Question

You place a sphere of radius R=1.54 m and of density ρ = 0.75g/cm3 in a pool of clean water. The sphere begins to float and a portion of the sphere begins to show. (a) What is the height of the portion that shows above the water? What minimum force should you press on the sphere to make it sink completely in the pool? Show your complete work step by step.

Answer 1

part(a)

from newtons second law

In equilibrim Fnet = 0

Fb - W = 0

Fb = boyancy force = Dw*Vimmeresed*g

V = immeresed = pi*r^2*(R- x/3)

W = weight od the sphere = Dsphere*Vsphere*g

Dw*Vimmeresed = Dsphere*Vsphere

pi*r^2*(R- r/3) = (Dsphere/Dw)*Vsphere

pi*r^2*(1.25- r/3)   = (750/1000)*(4/3)pi*1.25^3

r = 1.68 m

height of the ball above water = 2R - r = (2*1.25)-1.68 = 0.82 m <<<==========ANSWER

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part(b)

Fb - W - F = 0

F = Fb - W

F = Dw*Vsphere*g - Dsphere*Vsphere*g

F = (Dw-Dsphere)*Vsphere*g

F = (1000-750)*(4/3)*pi*1.25^3*9.81

F = 20064.5 N <<<<=======ANSWER