Finding acceleration of rod and wedge with friction negligible

[sArGe99]

Homework Statement

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.

 

[sArGe99]

Someone, please?

 

[tiny-tim]

Welcome to PF!

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!

conservation of energy?

 

[sArGe99]

Yes. I did try that out. Don't think I have got enough information to use that.

 

[sArGe99]

Normal reaction acts perpendicular to the wedge surface. Is that right, in this case?
I think that is where I might have gone wrong.

 

[tiny-tim]

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 ​

 

[sArGe99]

I think the kinematic relationship is Distance moved by Rod / Distance moved by wedge = tan (alpha)?

 

[tiny-tim]

just woke up :zzz: …

I think the kinematic relationship is Distance moved by Rod / Distance moved by wedge = tan (alpha)?

You know it is …

get on with it!​

 

[sArGe99]

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.