What is the acceleration of a ramp on a table when a body slides on it? I found an Olympiad problem:

Find the acceleration of a ramp on a table when a body slides on it. Assume there is no friction between the body and the ramp, and between the ramp and the table.

I found the final solution to this problem but I do not understand it:

*

*What is $m \vec{a}_1$, and (ii) why $m \vec{a}$ is parallel to the table in the free-body diagram?

*How do they come up with the equation in the solution?




 A: TL;DR The free-body diagram should not include the resultant forces, at least not in the way to make you think those are the forces acting on the body in question. The following vectors should be removed from diagrams in your question: (i) $m \vec{a}$ and $m \vec{a}_1$ for the body, and (ii) $M \vec{a}$ for the ramp.
Although it does not make any difference for the final solution, the free-body diagram should also include the normal force between the ramp and the table. Finally, the horizontal acceleration of the ramp is negative if the positive direction is to the right, which seems to be the case in your solution.

Detailed solution of the problem
Let the coordinate system be defined as follows:

*

*$\hat{\imath}$ is horizontal axis; positive direction points to the right

*$\hat{\jmath}$ is vertical axis; positive direction points upwards

Write equations of motion in vector form for the two bodies separately:
$$m \vec{a} = \vec{w} + \vec{n} \qquad \text{and} \qquad M \vec{A} = \vec{W} + \vec{N} - \vec{n}$$
where

*

*$m$, $\vec{a}$ and $\vec{w}$ are mass, acceleration and weight of the sliding body, respectively,

*$M$, $\vec{A}$ and $\vec{W}$ are mass, acceleration and weight of the ramp, respectively,

*$\vec{n}$ is normal force between the body and the ramp, and

*$\vec{N}$ is normal force between the ramp and the table.

The equations of motion for the ramp are
$$M A_x = -n \sin\alpha \qquad \text{and} \qquad M A_y = -Mg + N - n \cos\alpha$$
where $A_y = 0$ since the ramp does not move vertically, and $n$ and $N$ are magnitudes of normal force vectors $\vec{n}$ and $\vec{N}$, respectively.
The equations of motion for the sliding body are
$$m a_x = n \sin\alpha \qquad \text{and} \qquad m a_y = -mg + n \cos\alpha$$
From these equations it follows
$$\boxed{a_x = -\frac{M}{m} A_x} \qquad \text{and} \qquad \boxed{m a_y = -m g - A_x \frac{M}{\tan\alpha}} \tag 1$$
With $\Delta y = -\Delta x \tan\alpha$, where $\Delta x$ and $\Delta y$ are horizontal and vertical displacement of the sliding body relative to the ramp, it follows
$$\Delta\ddot{y} = -\Delta\ddot{x} \tan\alpha \qquad \text{and} \qquad a_x = \Delta\ddot{x} + A_x$$
where $a_y = \Delta\ddot{y}$ since the ramp does not move vertically. From this it follows
$$\boxed{\tan\alpha = \frac{a_y}{A_x - a_x}} \tag 2$$
From identities in Eq. (1) and Eq. (2) it follows
$$A_x \bigl(m + M \bigr) \tan\alpha = -mg - A_x \frac{M}{\tan\alpha}$$
and this finally leads to the acceleration at which the ramp slides on the table:
$$\boxed{A_x = \frac{-mg}{(m+M)\tan\alpha + M/\tan\alpha}}$$
After some basic trigonometry, the above expression can be written as
$$A_x = \frac{-mg \sin\alpha \cos\alpha}{M + m \sin^2\alpha}$$
which equals your solution. It is now trivial to find $a_x$ and $a_y$ acceleration components for the sliding body
$$a_x = \frac{M g \sin\alpha \cos\alpha}{M + m \sin^2\alpha} \qquad \text{and} \qquad a_y = \frac{-(m+M)g \sin^2 \alpha}{M + m \sin^2\alpha}$$
The resultant acceleration for the sliding body is
$$\vec{a} = a_x \hat{\imath} + a_y \hat{\jmath} = \frac{(m + M) g \sin\alpha}{M + m \sin^2 \alpha} \Bigl( \frac{M}{m + M} \cos\alpha \hat{\imath} - \sin\alpha \hat{\jmath} \Bigr)$$
However, this does not equal $\vec{a}_1$ from the solution (free-body diagram) in your question, which is defined relative to the ramp
$$\vec{a}_1 = \Delta \ddot{x} \hat{\imath} + \Delta \ddot{y} \hat{\jmath} = \frac{(m + M) g \sin\alpha}{M + m \sin^2 \alpha} \bigl( \cos\alpha \hat{\imath} - \sin\alpha \hat{\jmath} \bigr)$$
A: There is no $ma$ in the diagram or in the text. There is the term $Ma$.
You have two bodies in the problem:

*

*the block on the incline, with mass m and acceleration $a_1$ and

*the inclined plane with mass M and acceleration a.
You already have the free body diagrams for each of the two objects.
The inclined plane can only move horizontally, it will not fly up and it will not dive into the ground. The block moves both horizontally and vertically.

