I want to know the acceleration of an object down an inclined plane without breaking the gravitational force vector into its components that are parallel and perpendicular to the plane.
Consider the following situation where the mass of the object is $\text{m}$, the angle of the inclined plane from the horizontal is $\theta$ and the normal force is represented as $N$
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In this case, $\vec{N}$ can be broken down into its components : $N_x\hat{i}$ and $N_y\hat{j}$.
Here, $N_y=N\cos\theta=\text{m}g$.
From this, we find that, $N_x=N\sin\theta=\text{m}g\tan\theta$ and $N=\frac{\text{m}g}{\cos\theta}$.
Thus, the net force acting on the object at any instant is $\text{m}g\tan\theta$ in the direction of positive $x$ axis, but the object actually happens to slide because the gravitational force pulls the object in the direction of negative $y$ axis with the magnitude $\text{m}g$ as it happens to move in the direction of positive $x$ axis.
So how would we calculate the actual acceleration of the object down the plane in such an instance without breaking the gravitational force vector into its components that are parallel and perpendicular to the plane?
Additional question : Why does the magnitude of the normal force differ in the case discussed above and when breaking the gravitational force vector into its components that are parallel and perpendicular to the plane?
Talking about our case, we get the magnitude of the normal force ($N$) to be $\frac{\text{m}g}{\cos\theta}$ while in the latter case, the magnitude ($N$) turns out to be $\text{m}g\cos\theta$.
I've learnt that the magnitude of a vector doesn't change even if we change our inertial frame of reference, so why does it differ here? Am I thinking something wrong?
Best Answer
The mistake in your thinking is this $N_y=N\cos\theta=\text{m}g$.
It assumes no vertical acceleration, but when the object slides down the plane, it is accelerating downwards.
An alternative way is to think of the mass sliding down the plane a distance $d$, the accelerating force $F$ down the plane does work $Fd$
The gravitational potential energy lost is $mgd\sin\theta$ (as the drop in height is $d\sin\theta$), and the normal force does no work on the mass, as it doesn't move in the direction of that force...
So
$mgd\sin\theta = Fd$
the accelerating force is $F =mg\sin\theta$ and the acceleration is $g\sin\theta$.