When a load is resting on an inclined plane, there is force $mg \sin(x)$ that's vertical to the inclined plane and force $mg \cos(x)$ horizontal to the plane acting on it like this:
My textbook says the weight $mg$ has two components: $mg \sin(x)$ and $mg \cos(x)$.
But when adding up $mg \sin(x)$ to $mg cos(x)$ it does not equal to $mg$.
If mg and N are the only forces in this system, which means the net force should be 0, how should I explain the existence of $mg
\sin(x)$ and $mg \cos(x)$? Does $mg \sin(x)$ and $mg \cos(x)$ have their own reaction force to cancel them out and turn the net force to be 0 too or is it wrong the say the system's net force is 0?
And how are the formulas $mg \sin(x)$ and $mg \cos(x)$ that are used to describe these two forces derived?
Best Answer
Because $mg\cos x$ and $mg\sin x$ are orthogonal vectors, not colinear, and the norm of their sum is the Pythagorean sum. This kind of addition is pretty common in physics as well as in other places basic vector algebra shows up -- an elegant example is in statistics, where independent (orthogonal) random variables get added in a Pythagorean way, while multiple recordings of a variable get added up linearly.
The fact that we can do this is actually a very good indicator -- and part of the proof -- of the fact that physical forces are mathematical vectors, i.e. that it satisfies vector algebra. So the answer to the question is really an empirical one -- this is how forces get added, and can be mathematically explained as them being vectors.