Question about normal force acting on a mass on an inclined plane

Question

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.

Answer 1

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.

Answer 2

The normal force $\vec{N}$ is a vector, as indicated by my use of notation. As such, the expression does depend on the coordinate frame. However, the actual force is independent of the frame of reference.

1. In your first case, where you consider your $x-y$ axis to be parallel with the incline, the vector $\vec{N}$ has only a component in the $y$ direction. This means that

$$\vec{N} = 0\cdot \hat{x} + mg \cos \theta \hat{y}, $$

where the hatted vectors are unit vectors in the respective directions. The magnitude of $\vec{N}$ is, as you pointed out, $|\vec{N}| = mg \cos \theta$.

2. Now we consider the second case, where the $x-y$ axis is parallel to the ground. In this case, the normal force has to be decomposed in both the $x$ and $y$ directions, but now both components are non-zero. In this new coordinate system we call the vector $\vec{N}'=R(-\theta)\vec{N}$ to indicate that its algebraic expression looks differently, even though physically they are the same thing. Here $R(-\theta)$ (the minus sign is because the second coordinate system is rotated clockwise with respect to the first one) is a rotation matrix, which yields the following expression for $\vec{N}'$

$$\vec{N}'= -|\vec{N}| \sin\theta \, \hat{x} + |\vec{N}| \cos\theta \,\hat{y}\\ =-mg \cos \theta \sin \theta \,\hat{x} + mg \cos^2\theta \,\hat{y}.$$ You can see now that computing the magnitude of $\vec{N}'$ in this reference frame would yield the same result $$|\vec{N}'|= \sqrt{m^2 g^2 \cos^2 \theta (\cos^2 \theta +\sin^2\theta)}=\sqrt{m^2 g^2 \cos^2 \theta }=m g \cos \theta.$$

I hope that helps.

Note: You don't need matrices to do this. You can calculate the $x$ and $y$ components by pure geometry. I included the matrix to showcase a different point of view, but it is not essential. The point is that vectors look differently in different coordinate systems, despite being fundamentally the same object.