The angle between two forces of equal magnitude $\mathrm{R}$, if the magnitude of their resultant is…
The angle between two forces of equal magnitude $\mathrm{R}$, if the magnitude of their resultant is $\frac{R}{2}$
$\cos ^{-1}\left(-\frac{7}{8}\right)$
$\cos ^{-1}\left(-\frac{5}{7}\right)$
$\cos ^{-1}\left(-\frac{3}{7}\right)$
$\cos ^{-1}\left(-\frac{3}{4}\right)$
Solution
$\left(\frac{R}{2}\right)^{2}=R^{2}+R^{2}+2 R^{2} \cos \theta$
$\cos \theta=-\frac{7}{8}$
$\theta=\cos ^{-1}\left(\frac{-7}{8}\right)$
Alternative Solution:
Sure, let's break it down.
Given two forces of equal magnitude $R$ and the resultant of these two forces is $\frac{R}{2}$. We know the formula to find the magnitude of the resultant of two forces is
$R_r = \sqrt{R_1^2 + R_2^2 + 2R_1R_2\cos\theta}$,
where $R_1$ and $R_2$ are the magnitudes of the two forces and $\theta$ is the angle between them.
Since the forces have equal magnitude ($R_1 = R_2 = R$), we can simplify the equation to
$R_r = R\sqrt{2 + 2\cos\theta}$.
Given that $R_r = \frac{R}{2}$, we can substitute this into the equation:
$\frac{R}{2} = R\sqrt{2 + 2\cos\theta}$.
Divide both sides by $R$ and square both sides to get rid of the square root:
$\frac{1}{4} = 2 + 2\cos\theta$.
Rearrange to find $\cos\theta$:
$\cos\theta = \frac{1}{4} - 2 = -\frac{7}{8}$.
So, the correct answer is $\cos ^{-1}\left(-\frac{7}{8}\right)$, which