<p>Which pair of forces can produce a resultant of 15 newtons?
(A) 20 N, 20 N
(B) 25 N, 5 N
(C) 5 N, 5 N
(D) 7 N, 7 N
(E) 5 N, 3 N</p>
<p>Could someone please tell my why the book is saying that A is the correct answer? (Barrons).</p>
<p>Thanks.</p>
<p>C,D, and E cannot add up or subtract (referring to pythagorean theorem) to 15 in any way. With B, the closest you can get is by subtracting 5 from 25, which is 20. This is leaves A, 20 and 20. I assume that with the correct angle, you can subtract the vectors to equal 15. </p>
<p>Someone correct me if my reasoning is incorrect.</p>
<p>^Just right. Math people may also remember the rule about triangles: the length of the third side is greater than the difference and less than the sum of the other two sides. Here, the third side is the resultant.</p>
<p>Thanks a lot! I didn’t think of the possibility of there being angles between the forces other than 0 or 180 degrees! That rule about the third side of a triangle cleared things up as well. Thanks again.</p>
<p>Schoolisfun, I don’t think you mean “Pythagorean theorem.” I think you mean “triangle inequality.”</p>
<p>Not that that has any bearing on the correct answer, or how to find it.</p>
<p>^I was referrring to if you had them at 90 degree angles specifically and/or the other possibilities</p>
<p>look at law of cosines not pythag theorem:</p>
<p>a^2 + b^2 - 2absin(Phi) = c^2</p>
<p>2ab is (-800,800)</p>
<p>therefore c^2 is in (0, 1600)</p>
<p>resultant is in (0, 40)</p>
<p>15 is in this set so (A).</p>
<p>Greed, that’s a much bigger cannon, so to speak, than you need for this problem. By the triangle inequality, A is the only choice that could yield a third side with a measure of 15.</p>
<p>(As fignewton said, several posts and 26 hours ago.)</p>
<p>I agree Sikor, but I just like to show an alternate method that perhaps makes more sense to some people.</p>
<p>All of the ways are good, sorry if my reasoning was unclear.</p>