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A plank is resting on a horizontal ground in the northern hemisphere of the Earth at a $45^{\circ}$ latitude. Let the angular speed of the Earth be $\omega$ and its radius $r_{e}$. The magnitude of the frictional force on the plank will be
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The correct answer is:
$\frac{\mathrm{mr}_{\mathrm{e}} \omega^{2}}{2}$
$\mathrm{F}_{5}=\mathrm{m} \omega^{2} r \cos 45^{\circ}$
where $r=\operatorname{Rcos} 45^{\circ}$
$\mathrm{F}_{\mathrm{r}}=\frac{\mathrm{m} \omega^{2} \mathrm{R}}{2}$
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