ABCDE is a pentagon. Forces overline{AB}, ove | vedclass

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(B) Let the resultant of the given forces be $\vec{R}$.$\vec{R} = \overline{AB} + \overline{AE} + \overline{DC} + \overline{ED}$Using the triangle law

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$\overline{BC}$

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(B) Let the resultant of the given forces be $\vec{R}$.$\vec{R} = \overline{AB} + \overline{AE} + \overline{DC} + \overline{ED}$Using the triangle law of vector addition:$\overline{AE} + \overline{ED} = \overline{AD}$So,$\vec{R} = \overline{AB} + \overline{AD} + \overline{DC}$Since $\overline{AD} + \overline{DC} = \overline{AC}$,we have:$\vec{R} = \overline{AB} + \overline{AC}$We want the final resultant to be $2\overline{AC}$.Let the force to be added be $\vec{F}$.$\vec{R} + \vec{F} = 2\overline{AC}$$(\overline{AB} + \overline{AC}) + \vec{F} = 2\overline{AC}$$\vec{F} = 2\overline{AC} - \overline{AC} - \overline{AB}$$\vec{F} = \overline{AC} - \overline{AB}$Using the triangle law,$\overline{AB} + \overline{BC} = \overline{AC}$,which implies $\overline{BC} = \overline{AC} - \overline{AB}$.Therefore,$\vec{F} = \overline{BC}$.

$ABCDE$ is a pentagon. Forces $\overline{AB}, \overline{AE}, \overline{DC}, \overline{ED}$ act at a point. Which force should be added to this system to make the resultant $2\overline{AC}$?

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