Two forces,F_1 and F_2,are acting on a body. | vedclass

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(C) Let the smaller force be $F_1 = F$ and the greater force be $F_2 = 2F$.Given that the resultant force $R$ is equal to the greater force,so $R = 2F

Vedclass · Accepted Answer

$\cos^{-1}(-1/4)$

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(C) Let the smaller force be $F_1 = F$ and the greater force be $F_2 = 2F$.Given that the resultant force $R$ is equal to the greater force,so $R = 2F$.The formula for the resultant of two vectors is $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos 	heta}$.Substituting the given values: $2F = \sqrt{F^2 + (2F)^2 + 2(F)(2F) \cos 	heta}$.Squaring both sides: $(2F)^2 = F^2 + 4F^2 + 4F^2 \cos 	heta$.$4F^2 = 5F^2 + 4F^2 \cos 	heta$.Subtracting $5F^2$ from both sides: $-F^2 = 4F^2 \cos 	heta$.Dividing by $4F^2$: $\cos 	heta = -1/4$.Therefore,the angle $	heta = \cos^{-1}(-1/4)$.

Two forces,$F_1$ and $F_2$,are acting on a body. One force is double that of the other force and the resultant is equal to the greater force. Then the angle between the two forces is

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