First, I understand triangles and trig (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 0net 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 250xroot2 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.

  • $\begingroup$ That's because force is a VECTOR which means that when summing forces you don't simply sum the magnitudes of the forces, but you sum both the directions and the corresponding component of each force. This question is a misconception of what a force is. I cannot suggest a book talking about such trivialities alone, but I think that any book of Griffiths should be sufficient to cover such misconceptions in an introductory chapter. Otherwise, ask your teacher to explain to you what vectors are. $\endgroup$ – Panos C. Apr 28 '18 at 15:25
  • $\begingroup$ Thanks for the reply, though I could have done without the patronizing tone. To be clear, I understand vectors and magnitudes and the WHAT/HOW of it all. Perhaps this is too much of a foundational, conceptual question. At the end of the day, I get that the vectors and magnitudes together resolve neatly into the original force. There's still something missing in the description of WHY a single force can exert itself as magnitudes (in two vectors simultaneously) that collectively exceed the actual force applied. $\endgroup$ – David R Apr 30 '18 at 1:01
  • $\begingroup$ Maybe turn it around and say why adding the magnitudes of the vectors like that is physically meaningful: i.e. explain what aspect of the system that sum is a measure of. If you can't identify such an aspect then perhaps that calculation isn't valid. $\endgroup$ – PM 2Ring Apr 30 '18 at 12:42

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!

  • $\begingroup$ Thanks! I should have been more clear in my original post that I understand vectors and vector diagrams. I'm fine reproducing the actual, single force out of its component vectors. Going back to the original example with the 500-Newton block on the ramp, a scale embedded in the ramp would read 250root2 Newtons (representing the normal force) and, if the block was static, the force from friction would also be 250root2. In short, gravity is applying 500 Newtons of force but the block needs over 700 Newtons of force to counteract gravity. Any more thoughts? Thanks again! $\endgroup$ – David R Apr 30 '18 at 1:14
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    $\begingroup$ "the block needs over 700 Newtons of force to counteract gravity." That's your mistake. You're adding magnitudes of component forces, and that's not the way to add forces when they're not in the same direction. It's a little like me owing you £15 and paying you £20, while you give me £5 change. £20 and £5 have passed between us but adding these sums together to make £25 has little relevance. I don't suppose this has helped, but $I've\ expanded\ my\ original\ answer $ in the hope that you'll be persuaded. I like your persistence! $\endgroup$ – Philip Wood Apr 30 '18 at 9:40
  • $\begingroup$ This reminds me of the classic Missing dollar riddle. $\endgroup$ – PM 2Ring Apr 30 '18 at 12:57
  • $\begingroup$ And me ……………... $\endgroup$ – Philip Wood Apr 30 '18 at 13:16
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    $\begingroup$ I think that a light bulb $coming\ on\ $ is supposed to signify a moment of understanding, but who cares? $\endgroup$ – Philip Wood Apr 30 '18 at 22:12

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