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Question

(A) When the vessel is released on the inclined plane,it accelerates down the plane.The acceleration of the vessel is $a = g(\sin 	heta - \mu \cos

Vedclass · Accepted Answer

$	an ^{-1} \mu$

Vedclass · Answer

(A) When the vessel is released on the inclined plane,it accelerates down the plane.The acceleration of the vessel is $a = g(\sin 	heta - \mu \cos 	heta)$.In the frame of the vessel,a pseudo force $ma$ acts on the water particles in the upward direction along the incline.The effective acceleration $\vec{g}_{eff}$ experienced by the water is the vector sum of the gravitational acceleration $\vec{g}$ and the pseudo acceleration $-\vec{a}$.The surface of the water will be perpendicular to the effective acceleration vector $\vec{g}_{eff}$.Let $\alpha$ be the angle that the water surface makes with the horizontal. The angle $\phi$ that the water surface makes with the incline is given by the ratio of the components of $\vec{g}_{eff}$ perpendicular and parallel to the incline.The component of $\vec{g}_{eff}$ parallel to the incline is $g_{||} = g \sin 	heta - a = g \sin 	heta - g(\sin 	heta - \mu \cos 	heta) = \mu g \cos 	heta$.The component of $\vec{g}_{eff}$ perpendicular to the incline is $g_{\perp} = g \cos 	heta$.The angle $\phi$ that the water surface makes with the incline is $	an \phi = \frac{g_{||}}{g_{\perp}} = \frac{\mu g \cos 	heta}{g \cos 	heta} = \mu$.Therefore,$\phi = 	an ^{-1} \mu$.

$A$ cylindrical vessel filled with water is released on an inclined surface of angle $\theta$ as shown in the figure. The friction coefficient of the surface with the vessel is $\mu (< \tan \theta)$. Then the contact angle made by the surface of water with the incline will be

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