The free surface of oil in a tanker,at rest,i | vedclass

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(A) When a tanker accelerates,a pseudo force is exerted on the oil,and hence its surface does not remain horizontal.Consider a small element of mass $

Vedclass · Accepted Answer

$	an 	heta = a/g$

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(A) When a tanker accelerates,a pseudo force is exerted on the oil,and hence its surface does not remain horizontal.Consider a small element of mass $dm$ on the surface of the oil.The forces acting on this element are:$1$. The weight $dm \cdot g$ acting vertically downwards.$2$. The pseudo force $dm \cdot a$ acting horizontally in the direction opposite to the acceleration.For the surface to be in equilibrium in the non-inertial frame of the tanker,the net force acting on the surface element must be perpendicular to the surface. Alternatively,resolving forces parallel to the tilted surface:The component of the pseudo force along the surface is $(dm \cdot a) \cos 	heta$.The component of the weight along the surface is $(dm \cdot g) \sin 	heta$.For equilibrium along the surface:$(dm \cdot a) \cos 	heta = (dm \cdot g) \sin 	heta$Dividing both sides by $(dm \cdot \cos 	heta)$:$	an 	heta = \frac{a}{g}$Thus,the slope of the free surface is $\frac{a}{g}$.

The free surface of oil in a tanker,at rest,is horizontal. If the tanker starts accelerating,the free surface will be tilted by an angle $\theta$. If the acceleration is $a \ m/s^2$,what will be the slope of the free surface?

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