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
KVPY 2013, Advanced
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(c)
As plank is resting, force of friction is just equal to centripetal force due to rotation of earth.
Force trying to slip plank on the ground is parallel component of centripetal force.
$\therefore$ friction, $f=F_{||}$
$\Rightarrow f =m \omega^2 r_e \sin 45^{\circ}$
$=m \omega^2 r_e \cos 45^{\circ} \sin 45^{\circ}=\frac{m(\omega)^2 r_e}{2}$
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