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## Homework Statement

I have been solving problems on my own and I came out with a problem I made up myself.

It's simple: There's a triangular big block of mass M with an angle of [itex]\theta [/itex] and on top of it there's a rectangular block of mass m. See figure attached.

When the triangular block is acted upon a horizontal force F, both will accelerate (I can, without loss of generality, assume the big block accelerates to the right). I just wanted to find the acceleration of each block with respect to the ground, which is frictionless and so is the interface between the blocks.

## Homework Equations

[itex] \sum \vec{F} = m \vec{a}

## The Attempt at a Solution

Ok, reposting here as this looks too much like a h/w problem, although it's not :D

I managed to solve the problem first isolating the triangular block and finding the horizontal acceleration. It was crucial to use the constraint of motion along the ground only. Likewise, When I go to the non inertial frame of reference of the accelerating block to analyze the FBD of the smaller block, I use the constraint of motion along the incline. From there, I can find the magnitude of the normal force between the blocks BUT also using the inertial *fictitious* force that needs to be considered for using a non inertial frame of reference. THIS IS WHAT BOTHERS ME.

Calling "n" the normal force between the blocks, and "[itex]n_g[/itex] the normal force between the ground and the big block, the FBD of the triangular block gives me

[itex]\sum F_x = F-n Sin(\theta) = M a_M [/itex]

[itex]\sum F_y = n_g-Mg-n Cos(\theta) = 0 [/itex]

The FBD of the small block *once I hop onto the non inertial frame of reference of the accelerating incline* gives me *see figure*

[itex]\sum F_x = mg Sin(\theta) - F_{inert}Cos(\theta) = m a_m [/itex]

[itex]\sum F_y = n-mgCos(\theta)- F_{inert}Sin(\theta) = 0 [/itex]

The last equation is the constraint I needed, but I use the inertial force with magnitude [itex]F_{inert} = \frac{m(F-nSin(\theta) )}{M}= m a_M[/itex]. Solving these equation is enough to find the value of n and from there, just plug it back into [itex]a_M[/itex], but I don't like this

From here, I can totally solve the problem (I already got the answer, which relied crucially on finding n, and I have 4 unknowns and 4 equations, I already did it).

**** I would like to work entirely with inertial frames of reference.*****

I was used to always work with an inertial frame of reference (like the ground), even when I had rotating bodies. In the latter case I could use the constraint that the centripetal force must be equal to [itex]\frac{m\cdot v^2}{R} [/itex]. However, in the problem I'm talking about, the constraint is not clear from the frame of reference of the ground. I need to figure out a constraint to solve for the normal foce.

Any thoughts of how I can avoid the non inertial frame of reference?

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