Atwood Machine on a Frictionless Inclined Plane: Components Resolution

[logearav]

Homework Statement

Revered Members,
Kindly see my attachments.

Homework Equations

This is Atwood machine dealing with motion on a frictionless inclined plane.

The Attempt at a Solution

To better visualize the forces involved, i isolated m₁ and m₂ and drawn free body diagrams for both. For mass m₁, there are three concurrent forces, that is, T,m₁g,and N, where T is the force in the string because of Tension and N is the normal force of the table on the block. The assumption is m₁ accelerates up the plane, which is taken in x direction. Now, the weight m₁g is broken down into components. The x component is in the assumed direction of acceleration , and the y component acts perpendicularly to the plane and is balanced by the normal force N.
I have shown the components in the attachment1. My question is
1) why m₁gcosθ and m₁gsinθ can't be interchanged, as given in my second attachment?
2) what is wrong in changing the angle θ, that lies between m₁g and m₁gcosθ( first attachment) to between m₁gsinθ and m₁g?

 

[grzz]

You can either work with θ or with (90 - θ), the latter angle you can call β. But, of course, you cannot call them both θ in the same problem.

 

[logearav]

But invariably in all the books i referred , the Tension is always balanced by the mgcosθ and not by mgsinθ. My teacher said, that 90 - θ should be the sine component always and θ should be cos component. Is that so?

 

[grzz]

See the attachment.

 

[logearav]

Oh! Beautifully explained grzz. Thanks a lot. I got it now.