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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 a m/s2, what will be the slope of the free surface?

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Let us assume the tanker is pulled by a force F and it produces an acceleration in the forward direction. Let \deltam be the mass of a small element.  As the tanker is pulled forward and accelerates, the small element also experiences a force in the same direction. The inertia of rest makes the element to remain at rest and hence as the tank moves forward, the element moves backwards as a result of inertia.

Force acting on \deltam, F1 = \deltama in the backward direction

F2 = \deltamg which is in a vertically downward direction.

The normal reaction gets balanced by various components: \deltama \sin \theta from F1

The situation when there is a maximum angle inclination by the surface,

\delta ma \cos \theta = \delta mg \sin \theta

So required slope, \frac{\sin \theta }{\cos \theta}=\frac{a}{g}

So, \tan \theta=\frac{a}{g}

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