Box sliding down an inclined plane

  • Thread starter diegocas
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Homework Statement

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?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
ehild
Homework Helper
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You are on the right track. Go ahead.

ehild
 

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