4

For example, Bob pulls with a force of $250$ Newtons due West and Tim pulls with a force of $500$ Newtons due North. What is the resultant vector?

I use the Pythagorean theorem on $250$ and $500$. Then I find $~\tan^{-1}\left(\frac{500}{250}\right)$, which equals $27^{\circ}$. But then I have to add $27^{\circ}$ to $90^{\circ}$. The answer is $117^{\circ}$. But how do I know when to add or subtract to/from $90^{\circ}$. What about $180^{\circ}$, $270^{\circ}$, or $360^{\circ}$?

nmasanta
  • 9,222
Dylan
  • 73
  • If I understand the question correctly, I believe the answer is: it depends. Because these events are not occurring on a pre-defined coordinate plane, isn't the origin's position arbitrary? For example, the origin could be placed on Bob, or it could be placed on Tim. – N. Bar Sep 04 '19 at 01:49
  • Instead of doing that, compute the “east-west” and “north-south” components of the two vectors and add them, then convert back. There’s still an ambiguity, of course, but it’s easily resolved either by examining the signed of the resultant’s components (which is what an ATAN2-like function as suggested in one of the answers does), or by observing that the resultant is the diagonal of the parallelogram formed by the two input forces, and so is “between” them. – amd Sep 04 '19 at 02:03

2 Answers2

1

You can use the signal for the quadrants of the sum of two vectors in the following way:

  • N and E will result in a vector in the first quadrant (both coordinates positive);

  • N and W will result in a vector in the second quadrant (the horizontal coordinate is negative);

  • S and W will result in a vector in the third quadrant (both coordinates negative);

  • S and E will result in a vector in the fourth quadrant (the vertical coordinate is negative).

And you may also use the magnitude for the octants: If the absolute value of the vertical coordinate is greater than the absolute value of the horizontal coordinate, then you are in the second (NNE), third (NNW), sixth (SSW) or seventh (SSE) octant.

Edit:

If you compute the angle using the absolute value of the horizontal coordinate $x$ and the absolute value of the vertical coordinate $y$:

$$ \theta = \arctan{\frac{|y|}{|x|}} $$

then the "true" angle is:

  • $\theta$ if you are in the first quadrant;

  • $90^{\circ} - \theta$ if you are in the second quadrant (if $x$ is positive but very close to zero, that angle will be the vertical symmetric of the angle if the horizontal coordinate was $-x$);

  • $180^{\circ} + \theta$ if you are in the first quadrant (if you switch the signal of both coordinates, the angle "points" in the opposite direction, creating a straight line);

  • $270^{\circ} - \theta$ (for a similar reason to an angle in the second quadrant).

Ertxiem - reinstate Monica
  • 1,664
  • 3
  • 11
  • 15
  • Thank you, but how do I know if I am to subtract or add to 90, 180, 270, or 360? For example, the vector is in the second quadrant. Therefore, it involves 180 degrees. But do I subtract from 180, or do I add to 180? How do I determine that? – Dylan Sep 04 '19 at 01:55
  • @Dylan perhaps the definition of the atan2 computer function may help you. – Ertxiem - reinstate Monica Sep 04 '19 at 01:59
  • I haven't taken calculus or learned about trigonometry yet. I'm in physics and my teacher said to use arctan. I only know SOHCAHTOA. Anyways, what degree is the first, second, third, and fourth quadrant and how do I know when to subtract or add? – Dylan Sep 04 '19 at 02:01
  • @Dylan: I've edited my post to include the explanation of the computation of the angle. – Ertxiem - reinstate Monica Sep 04 '19 at 02:12
0

Draw a force diagram. You have a vector of length $500$ going north and one of length $250$ going east.

Complete the parallelogram and draw the resultant vector. It will naturally be in the correct quadrant already.

Now you just need to measure the angle from the correct starting direction. When I see north and east components I'm usually in an actual geographic system and want to measure the compass direction, which is measured clockwise from due north. But you seem to be working in a system where you have to measure counterclockwise from due east.

In more advanced physics classes you will rarely see north, south, east, or west. You will just have $x$ and $y$ components (plus a $z$ component for a three-dimensional problem) and you will almost never need to ask what quadrant you're in.

David K
  • 98,388