Two forces having magnitude A and frac{A}{2} | vedclass

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(D) Let the two forces be $\vec{F}_1$ and $\vec{F}_2$,where $|\vec{F}_1| = A$ and $|\vec{F}_2| = \frac{A}{2}$.Since the forces are perpendicular to ea

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$\frac{\sqrt{5}A}{2}$

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(D) Let the two forces be $\vec{F}_1$ and $\vec{F}_2$,where $|\vec{F}_1| = A$ and $|\vec{F}_2| = \frac{A}{2}$.Since the forces are perpendicular to each other,the angle between them is $	heta = 90^{\circ}$.The magnitude of the resultant force $\vec{F} = \vec{F}_1 + \vec{F}_2$ is given by:$|\vec{F}| = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos 90^{\circ}}$Since $\cos 90^{\circ} = 0$,the expression simplifies to:$|\vec{F}| = \sqrt{F_1^2 + F_2^2}$Substituting the given values:$|\vec{F}| = \sqrt{A^2 + \left(\frac{A}{2}ight)^2} = \sqrt{A^2 + \frac{A^2}{4}}$$|\vec{F}| = \sqrt{\frac{4A^2 + A^2}{4}} = \sqrt{\frac{5A^2}{4}}$$|\vec{F}| = \frac{\sqrt{5}A}{2}$

Two forces having magnitude $A$ and $\frac{A}{2}$ are perpendicular to each other. The magnitude of their resultant is

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