GOAL Apply Archimedes\' principle to an object floating in a fluid having two la

Question

GOAL Apply Archimedes' principle to an object floating in a fluid having two layers with different densities BugT Bwater water PROBLEM A 1.00 x 103-kg cube of aluminum is placed in a tank. Water is then added to the tank until half the cube is immersed. (a) What is the normal force on the cube? (See Figure (a) (b) Mercury is now slowly poured into the tank until the normal force on the cube goes to zero. (See Figure (b).) How deep is the layer of mercury? Assume a very thin layer of fluid is underneath the block in both parts of the figure, due to imperfections between the surfaces in contact. MAIg MAIg STRATEGY Both parts of this problem involve applications of Newton's second law for a body in equilibrium, together with the concept of a buoyant force. In part (a) the normal, gravitational, and buoyant forces of water act on the cube. In part (b) there is an additional buoyant force of mercury, while the normal force goes to zero. Using VHgAh, solve for the height of mercury, h

Explanation / Answer

m =1.2*10^3 kg

(A) APply nweton second law F =0

n +Fb = mcube*g ..(1)

Fb = mdw*g = dw*Vdw*g = dw*(Vcube/2)*g ..(2)

dAL = mcube/Vcube ... (3)

from (1) ,(2) and (3)

n = mcube*g -Fb

n = mcube*g(1- dw/2dAL) = 1.2*10^3(1-(1000/(2*2.7*1000)))

n = 977.8 N

(b) Apply newton second law F =0

There is buyoyancy from HG and water

Fb.Hg +Fb.w = mcube*g

dHg*A*hHg*g +dw*A*hw*g = mcube*g = dAL*A*h*g ..(4)

hw = h/2 ..(5)

and from (3)

Vcube = h^3 = (mcube/dAL)

h = (1.2*10^3/(2.7*10^3))^1/3 = 0.763 m ..(6)

from(4),(5) and (6)

dHg*hHg +dW*h/2 = dAL*h

hHg = dAL*h - dWh/2/dHg

Hhg = 0.763((2.7*10^3) -(1.2*10^3)(0.5))/(13.6*10^3)

= 0.118 m

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