Buoyancy + Force +Pressure!

Classical Mechanics Level 3

A solid sphere of radius r r is floating at the interface of two immiscible liquids of densities ρ 1 \rho_1 and ρ 2 \rho_2 where ρ 2 > ρ 1 \rho_2 >\rho_1 , half of its volume lying in each zone. The height of the upper liquid column from the interface of the two liquids is h h .

Find the force exerted on the sphere by contact with the upper liquid layer.

Details

  • atmospheric pressure = p 0 p_0
  • acceleration due to gravity is g g
p 0 π r 2 p_0 \pi r^2 ( h − 2 3 r ) π r 2 ρ 1 g (h-\frac{2}{3}r) \pi r^2 \rho_1 g p 0 π r 2 + ( h − 2 3 r ) π r 2 ρ 1 g p_0 \pi r^2 + (h-\frac{2}{3}r) \pi r^2 \rho_1 g 2 3 π r 3 ρ 1 g \frac{2}{3} \pi r^3 \rho_1 g

Nishant Rai by Nishant Rai , 25, India

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1 solution

Nishant Rai
May 13, 2015

4 Helpful 1 Interesting 1 Brilliant 1 Confused

Ahhhhh of course... I did it the hard way:

F = 2 π ∫ h − r h ( P o + ρ 1 g x ) ( h − x ) d x F=2\pi\int\limits_{h-r}^h(P_o+\rho_1gx)(h-x)dx

Nathanael Case - 6 years ago

Shoot, this one got me; the second fluid threw me off. But at least I know now.

Andrew Song - 5 years, 9 months ago

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