A heavy hollow cone of radius R and height h | vedclass

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Question

(A) Consider the equilibrium of the water inside the cone.The forces acting on the water are:$1$. The weight of the water acting downwards: $W = V \rh

Vedclass · Accepted Answer

$(2/3) \pi R^2 h \rho g$

Vedclass · Answer

(A) Consider the equilibrium of the water inside the cone.The forces acting on the water are:$1$. The weight of the water acting downwards: $W = V ho g = (\frac{1}{3} \pi R^2 h) ho g$.$2$. The upward force exerted by the table on the water: $F_{	ext{table}} = P 	imes A = (ho g h) 	imes (\pi R^2) = \pi R^2 h ho g$.$3$. The downward force exerted by the slanted walls of the cone on the water: $F_{	ext{walls}}$.For the water to be in equilibrium,the net vertical force must be zero:$F_{	ext{table}} - W - F_{	ext{walls}} = 0$By Newton's third law,the force exerted by the water on the cone walls is equal and opposite to the force exerted by the walls on the water. Thus,the upward force exerted by the water on the cone is equal to the downward force exerted by the walls on the water:$F_{	ext{upward}} = F_{	ext{walls}} = F_{	ext{table}} - W$$F_{	ext{upward}} = \pi R^2 h ho g - \frac{1}{3} \pi R^2 h ho g$$F_{	ext{upward}} = \frac{2}{3} \pi R^2 h ho g$.

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