Question

A cylindrical vessel filled with water is released on a fixed inclined surface of angle $\theta$ as shown in figure. The friction coefficient of surface with vessel is $\mu$. Then the constant angle made by the surface of water with the incline will be (Neglect the viscosity of liquid) :-

• A. $tan^{-1}\mu$ if $\mu < tan \theta$
• B. 0 if $\mu = 0$
• C. $\theta$ if $\mu > tan \theta$
• D. $\theta$ if $\mu < tan \theta$

Answer 1

Answer: (A), (C)

A, C

Answer 2

The effective gravitational acceleration in the vessel's frame is $\vec{g}_{eff} = \vec{g} - \vec{a}$. The acceleration of the vessel down the incline is $a = g(\sin \theta - \mu \cos \theta)$.

The components of $\vec{g}_{eff}$ along the incline ($x'$) and perpendicular to it ($y'$) are: $g_{eff, x'} = g \sin \theta - a = g \sin \theta - g(\sin \theta - \mu \cos \theta) = g \mu \cos \theta$ $g_{eff, y'} = -g \cos \theta$ (perpendicular to incline, downwards)

The surface of the water is perpendicular to $\vec{g}_{eff}$. The angle $\alpha$ the water surface makes with the incline is given by $\tan \alpha = \frac{g_{eff, x'}}{-g_{eff, y'}} = \frac{g \mu \cos \theta}{g \cos \theta} = \mu$. Thus, $\alpha = \tan^{-1} \mu$.

This is valid when the vessel is sliding, i.e., when the net force down the incline is positive or when there is kinetic friction. The condition for sliding down is $\sin \theta > \mu \cos \theta$, or $\tan \theta > \mu$. So, if $\mu < \tan \theta$, the angle is $\tan^{-1} \mu$. This matches option (A).

If $\mu \ge \tan \theta$, and we assume $\mu$ is the static friction coefficient, the vessel remains at rest. In this case, the water surface is horizontal, making an angle $\theta$ with the incline. This matches option (C) for $\mu > \tan \theta$. If $\mu = \tan \theta$, the acceleration is zero. If moving, it's with constant velocity, and the angle is $\tan^{-1} \mu = \tan^{-1}(\tan \theta) = \theta$. If at rest, the angle is $\theta$. Therefore, if $\mu \ge \tan \theta$, the angle is $\theta$. Option (C) covers the case $\mu > \tan \theta$.