0
$\begingroup$

I know vectors additions, components, triangle inequalities, also law of cosine when length of vectors components and angle given and any math involved in it, but my question is about physical quantities . If force $ |\vec F |$ (like in 2-D) has some value "strength / Length of vector" how it can produce two components ($ | \vec F_x | $ , $ | \vec F_y | $ ) their sum is larger the $ | \vec F |$ . e.g : $$ | \vec F | \le | \vec F_x | + | \vec F_y| $$ $$ 5 < 3+4 $$ how strength of $ ``\vec F = 5" $ can produce $ `` \vec F_x=3" $ & $ `` \vec F_y =4 " $ strength bigger than "5" in term of their components. If someone is applying force $ \vec F = 5N $ how it can effect $ \vec F_x =3N $ in x-component direction & $ \vec F_y = 4N $ in y-component direction bigger than it has. It is applicable on any vector.

$\endgroup$
  • 1
    $\begingroup$ 3 meters east, 4 meters north, would you not have shorter distance from the initial point to final? Pythagoras theorem? $\endgroup$ – exp ikx May 16 at 15:29
  • 2
    $\begingroup$ "Adding the magnitude of components in different directions" doesn't mean anything physically. Mathematically, this is just the triangle equality, of course. Your question is no different from asking why (+3) + (-2) = 1, when |+3| + |-2| = 5. $\endgroup$ – alephzero May 16 at 15:46
  • 1
    $\begingroup$ Well, if the sum or orthogonal components does have any physical relevance, it's geometrically obvious what is happening. If you are at one corner of a rectangular field and you want to get to the diagonally opposite corner, it's obvious that walking along two edges (orthogonal components) is a longer distance than walking in a straight line across the middle of the field (the magnitude of the resultant vector). $\endgroup$ – alephzero May 16 at 15:50
  • $\begingroup$ What alephzero said. You can't just add stuff together without justification. This is like the classic Missing Dollar riddle. $\endgroup$ – PM 2Ring May 16 at 15:50
  • $\begingroup$ Your $<$ should be a $\le$. The vector might be directed along the $x$ or $y$ axis. $\endgroup$ – G. Smith May 16 at 16:23
1
$\begingroup$

Vectors have direction. You can only add the horizontal components together and same for vertical components. You cannot add different directions. Your inequality looks like the famous triangle inequality which basically states that the sum of any two sides is greater than the third side. You do not seem to be adding independent components, as some comments suggest, but you do seem to be confusing this inequality with vector addition results. The components of a vector, referenced to Cartesian coordinates, will always form the legs of a right triangle. The magnitude of the vector (force, velocity, etc.) will be given by the Pythagorean theorem, $$ F = \sqrt{F_x^2 + F_y^2} \,. $$ The inequality you are asking about simply says that $$ \sqrt{F_x^2 + F_y^2} \le \left|F_x\right| + \left|F_y\right| $$ which is a fairly easy identity to derive. Square both sides.

$\endgroup$
  • $\begingroup$ Thanks @ggcg for your reply . Ahaaa...! i think now i know my answer. Pls correct if i am wrong. If force has some strength at particular point let say $ \vec F =5N $ it becomes lesser and lesser strength as move along. It depends how far you are. If $ \vec F $ has far from x-axis it strength become less compare to y-axis. $\endgroup$ – 123 May 16 at 17:03
  • $\begingroup$ It does not relate to the addition. Pythagoras was just a way of calculation . we need two vector components to calculate the original vector. $\endgroup$ – 123 May 16 at 17:20
  • $\begingroup$ I am more confused by both your comments. What are you trying to do? Perhaps you can explain it more clearly. But I believe my answer addresses everything in your post. You seem to be adding independent directions to get a net force and this is an illegal operation. You must get components and then use vector addition. The magnitude of the resultant vector is in fact given by the Pythagorean theorem. $\endgroup$ – ggcg May 16 at 17:26
  • $\begingroup$ thanks @ggcg i want to understand physical strength at that point in terms of physics not in terms of mathematics. but it is a matter of thinking in what way we think. of course you are right. Once again thanks for support all of you guys. $\endgroup$ – 123 May 16 at 17:34
  • $\begingroup$ The math and physics come hand-in-hand. $\endgroup$ – ggcg May 16 at 17:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.