I had the following doubt regarding normal forces and its resolution on inclined plane. (Also I am quite new to SE, so I have no idea how do we upload diagrams to better represent my arguement).
Consider two cases, in both cases there is a inclined plane with some angle of inclination ($\theta$), on which there is a block of mass m that is kept. In Case I, we fix our coordinate axes such that the x-axis lies on the base of the inclined plane, and the y-axis perpendicular to x-axis, and in the next case the x-axis is placed along the inclined plane and the y-axis is perpendicular to the x-axis.
The problem I was facing was that if we consider the normal force in case I and say that the normal force is $mg cos(\theta)$ then this is usually accompanied by the reasoning that there is no acceleration in the direction of normal force, therefore net force in that direction is 0, and hence acceleration is 0 as well.
But in this coordinate plane,saying that acceleration normal to the inclined plane is 0, is just like saying that its rectangular components is also 0, because if we have a vector whose magnitude is 0, then its rectangular components must also have a 0 magnitude. So in one sense we will be saying that the horizontal component of the block's acceleration is also 0, which is not true.
But in the second case, where we have a inclined coordinate system, we can easily make such a assumption that normal force is mgcos($\theta$) because in the inclined coordinate system, due to the normal force being perpendicular to the x-axis (which we placed on the inclined plane), it doesn't have any rectangular x-component and thus we can easily conclude the normal force to be mgcos($\theta$).
I don't understand how different coordinate system can predict different normal forces? Where exactly am I going wrong?