(adsbygoogle = window.adsbygoogle || []).push({}); The problem statement, all variables and given/known data

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

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:

[tex] N_1 \sin \theta = m_1 a_{1x} [/tex]

[tex] N_1 \cos \theta - m_1 g = m_1 a_{1y} [/tex]

[tex] - N_1 \sin \theta = m_2 a_{2x} [/tex]

[tex] N_2 - N_1 \cos \theta - m_2 g = 0 [/tex]

where [tex] N_1 [/tex] is the contact force between the box and the inclined plane and [tex] N_2 [/tex] is the contact force between the inclined plane and the floor.

These are four equations in the unknowns [tex] N_1, N_2, a_{1x}, a_{1y}, a_{2x} [/tex].

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

[tex] \vec{a}_{box/plane} = \vec{a}_{box} - \vec{a}_{plane} [/tex]

should be in the direction of the plane. Thus

[tex] \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}}. [/tex]

Is that right?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Box sliding down an inclined plane

**Physics Forums | Science Articles, Homework Help, Discussion**