How can sum of component forces of gravity on inclined plane be greater than actual force of gravity? First, I understand triangles and trigonometry(I think!), so that's not the issue. The question is more of a conceptual one. How can it be that a force has components whose sum exceeds the actual force? Specifically, on an inclined plane, when the force of gravity on the object (i.e., its weight) is split into its component vectors (parallel and perpendicular to the incline), those two force vectors add up to more than the force of gravity as a single vector.
For example, let's say a 500-Newton objects is on a 45-degree plane (we can assume it's at rest, so there is no net force). The force of gravity is straight down (the hypotenuse), and the parallel and perpendicular vectors of that force (the two legs of the triangle) would be $250\sqrt2$ Newtons. That equals 353.55 Newtons for each component. How can gravity apply 500 Newtons on a system, yet its component force vectors apply over 700 Newtons on that system? What gives?!
This has been bugging me for years. Any illumination would be greatly appreciated.
 A: You're adding the magnitudes of the component forces. This gives an answer that is too large, because the component forces aren't pulling in the same direction as each other. The component forces, $\vec{A}$ and $\vec{B}$, say, need to be added as vectors, that is represented as arrows of length proportional to the magnitude of the components and pointing in the direction of these components. Arrange them tail of $\vec{B}$ to head of $\vec{A}$ and the resultant force (the vector sum of $\vec{A}$ and $\vec{B}$) os represented by an arrow going from the tail of $\vec{A}$ to head of $\vec{B}$.
You can do this using a diagram drawn to scale, or a sketched diagram to which you apply trigonometry. When you've done this sort of thing two or three times you don't even need a sketched vector diagram!
A couple of additional remarks that may (or may not) help…
(1) You've resolved the weight into a force of $500\ \frac{1}{\sqrt{2}}$ N normally into the slope, and a force of $500\ \frac{1}{\sqrt{2}}$ N 'down' along the slope. While the downward components (each of $500\ \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}$ N) of these two component forces add up as scalars to make the original downward force, their horizontal components, each of $500\ \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}$ N but in opposite directions, cancel.
(2) Force is one of a 'family' of Euclidian vectors. The archetypal Euclidian vector is displacement. 500 m North East is an example of a displacement. Suppose that we wanted to get from A to B, a displacement of 500 m North East. Suppose also that roads and rail were only available in a grid pattern, going East-West (and West-East) and North South (and South North). To get from A to B by road or rail you'd have to go $500\ \frac{1}{\sqrt{2}}$ m North and $500\ \frac{1}{\sqrt{2}}$ m East. That means you've gone a total distance of $500\ \sqrt{2}$ m. We've just added the magnitudes (scalars) of the two component vectors, and found a figure that is greater than the magnitude (500 m) of our displacement. But the two component displacements ($500\ \frac{1}{\sqrt{2}}$ m North and $500\ \frac{1}{\sqrt{2}}$ m East) add as vectors to give the required displacement of 500 m North East!
