The angle between two forces of equal magnitu | vedclass

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Question

(A) Let the two forces be $\vec{F_1}$ and $\vec{F_2}$ such that $|\vec{F_1}| = |\vec{F_2}| = R$. The magnitude of the resultant $\vec{R_{res}}$ is giv

Vedclass · Accepted Answer

$\cos ^{-1}\left(-\frac{7}{8}\right)$

Vedclass · Answer

(A) Let the two forces be $\vec{F_1}$ and $\vec{F_2}$ such that $|\vec{F_1}| = |\vec{F_2}| = R$. The magnitude of the resultant $\vec{R_{res}}$ is given by the formula: $|\vec{R_{res}}| = \sqrt{R^2 + R^2 + 2R^2 \cos 	heta}$.Given that $|\vec{R_{res}}| = \frac{R}{2}$,we square both sides: $\left(\frac{R}{2}ight)^2 = R^2 + R^2 + 2R^2 \cos 	heta$.$\frac{R^2}{4} = 2R^2 + 2R^2 \cos 	heta$.Dividing by $R^2$: $\frac{1}{4} = 2 + 2 \cos 	heta$.$\frac{1}{4} - 2 = 2 \cos 	heta$.$-\frac{7}{4} = 2 \cos 	heta$.$\cos 	heta = -\frac{7}{8}$.Therefore,$	heta = \cos ^{-1}\left(-\frac{7}{8}ight)$.

The angle between two forces of equal magnitude $R$,if the magnitude of their resultant is $\frac{R}{2}$,is:

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