Box sliding down an inclined plane

[diegocas]

Homework Statement

A box is placed on an inclined plane of angle $$\theta$$. There is no fiction between the box and the plane nor between the inclined plane and the floor. The mass of the box is $$m_1$$ and the mass of the inclined plane is $$m_2$$.

Find the acceleration of the box and that of the inclined plane.


The attempt at a solution

It is easy to write down Newton's equations for both the box and the inclined plane.
Here they are:

$$N_1 \sin \theta = m_1 a_{1x}$$

$$N_1 \cos \theta - m_1 g = m_1 a_{1y}$$

$$- N_1 \sin \theta = m_2 a_{2x}$$

$$N_2 - N_1 \cos \theta - m_2 g = 0$$

where $$N_1$$ is the contact force between the box and the inclined plane and $$N_2$$ is the contact force between the inclined plane and the floor.

These are four equations in the unknowns $$N_1, N_2, a_{1x}, a_{1y}, a_{2x}$$.

How am I supposed to solve them?

My idea is the following: since the box is sliding down the inclined plane, the motion with respect to the inclined plane is easier. I mean the acceleration of the box relative to the inclined plane must be in the direction of the inclined plane. That is, the vector

$$\vec{a}_{box/plane} = \vec{a}_{box} - \vec{a}_{plane}$$

should be in the direction of the plane. Thus

$$\tan \theta = \frac{a_{box/plane,y}}{a_{box/palne,x}} = \frac{a_{box,y}-a_{plane,y}}{a_{box,x}-a_{plane,x}} = \frac{a_{1y}}{a_{1x}-a_{2x}}.$$

Is that right?

 

[ehild]

You are on the right track. Go ahead.

ehild