Problem with Dynamic blocks on a fixed inclined plane

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Homework Help Overview

The discussion revolves around a problem involving dynamic blocks on a fixed inclined plane, specifically focusing on the interactions between a block and a wedge of equal mass. Participants are exploring the forces acting on both the block and the wedge, particularly the normal forces and their components.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law and the resolution of forces, particularly the normal forces acting between the block and the wedge, as well as the wedge and the incline. There are questions about the correctness of initial calculations and the need for free body diagrams (FBDs) to clarify the forces involved.

Discussion Status

There is an ongoing exploration of the problem with participants providing guidance on how to approach the solution. Some participants have pointed out potential oversights in the original poster's reasoning, while others emphasize the importance of drawing FBDs and resolving forces correctly. The discussion is productive, with participants encouraged to apply the advice given.

Contextual Notes

Participants note that the surfaces are assumed to be smooth, and there is a focus on understanding the dynamics without directly providing solutions. The original poster expresses confusion about their calculations, indicating a need for clarification on the role of normal forces.

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Homework Statement
Block a of mass m is placed over a wedge of the same mass m. Both the block and wedge are placed on a fixed inclined plane. assuming all surfaces to be smooth calculate the displacement of block in ground frame in 1 s
Relevant Equations
the homework question.
This is what I thought. since the y component of both their accelerations will be same. we can do this. Mg-N=ma where N is the normal force. and for the wedge it is N+Mgsin^2(theta)=ma using second law of motion where N is the normal force. sin^2(theta) because i resolved gravity twice. now if you elminate N from the equation you get g(1+sin^2(theta))/2 and if you put it in the law of kinematics with t=1 it becomes g(1+sin^2(theta))/4. But apparently thats not the correct answer and I have no idea why.
 

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physicsissohard said:
Homework Statement: Block a of mass m is placed over a wedge of the same mass m. Both the block and wedge are placed on a fixed inclined plane. assuming all surfaces to be smooth calculate the displacement of block in ground frame in 1 s
Relevant Equations: the homework question.

This is what I thought. since the y component of both their accelerations will be same. we can do this. Mg-N=ma where N is the normal force. and for the wedge it is N+Mgsin^2(theta)=ma using second law of motion where N is the normal force. sin^2(theta) because i resolved gravity twice. now if you elminate N from the equation you get g(1+sin^2(theta))/2 and if you put it in the law of kinematics with t=1 it becomes g(1+sin^2(theta))/4. But apparently thats not the correct answer and I have no idea why.
Do you have an accompanying diagram by any chance?
 
1693445210093.png
 
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Write Newtons 2nd for the wedged shaped block in the vertical direction. A FBD would be good practice.

Another tip would be to designate the normal forces by double subscripts. i.e. the normal force acting on block B from the fixed incline might be ##N_{IB}##, and the normal force from block A acting on B might be designated as ##N_{AB}##.

Try to take a minute to learn LaTeX Guide. We use it to make the math easily readable and quotable.
 
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physicsissohard said:
for the wedge it is N+Mgsin^2(theta)=ma
You have overlooked that N, the normal force between block and wedge, increases the normal force between wedge and ramp, and that this last has a vertical component.
 
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erobz said:
Write Newtons 2nd for the wedged shaped block in the vertical direction. A FBD would be good practice.

Another tip would be to designate the normal forces by double subscripts. i.e. the normal force acting on block B from the fixed incline might be ##N_{IB}##, and the normal force from block A acting on B might be designated as ##N_{AB}##.

Try to take a minute to learn LaTeX Guide. We use it to make the math easily readable and quotable.
 
haruspex said:
You have overlooked that N, the normal force between block and wedge, increases the normal force between wedge and ramp, and that this last has a vertical component.
can u complete the solution
 
physicsissohard said:
can u complete the solution
We aren't permitted to hand out solutions right away. You need to make an effort into applying the advice given to you, and we will help you along the way if you get stuck. The objective is for you to learn. If you have specific questions about what was asked of you please let us know.
 
physicsissohard said:
can u complete the solution
@erobz told you how to proceed in post #4. Draw separate FBDs for the two bodies and write their F=ma equations. I merely pointed out where your attempt went wrong.
 
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  • #10
erobz said:
We aren't permitted to hand out solutions right away. You need to make an effort into applying the advice given to you, and we will help you along the way if you get stuck. The objective is for you to learn. If you have specific questions about what was asked of you please let us know.

Thanks, I re-solved it. I am getting the solution. yeah the normal force exerted on the wedge by the block also needs to be resolved into components. im so dumb
 
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  • #11
physicsissohard said:
Thanks, I re-solved it. I am getting the solution. yeah the normal force exerted on the wedge by the block also needs to be resolved into components. im so dumb
Inexperienced is not dumb. Don't be hard on yourself, it takes time and sometimes struggle.
 
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