Solve FBD Problem: M.asinθ=N, N.sinθ=Ma

[hav0c]

Homework Statement

see attached image

Homework Equations

F=ma

The Attempt at a Solution

I have the relations
1)M.asinθ=N
2)N.sinθ=Ma
Which one is correct and why?
3)if (1) is incorrect is the correct relation --M.asinθ+M.gcosθ=N?

 

[voko]

Forces and accelerations are vectors. Once you have Newton's second law in the vector form, you obtain equations for the components of forces and accelerations. In this case, you only care about the horizontal components. What are they?

 

[hav0c]

horizontal components
of acc. are-- a and (nsinθ)/m
and force are-- ma and nsinθ
but my main question is that is my equation 3 correct?

 

[voko]

horizontal components
of acc. are-- a and (nsinθ)/m
and force are-- ma and nsinθ

It is not that "a and (nsinθ)/m" are "acceleration components". a = (nsinθ)/m is the horizontal acceleration component.

but my main question is that is my equation 3 correct?

I do not see any "g" in the picture so based solely on the picture, that equation is definitely wrong.

 

[hav0c]

I do not see any "g" in the picture so based solely on the picture, that equation is definitely wrong.

I have taken g to be the acceleration downwards due to gravity

 

[voko]

Perhaps you need to describe your problem in full detail. What you have now makes little sense.

 

[hav0c]

from here i have got the relations
1)N=M.g.cosθ+M.a.sinθ
2) a.cosθ=g.sinθ

Edit:i have forgotten to consider the normal force from underneath , I'm going to re-write the equations
Edit: nevermind these equations are totally useless

 

[haruspex]

Edit: nevermind these equations are totally useless

As are the diagrams. Please state the question exactly as it was given to you.

 

[hav0c]

I wasnt having any actual problem in solving questions it was just that i wanted to check whether or not i could derive relations by using different x-y directions, but the new general equations turned out to be useless.