Solution 14.6a

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The diagram shows the forces acting on the ladder.

Image:14.6temp.gif

Taking moments about the base of the ladder:

\displaystyle \begin{align} & 5\sin 70{}^\circ \times S=2\textrm{.}5\cos 70{}^\circ \times 245 \\ & S=\frac{2\textrm{.}5\cos 70{}^\circ \times 245}{5\sin 70{}^\circ }=44\textrm{.}59\text{ N} \\ \end{align}

Considering the horizontal forces \displaystyle F=S=44\textrm{.}59\text{ N}

Considering the vertical forces \displaystyle R=245\text{ N}