The free surface of oil in a tanker, at rest, is horizontal. If the tanker starts accelerating the free surface will be titled by an angle $\theta $. If the acceleration is $\mathrm{a}$ $\mathrm{ms}^{-2}$, what will be the slope of the free surface ?
The free surface of oil in a tanker, at rest, is horizontal. If the tanker starts accelerating the free surface will be titled by an angle $\theta $. If the acceleration is $\mathrm{a}$ $\mathrm{ms}^{-2}$, what will be the slope of the free surface ?
When a tanker accelerates a pseudo force exerted on oil and hence its surface does not remain horizontal.
Consider a small element of mass $d m$. The forces on it are shown as in figure.
For equilibrium, the forces parallel to the slope, $(d m) a \cos \theta=(d m) g \sin \theta$ where the pseudo force exerted opposite to the acceleration $(d m) a$
$\therefore a \cos \theta=g \sin \theta$
$\therefore \tan \theta=\frac{a}{g}=\text { slope }$
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