D'Alembert's principle on inclined plane problem

[PhysicsRock]

Homework Statement

Examine the motion of a particle with mass $m$ under the influence of the gravitational force $\vec{F}_g = -mg \vec{e}_y$ sliding down an inclined plane with angle $\alpha$. The particle is initially positioned at $\vec{r}(0) = (D,H)$.
Derive the equations of motions using d'Alemberts principle.

Relevant Equations

d'Alembert's principle $\left( m \ddot{\vec{r}} - \vec{F}_i \right) \delta \vec{r} = 0$

The virtual displacement should be given by

$$
\delta\vec{r} = \begin{pmatrix} \cos(\alpha) \\ \sin(\alpha) \\ \end{pmatrix} \delta s
$$

where $\delta s$ is a displacement parallel to the plane. The relevant force should be the gravitational force, as given above. Thus, the equations of motion are ought to be

$$
\left[ \begin{pmatrix} m \ddot{x} \\ m \ddot{y} \end{pmatrix} - \begin{pmatrix} 0 \\ -mg \\ \end{pmatrix} \right] \begin{pmatrix} \cos(\alpha) \\ \sin(\alpha) \\ \end{pmatrix} \delta s = m \begin{pmatrix} \ddot{x} \\ \ddot{y} + g \end{pmatrix} \begin{pmatrix} \cos(\alpha) \\ \sin(\alpha) \\ \end{pmatrix} = 0
$$

Doing the multiplications I get

$$
\ddot{x} \cos(\alpha) + (\ddot{y} + g) \sin(\alpha) = 0
$$

Separating that I obtain

$$
\ddot{y} = -g \, , \, \ddot{x} = 0
$$

This, however, describes a free fall with no horizontal acceleration. I must've done something wrong obviously, I just cannot figure out what that is.

Any type of help is highly appreciated. Thank you in advance.

 

[erobz]

Homework Statement: Examine the motion of a particle with mass $m$ under the influence of the gravitational force $\vec{F}_g = -mg \vec{e}_y$ sliding down an inclined plane with angle $\alpha$. The particle is initially positioned at $\vec{r}(0) = (D,H)$.
Derive the equations of motions using d'Alemberts principle.
Relevant Equations: d'Alembert's principle $\left( m \ddot{\vec{r}} - \vec{F}_i \right) \delta \vec{r} = 0$

The virtual displacement should be given by

$$
\delta\vec{r} = \begin{pmatrix} \cos(\alpha) \\ \sin(\alpha) \\ \end{pmatrix} \delta s
$$

where $\delta s$ is a displacement parallel to the plane. The relevant force should be the gravitational force, as given above. Thus, the equations of motion are ought to be

$$
\left[ \begin{pmatrix} m \ddot{x} \\ m \ddot{y} \end{pmatrix} - \begin{pmatrix} 0 \\ -mg \\ \end{pmatrix} \right] \begin{pmatrix} \cos(\alpha) \\ \sin(\alpha) \\ \end{pmatrix} \delta s = m \begin{pmatrix} \ddot{x} \\ \ddot{y} + g \end{pmatrix} \begin{pmatrix} \cos(\alpha) \\ \sin(\alpha) \\ \end{pmatrix} = 0
$$

Doing the multiplications I get

$$
\ddot{x} \cos(\alpha) + (\ddot{y} + g) \sin(\alpha) = 0
$$

Separating that I obtain

$$
\ddot{y} = -g \, , \, \ddot{x} = 0
$$

This, however, describes a free fall with no horizontal acceleration. I must've done something wrong obviously, I just cannot figure out what that is.

Any type of help is highly appreciated. Thank you in advance.

I think you are doing your matrix multiplication incorrectly.

Never mind.

 

[PhysicsRock]

I think you are doing your matrix multiplication incorrectly.

It's just basic vector multiplication, isn't it? Multiply componentwise and add them up.

 

[erobz]

It's just basic vector multiplication, isn't it? Multiply componentwise and add them up.

Sorry, my bad.

 

[haruspex]

The virtual displacement should be given by

Should it? Surely $\vec r$ is three dimensional, and there is a choice of directions to move within the plane.

 

[PhysicsRock]

Should it? Surely $\vec r$ is three dimensional, and there is a choice of directions to move within the plane.

Excuse me, I should've mentioned that above, but together with the text comes a schematic that shows that this problem is 2-dimensional, i.e. the displacement only has to account for displacement in the $x$ and $y$ direction.

 

[erobz]

I don't know why you set each term equal to zero. I believe you need to use the constraint that:

$\ddot y = \ddot x \tan \alpha$

sub that in and see what you get for $\ddot x$?

Also, Wiki has the principle quoted as the negative of what you have used.

https://en.wikipedia.org/wiki/D'Alembert's_principle#Special_case_with_constant_mass

I don't know if that matters fundamentally, but it might matter for making sense of the constraints.

 

[PhysicsRock]

I don't know why you set each term equal to zero. I believe you need to use the constraint that:

$\ddot y = \ddot x \tan \alpha$

sub that in and see what you get for $\ddot x$?

Also, Wiki has the principle quoted as the negative of what you have used.

https://en.wikipedia.org/wiki/D'Alembert's_principle#Special_case_with_constant_mass

I don't know if that matters fundamentally, but it might matter for making sense of the constraints.

What you said worked. I had the constraint written down, just in the form that it's usually written in, i.e. $g(\vec{r})=0$ and for some reason I just didn't see that I could plug in $\ddot{x}$ in terms of $\ddot{y}$ or vise versa. Thank you a lot.

 

[haruspex]

Excuse me, I should've mentioned that above, but together with the text comes a schematic that shows that this problem is 2-dimensional, i.e. the displacement only has to account for displacement in the $x$ and $y$ direction.

ok, so it should have said sliding along a line, not in a plane.
Fwiw, in a plane (standard x, y, z coordinates, x axis lying in the plane) you would have $\vec{\delta r}=\delta s(\cos(\theta), \sin(\theta)\cos(\alpha), \sin(\theta)\sin(\alpha))$, $\ddot z=\ddot y\tan(\alpha)$, etc.