Solution 14.7b
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(New page: Image:14.7.gif Using part a, we see from the figure that <math>F=S=\frac{98}{\tan \theta }</math> and <math>R=196</math> <math>\begin{align} & F\le \mu R \\ & \frac{98}{\tan \the...)
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Using part a, we see from the figure that
\displaystyle F=S=\frac{98}{\tan \theta } and \displaystyle R=196
\displaystyle \begin{align}
& F\le \mu R \\
& \frac{98}{\tan \theta }\le 0\textrm{.}6\times 196 \\
& \tan \theta \ge \frac{98}{0\textrm{.}6\times 196} \\
& \theta \ge 39\textrm{.}8{}^\circ \\
\end{align}
