Acceleration of the wedge: What forces act on the wedge?

[teodora]

Homework Statement

The problem text and picture are in the appendix. I am not sure whether all the forces that I drew in are actually contributing to the wedge's motion. I read somewhere that each rope segment that touches the pulley exerts a force of that pulley in the direction of the applied force (even when pulley is ideal and all frictions are neglected). I'm also skeptical whether the horizontal component of tension (F) contributes to the wedges motion. I am only sure for normal force

Relevant Equations

Normal force on the block: N=ma*sinα - mg*cosα
for wedge I just take the opposite signs
Forces on the wedge would then be: Ma=F+Fcosα+Nsinα

 

[haruspex]

The simplest way to think of it is to include the portion of the rope in contact with the pulley as being part of the wedge+pulley system. That way, you can take the forces, both magnitude F but at different angles, from the straight segments of the rope as being external forces on that system.

 

[teodora]

Thanks. That would make my solution correct I guess? (except for the a in the denominator, I wrote it there by accident)

 

[haruspex]

Thanks. That would make my solution correct I guess? (except for the a in the denominator, I wrote it there by accident)

No.
You have confused yourself by drawing the forces on the wedge and the block in the same diagram. Draw two separate FBDs.