If $F_1$ has magnitude 30 N and $F_2$ has magnitude 40 N, what is the exact magnitude and direction of the resultant vector?
I think I should apply the Pythagorean theorem to get the magnitude, so is $M=\sqrt{30^2+40^2}=50$? Is the direction $\tan^{-1}\frac{40}{30}=53^\circ$?
Best Answer
No, that is not correct. The Pythagorean theorem applies only to right triangles. Here, the simplest way to do the problem is to use the "cosine rule". Move vector F2 to the tip of vector F1 and draw the line connecting the base of F1 to the tip of F2. That gives a triangle in which two sides have "length" 30 and 40. The angle between them, inside the triangle, is 180- 60= 120 degrees. By the "cosine law", if a triangle has two sides of length a and b, and the angle between them is C, then the length of the third side is given by $c^2= a^2+ b^2- 2ab cos(C)$ (In the case that C is a right triangle, cos(C)= 0 so that becomes the Pythagorean theorem. In the case that C is a straight angle (180 degrees), cos(C)= -1 so that becomes $c^2= a^2+ b^2+ 2ab= (a+ b)^2 or c= a+ b.).
Here, taking a= 30, b= 40, and C= 120 degrees, $c^2= 900+ 1600- 2400 (-1/2)= 900+ 1600+ 1200= 3700$ so $c= \sqrt{3700}= 10\sqrt{37}$.
(Paul, no. The direction of the two vectors is as given in the picture. "N here is "Newtons", not "North".)