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

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Homework Help Overview

The discussion revolves around the forces acting on a wedge in a physics problem, specifically focusing on the dynamics of a wedge-pulley system. Participants are exploring the implications of including different components of the system in their free body diagrams (FBDs).

Discussion Character

  • Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants are discussing the inclusion of the rope in contact with the pulley as part of the wedge-pulley system and how this affects the analysis of forces. There is also a mention of confusion regarding the representation of forces in a single diagram versus separate free body diagrams.

Discussion Status

The discussion is active, with participants providing insights and questioning the clarity of their diagrams. Some guidance has been offered regarding the separation of forces into distinct free body diagrams, indicating a productive direction in the analysis.

Contextual Notes

There is a mention of an accidental inclusion of a variable in a formula, which may indicate a misunderstanding or misrepresentation of the problem setup. Participants are navigating through these potential errors in their reasoning.

teodora
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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
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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.
 
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Thanks. That would make my solution correct I guess? (except for the a in the denominator, I wrote it there by accident)
 
teodora said:
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.
 
It is interesting to write the Lagrange equations and especially to calculate the generalized forces.
 
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