"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." [closed]

Question

"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.

Answer 1

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 $\theta \approx 77.31^\circ$.

Answer 2

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 $\alpha = 77.3^\circ$.

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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:

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