Solving Dynamics of Rod A & Wedge B with Negligible Friction

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The discussion revolves around calculating the acceleration of rod A and wedge B in a system with negligible friction and a specific mass ratio. Participants express uncertainty about the forces acting on the wedge and the normal reaction's role in accelerating it. The normal force is confirmed to act perpendicular to the wedge, but its components need clarification to determine the wedge's acceleration. A kinematic relationship is suggested, linking the distances moved by the rod and wedge. The conversation emphasizes the need for a clearer understanding of the forces involved to solve the problem effectively.
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Homework Statement


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Find out the acceleration of the rod A and the wedge B in the arrangement if the ratio of mass of the wedge to that of the rod equals \eta, and the friction between all contact surfaces is negligible

Homework Equations


Mass of rod = m, Mass of wedge = M

mg - N cos \alpha = ma, downwards

The Attempt at a Solution


Normal reaction is perpendicular the wedge surface and I don't know if it has a component that can accelerate the wedge to the right. If the force is moving to the left, surely the wedge can't be accelerated to the left.

I need to know which force accelerates the wedge to the right. I'm thinking that the normal reaction has the component N sin \alpha directed to the left.
If I knew which component of force was directed to the right, I could solve the remaining part of it.
 
Last edited:
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Someone, please?
 
Welcome to PF!

sArGe99 said:
Find out the acceleration of the rod A and the wedge B in the arrangement if the ratio of mass of the wedge to that of the rod equals \eta, and the friction between all contact surfaces is negligible

Hi sArGe99! Welcome to PF! :smile:

conservation of energy? :wink:
 
Yes. I did try that out. Don't think I have got enough information to use that.
 
Normal reaction acts perpendicular to the wedge surface. Is that right, in this case?
I think that is where I might have gone wrong.
 
sArGe99 said:
Yes. I did try that out. Don't think I have got enough information to use that.

Yes you have …

try again (use x and x', where x is horizontal displacement), and show us what you get :smile:​
 
I think the kinematic relationship is Distance moved by Rod / Distance moved by wedge = tan (alpha)?
 
just woke up :zzz: …
sArGe99 said:
I think the kinematic relationship is Distance moved by Rod / Distance moved by wedge = tan (alpha)?

You know it is …

get on with it!​
 
Oh.. I could do with this one, I guess.
It can be solved using force equations, right? I actually wanted to try that method out first.
 

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