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(D) Let the two forces be $\vec{F_1}$ and $\vec{F_2}$ with magnitudes $F_1$ and $F_2$,where $F_1 > F_2$. The magnitude of the resultant force $R$ is g

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Point in opposite directions

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(D) Let the two forces be $\vec{F_1}$ and $\vec{F_2}$ with magnitudes $F_1$ and $F_2$,where $F_1 > F_2$. The magnitude of the resultant force $R$ is given by $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos 	heta}$,where $	heta$ is the angle between the forces. We are given that $R Squaring both sides,we get $F_1^2 + F_2^2 + 2F_1F_2 \cos 	heta This simplifies to $F_2^2 + 2F_1F_2 \cos 	heta Dividing by $F_2$ (since $F_2 > 0$),we get $F_2 + 2F_1 \cos 	heta Since $F_1 > F_2$,the value of $\frac{F_2}{2F_1}$ is between $0$ and $0.5$. Thus,$\cos 	heta$ must be negative,meaning the angle $	heta$ is obtuse $(90^\circ Among the given options,the condition that the forces point in opposite directions (or generally have an obtuse angle between them) is the only one that satisfies the requirement for the resultant to be smaller than the larger force.

If the resultant of two forces has a magnitude smaller than the magnitude of the larger force,the two forces must be

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