1
$\begingroup$

Given two vectors $\mathbf{F_1}, \mathbf{F_2}$ and an angle $\alpha$ between two vectors we can derive the resulting force $F_R:=\Vert \mathbf{F_R}\Vert$.

$$F_R^2=\Vert\mathbf{F_2}-\mathbf{F_1}\Vert^2=(\mathbf{F_2}-\mathbf{F_1})^2=\Vert\mathbf{F_1}\Vert^2+\Vert\mathbf{F_2}\Vert^2-2\mathbf{F_1}\mathbf{F_2}=F_1^2+F_2^2-2F1F2\cos\alpha$$

such that $F_R=\sqrt{F_1^2+F_2^2-2F1F2\cos\alpha}$.

If we now leave the mathematical world towards the mechanical/engineering world we often find:

enter image description here

Thus here $F_R$ is given with a positive sign in front of $2F_1F_2\cos\alpha$.

How can there be two different formulas when the angle should be the same?

$\endgroup$
  • $\begingroup$ There are indeed two formulas, which both apply to parallelograms: one for each of the diagonals. See answer below. $\endgroup$ – Peter Diehr Mar 19 '16 at 19:54
3
$\begingroup$

The first derivation is correct, but only if you mean to take the difference between the two vectors, $\mathbf{F_1} - \mathbf{F_2}$; the figure would then show $\mathbf{F_D}$ running from the tip of one vector to the tip of the other, across the parallelogram. This is the Law of Cosines, which refers to the angle enclosed by the two sides of the triangle: $$F_D^2=\Vert \mathbf{F_2} - \mathbf{F_1}\Vert^2=\Vert\mathbf{F_1}\Vert^2+\Vert\mathbf{F_2}\Vert^2-2\mathbf{F_1}\mathbf{F_2}=F_1^2+F_2^2-2F_1F_2\cos(\alpha )$$

The second derivation obtains the correct result, but is flawed. Here we follow the figure in order to obtain the longer diagonal of the parallelogram by moving $\mathbf{F_1}$ to the tip of $\mathbf{F_2}$, the enclosed angle becomes the supplement, $(\pi - \alpha )$; and the vector $\mathbf{F_1}$ should be pointing away from the new vertex.

Now when we evaluate the law of cosines we get $$F_R^2=\Vert-\mathbf{F_2} + \mathbf{F_1}\Vert^2=\Vert\mathbf{F_1}\Vert^2+\Vert\mathbf{F_2}\Vert^2+2\mathbf{F_1}(-\mathbf{F_2})=F_1^2+F_2^2-2F_1F_2\cos(\pi - \alpha )$$

But $cos(\pi - \alpha ) = -cos(\alpha )$, which recovers the second formula, from the first: $$F_R^2=\Vert\mathbf{F_2}-\mathbf{F_1}\Vert^2=\Vert\mathbf{F_1}\Vert^2+\Vert\mathbf{F_2}\Vert^2-2\mathbf{F_1}\mathbf{F_2}=F_1^2+F_2^2+2F_1F_2\cos\alpha$$

You can also obtain the second formula from Euclid's Law of Parallelograms, which states that the the sum of the squares of the four sides is equal to the sum of the squares of the two diagonals. Subtract the first formula, the Law of Cosines, which gives the square of the length of the short diagonal, from the sum of the squares of the four sides, $$ F_R^2 + F_D^2=(2\Vert\mathbf{F_1}\Vert^2+2\Vert\mathbf{F_2}\Vert^2)$$ and you are left with the second formula, which gives the length squared of the long diagonal of the parallelogram.

$\endgroup$
3
$\begingroup$

The first formula you derived is actually not correct. In the first formula, you have actually written, resultant force is the difference between the two forces. It's wrong... enter image description here

By triangle law of vector addition, OA + AC = OC, OA = P, AC = OB = Q, OC = R. Thus, P + Q =R but not P - Q = R that implies, FR=F21+F22+2F1F2cosα.. Thus the both formulae are the same...

$\endgroup$
  • $\begingroup$ my reading of the OP is slightly different, and resolves the apparent discrepancy by finding two flaws. $\endgroup$ – Peter Diehr Mar 19 '16 at 19:53

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.