{1}{2}m_A(2 \dot x_C-\dot x_B)^2 +\frac{1}{2}m_B\dot{x_B}^2+\frac{1}{2}m_C\dot{x_C}^2+m_Cgx_C$
Putting the values of x_C in terms of x_A and x_B
$L_3=\frac{1}{2}m_A\dot m_A^2 +\frac{1}{2}m_B\dot{x_B}^2+\frac{1}{2}m_C\dot{\frac{\dot x_A+\dot x_B}{2}}^2+m_Cg(\frac{x_A+x_B}{2}+c)$

Now going to use first and second Lagrangian.

Using Euler-Lagrange on first Lagrangian for x_B.

$m_B\ddot x_B=T \label{1} \tag{1}$

Using Euler-Lagrange on second Lagrangian for x_C's.

$m_A (2\ddot{x_C}-\ddot x_B)+m_C\ddot x_C=m_C g \label{2}\tag{2}$
Put value of \ddot x_B from equation (1) in equation (2).

$\ddot{x_C}(2m_A+m_C)-\frac{Tm_A}{m_B}=m_C g \label{3}\equation{3}$

Using Euler-Lagrange on second Lagrangian for x_B's.

$m_A (2\ddot x_C-\ddot x_B)+m_B \ddot x_B=0$

Now put value of \ddot{x_B} on the last equation and solve for \ddot{x_C} then put it on 3rd equation then solve for T
$T=\frac{2m_Am_Bm_C g}{m_A m_C-2m_Am_B-m_Bm_C}$

it's wrong answer..

-----------------------------------------------------------------------

Now let me try using third Lagrangian.

From first Lagrangian we get that $m_B\ddot x_B=T$ and $m_A\ddot x_A=T$. (solved Euler Lagrange for x_A once and x_B next).

From third Lagrangian we get that $m_B\ddot x_B+m_C (\frac{\ddot x_A+\ddot x_B}{4})=\frac{m_Cg}{2}$

Keep putting values of \ddot x_A and \ddot x_B then solve for T, I get that
$T=\frac{2m_Am_Bm_Cg}{4m_Am_B+m_Am_C+m_Bm_C}$

it's the correct answer. But why did I get wrong answer for the second Lagrangian? No matter how you do, you will get wrong answer for second ones.

 

[erobz]

I don't know if I'm actually able to help, but perhaps while you wait for someone who knows what they are doing, if you help me understand how this is supposed to work something will jump out at you? Maybe not though.

That last term in your Lagrangian. Is that the total work done by all the tensile forces added together?

If it is:

For block A:

$$ W_A = T \left( x_A - l_A \right)$$

Block B:

$$ W_B = T \left( x_B - l_B \right)$$

Pulley:

$$ W_{pulley} = T' \left( x_P - l_P \right) = 2T \left( x_P - l_P \right) $$

Mass C:

$$ W_C = -T' x_C = -2T x_C $$

So if you sum all that I get:

$$ T x_A + T x_B + T 2 x_P - T2 x_C - c = T\left( x_A + x_B + c \right) $$Then using your definition:

$$ L = \frac{1}{2}m_A { \dot x_A }^2 + \frac{1}{2}m_B { \dot x_B }^2 + \frac{1}{2}m_C { \dot x_C }^2 + mgx_C + T\left( x_A + x_B + c \right) $$

What did I mess up?

 

[TSny]

$L_1=\frac{1}{2}m_A\dot{x_A}^2+\frac{1}{2}m_B\dot{x_B}^2+\frac{1}{2}m_C\dot{x_C}^2+m_Cgx_C+T(x_A+x_B+2x_C-c)$
whereas c is just a constant.

The last term appears to me to have a sign error. Shouldn't the $+2x_C$ be $- 2x_C$?

$L_2=\frac{1}{2}m_A(2 \dot x_C-\dot x_B)^2 +\frac{1}{2}m_B\dot{x_B}^2+\frac{1}{2}m_C\dot{x_C}^2+m_Cgx_C$
Putting the values of x_C in terms of x_A and x_B
$L_3=\frac{1}{2}m_A\dot m_A^2 +\frac{1}{2}m_B\dot{x_B}^2+\frac{1}{2}m_C\dot{\frac{\dot x_A+\dot x_B}{2}}^2+m_Cg(\frac{x_A+x_B}{2}+c)$

I agree with the expressions for $L_2$ and $L_3$.

Now going to use first and second Lagrangian.

Using Euler-Lagrange on first Lagrangian for x_B.

$m_B\ddot x_B=T$

OK. This shows that the Lagrange multiplier "$T$" for the last term in $L_1$ does turn out to be the tension in the string connecting $m_A$ and $m_B$.

Using Euler-Lagrange on second Lagrangian for x_C's.

$m_A (2\ddot{x_C}-\ddot x_B)+m_C\ddot x_C=m_C g$

I think you are missing an overall factor of 2 (in front of $m_A$) in the first term on the left side.

Using Euler-Lagrange on second Lagrangian for x_B's.

$m_A (2\ddot x_C-\ddot x_B)+m_B \ddot x_B=0$

I believe there is a sign error in this equation.

Now put value of \ddot{x_B} on the last equation and solve for \ddot{x_C} then put it on 3rd equation then solve for T
$T=\frac{2m_Am_Bm_C g}{m_A m_C-2m_Am_B-m_Bm_C}$

it's wrong answer..

If you make the corrections, I think you will get the correct expression for $T$.

Overall, I'm puzzled by the way you did the analysis. It seems odd to me to write the Lagrangian in two different ways ($L_1$ and $L_2$) and then combine equations of motion derived from the two Lagrangians. It's not wrong, but it's just strange to me. If you want to get the accelerations of the masses and the tension $T$, then all that you need is $L_1$. There is no need to bring in $L_2$. Or, you can solve for the accelerations by just using $L_2$ and the constraint (without using $L_1$).

 

[erobz]

I'm taking it I've "overdone it" by adding in the work on from $T'$ at the pulley? I suspect it gets canceled by the negative work that would be done by $T$ that I missed:

$$ W_{pulley} = T'( x_P - l_P ) - 2T ( x_P - l_P) = 0 $$Correcting that:

$$ L = \frac{1}{2}m_A { \dot x_A }^2 + \frac{1}{2}m_B { \dot x_B }^2 + \frac{1}{2}m_C { \dot x_C }^2 + mgx_C + T\left( x_A + x_B - 2x_C - c \right) $$

where the constant $c = l_A + l_B + l_C$

 

[mcconnellmelany]

The last term appears to me to have a sign error. Shouldn't the +2xC be −2xC?

My constraint was $x_p-x_a+x_p-x_b=c$
$2x_p-c=x_a+x_b$
Ohhh there's a sign error. It should be $T(x_a+x_b-2x_c+c)$

 

[mcconnellmelany]

Overall, I'm puzzled by the way you did the analysis. It seems odd to me to write the Lagrangian in two different ways (L1 and L2) and then combine equations of motion derived from the two Lagrangians. It's not wrong, but it's just strange to me. If you want to get the accelerations of the masses and the tension T, then all that you need is L1. There is no need to bring in L2. Or, you can solve for the accelerations by just using L2 and the constraint (without using L1).

I learned it from [here](https://www.physicsforums.com/threads/is-it-possible-to-find-tensional-force-from-lagrange.1016372/)