- #1

mcconnellmelany

- 21

- 5

- 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.