Find equation of motion of an inclined plane when there's friction

[Istiak]

Homework Statement

Find equation of motion of a incline plane when there's friction using Lagrangian

Relevant Equations

L=T-V $\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})-\frac{\partial L}{\partial x}+\frac{\partial f}{\partial \dot{x}}=0$

It's the body. So there's friction on that plane and there's tension also.

$$L=\frac{1}{2}m_1\dot{x}^2+\frac{1}{2}m_2\dot{x}^2-m_2g(l-x)-m_1gx\sin\theta$$
$$f=\mu N=-\mu m_1 g\dot{x}\cos\theta$$
I had found the frictional force's equation from [the class](https://www.youtube.com/watch?v=5UE9kzVcFao).

Using Euler Lagrange :

$$m_1\ddot{x}+m_2\ddot{x}-m_g+m_1g\sin\theta-\mu m_1 g\cos\theta=0$$
After rearranging the equation :

$$\ddot{x}=\frac{m_2+\mu m_1\cos\theta-m_1\sin\theta}{m_1+m_2}g$$

But my book says little bit different thing (only differences in sign) I had tried by taking negative common. Although my answer didn't match. Why?

 

[bob012345]

There is a note with the video that there was a sign error in part one which was corrected in part 2. Also note that the problem in this video is similar to but not the same as your problem so you have to be careful.

 

[TSny]

$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})-\frac{\partial L}{\partial x}+\frac{\partial f}{\partial \dot{x}}=0$

Here it looks like you are using the notation $f$ for the Rayleigh dissipation function.

$$f=\mu N=-\mu m_1 g\dot{x}\cos\theta$$

This is confusing. The first equality seems to be giving the friction force and not the Rayleigh dissipation function. But, the expression on the far right appears to be the dissipation function.

Using the Rayleigh dissipation for dry kinetic friction is awkward due to the fact that the direction of the friction force depends on the direction of motion of $M_1$. If you take the positive direction for $x$ to be up the incline, then you need to write the force of friction as $F_f = -\mu N$ when $M_1$ is sliding up the slope and as $F_f = +\mu N$ for $M_1$ sliding down the slope.

The Rayleigh dissipation function $f$ is defined such that $F_f = -\frac {\partial f}{\partial \dot x}$. So, if $M_1$ is sliding up the slope, $f = +\mu N \dot x$. If $M_1$ is sliding down, then $f = -\mu N \dot x$. Note that in both cases, $f$ is positive overall ($\dot x$ is negative for sliding down the slope). $f$ should always be positive since $f$ represents the magnitude of the rate at which energy is being dissipated.

Your result is good for $M_1$ moving down the slope. The answer given to you is good for $M_1$ sliding up the slope.

If you are dealing with fluid friction of the form $F_f = -b \dot x$ for some positive constant $b$, then the annoying sign problem does not occur since the sign of $\dot x$ automatically takes care of the direction of the friction force. In this case the dissipation function would be $f = \frac 1 2 b \dot x^2$ and the signs will take care of themselves. Also, now $f$ equals 1/2 the rate at which energy is dissipated.