Suppose we are given the free-body diagram above, with a mass on an incline at an angle $\theta$. If my coordinate axes are taken to be the regular $x-y$ plane rotated an angle of $\theta$ (i.e. the x axis is parallel to the direction of $mg \sin \theta$ and the $y$ axis is parallel to $mg \cos \theta$), then if there is no vertical motion, $N = mg \cos \theta$.
Now suppose I take my coordinate axes to be the $x-y$ plane but NOT rotated by any angle (i.e. the $x$ axis is parallel to the base of the triangle and the $y$ axis is parallel to the adjacent side), then if I want to resolve my weight $mg$ in the direction of $N$, wouldn't I need to make it $\frac{mg}{\cos \theta}$? (since in this case I would have to find the force whose vertical component is $mg$) If this is the case, no vertical motion would imply that $N = \frac{mg}{\cos \theta}$.
Does this mean that $N$ depends on my choice of coordinate axes? I might be making a mistake, but I'd appreciate if someone could point out the mistake in my logic.
Best Answer
Changing coordinate axis is not a physical change. The net value of normal reaction remains same.
If coordinate axis are chosen to be parallel to the incline - Now as the block is at rest i am assuming it is due to friction. Let's say that the friction force is $f$
X-direction :$ \ \ N=mg \text { cos }\theta$
Y- direction :$ \ \ mg \text { sin } \theta=f$
Squaring and adding above two we get : $$N^2+f^2=mg^2$$ Rearranging
$$N=\sqrt{mg^2-f^2}$$
Now when coordinate axes are parallel to the base of wedge
Y direction: $N \text { cos } \theta+f \text { sin }\theta=mg$
Squaring it :
$$(N \text { cos }\theta)^2+(f \text { sin }\theta)^2 + 2Nf\text { sin }\theta \text { cos }\theta=mg^2$$
X direction :$N \text { sin }\theta-f \text { cos }\theta=0$
Squaring it : $$(N \text { sin }\theta)^2+(f \text { cos }\theta)^2 -2Nf\text { sin }\theta \text { cos }\theta=0$$
Adding the two squaring term we get :
$$N^2+f^2=mg^2$$
$$ \text {or} $$ $$N=\sqrt{mg^2-f^2}$$
Hence proved.