"The resultant of two forces of equal size, that form an angle, is lowered by 20% when one of the forces is turned in the opposite direction."

Does anyone know how one would go about trying to find the angle where this happens? I've been reading an old textbook on Mechanics and it has stunped me for quite some time now.

The book is originally written in Swedish so forgive my bad translation skills.

  • 2
    $\begingroup$ What are you referring to, the magnitude of the resultant force?? Take the example to extreme - two forces that are perpendicular. If you change direction of one of the forces, the resultant force has the same magnitude. Please clarify what do you exactly mean. $\endgroup$ Jan 7, 2022 at 15:44
  • $\begingroup$ I'd draw a rhombus... $\endgroup$
    – PM 2Ring
    Jan 7, 2022 at 16:13
  • $\begingroup$ Write the magnitude of the resultant for the two cases and use the relationship between the magnitudes. $\endgroup$
    – nasu
    Jan 7, 2022 at 16:18
  • $\begingroup$ Nice question! I drew a head-to-tail vector addition diagram for the first case and superimposed the modified diagram for the second case. Some simple geometry was then apparent ... Finding the answer then needed the use of just one trigonometric ratio and the troublesome task of multiplying by 2. $\endgroup$ Jan 7, 2022 at 17:15
  • $\begingroup$ You'll see at once that $\theta=2\arctan 0.8$. $\endgroup$ Jan 7, 2022 at 18:29

2 Answers 2


I think this problem can be solved the easiest without introducing coordinates. Say we have two forces $\mathbf{f}$ and $\mathbf{g}$, then the question statement can be written as $$ \alpha\sqrt{(\mathbf{f} + \mathbf{g})^2} = \sqrt{(\mathbf{f} - \mathbf{g})^2} \, , $$ with $\alpha = 0.8$. Squaring and using $\cos \theta = \frac{\mathbf{f} \cdot \mathbf{g}}{\vert\mathbf{f}\vert \, \vert\mathbf{g}\vert}$ directly leads to $$ \cos \theta = \frac{(\mathbf{f}^2 + \mathbf{g}^2)(1-\alpha^2)}{2\vert\mathbf{f}\vert \, \vert\mathbf{g}\vert(1 + \alpha^2)} \,. $$ If $\vert\mathbf{f}\vert = \vert\mathbf{g}\vert$, this reduces to $$ \theta = \arccos \left(\frac{1-\alpha^2}{1+\alpha^2}\right) \, , $$ which evaluates to $\theta \approx 77.31^\circ$.


In first case the two forces are defined as:

$$\vec{F}_1 = F \hat{\imath} \quad \text{and} \quad \vec{F}_2 = F \cos\alpha \hat{\imath} + F \sin\alpha \hat{\jmath}$$

and their resultant force is:

$$\vec{F}_R = F (\cos\alpha + 1) \hat{\imath} + F \sin\alpha \hat{\jmath}$$

When you take force $F_1$ to point in the opposite direction, i.e. $\vec{F}_1 = -F \hat{\imath}$, the resultant force is:

$$\vec{F}_R' = F (\cos\alpha - 1) \hat{\imath} + F \sin\alpha \hat{\jmath}$$

The ratio of resultant force magnitudes is:

$$\frac{|\vec{F}_R'|}{|\vec{F}_R|} = \frac{1 - \cos\alpha}{\sin\alpha} \quad \text{and} \quad |\vec{F}_R'| = p \cdot |\vec{F}_R|$$

From the above equation it follows that

$$1 - \cos\alpha = \frac{p^2 + 1}{2} \sin^2\alpha$$

and the final solution is

$$\boxed{\alpha = \arcsin \Bigl( \frac{2p}{p^2 + 1} \Bigr)}$$

For $p = 0.8$ the angle is $\alpha = 77.3^\circ$.

Here is the detailed expansion for the magnitudes ratio:

$$\frac{|\vec{F}_R'|}{|\vec{F}_R|} = \frac{F \sqrt{\bigl(\cos\alpha - 1\big)^2 + \bigl(\sin\alpha\bigr)^2}}{F \sqrt{\bigl(\cos\alpha + 1\big)^2 + \bigl(\sin\alpha\bigr)^2}} = \frac{\sqrt{1 - \cos\alpha}}{\sqrt{1 + \cos\alpha}} \cdot \frac{\sqrt{1 - \cos\alpha}}{\sqrt{1 - \cos\alpha}} = \frac{1 - \cos\alpha}{\sin\alpha} = p$$

Here is the detailed expansion for the above trigonometric equation:

$$1 - \cos\alpha = p \sin\alpha \rightarrow 1 + \cos^2\alpha - 2\cos\alpha = p^2 \sin^2\alpha \rightarrow 2 - 2\cos\alpha = (p^2 + 1) \sin^2\alpha$$

  • $\begingroup$ The y component of F2 changes sign too when you reverse direction, doesn't? $\endgroup$
    – nasu
    Jan 7, 2022 at 16:25
  • $\begingroup$ The problem says only one force changes direction. I decided to simplify the problem by taking one force along only one axis, and then I changed direction of that force. You would get the same result if you reversed the other force. $\endgroup$ Jan 7, 2022 at 16:26
  • $\begingroup$ Oh, right. It makes sense. But the ratio of the magnitudes does not look right, based on your previous equations for the forces. $\endgroup$
    – nasu
    Jan 7, 2022 at 16:31
  • $\begingroup$ I verified the result, seems correct to me. Note that I left out some trigonometry steps and wrote only the final result. $\endgroup$ Jan 7, 2022 at 16:37
  • 1
    $\begingroup$ Yes, I agree. I solved for $cos \alpha$ and it was not obvious at first sight that the two expressions are equivalent. $\endgroup$
    – nasu
    Jan 7, 2022 at 20:32

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