ABCDEF is a regular hexagon and forces repres | vedclass

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Question

(C) Let the resultant force be $\vec{R} = \overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF}$

Vedclass · Accepted Answer

$3\, \overrightarrow{AD}$

Vedclass · Answer

(C) Let the resultant force be $\vec{R} = \overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF}$.In a regular hexagon $ABCDEF$,the center $O$ is the midpoint of the diagonals $AD$,$BE$,and $CF$. Thus,$\overrightarrow{AB} + \overrightarrow{AE} = 2\overrightarrow{AO} = \overrightarrow{AD}$ (since $\overrightarrow{AD} = 2\overrightarrow{AO}$).Similarly,$\overrightarrow{AC} + \overrightarrow{AF} = 2\overrightarrow{AO} = \overrightarrow{AD}$.Substituting these into the expression for $\vec{R}$:$\vec{R} = (\overrightarrow{AB} + \overrightarrow{AE}) + (\overrightarrow{AC} + \overrightarrow{AF}) + \overrightarrow{AD}$$\vec{R} = \overrightarrow{AD} + \overrightarrow{AD} + \overrightarrow{AD} = 3\overrightarrow{AD}$.

$ABCDEF$ is a regular hexagon and forces represented in magnitude and direction by $\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}, \overrightarrow{AE}$ and $\overrightarrow{AF}$ act at point $A$. Their resultant is:

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