Midterm 2, Physics 6C, Fall 2014 David Schriver Name Problem 3 A solid cube with

Question

Midterm 2, Physics 6C, Fall 2014 David Schriver Name Problem 3 A solid cube with side length 0.415 m and mass density po floats at equil 0.324 m submerged in salt water (mass density: .-1025 kg/m') with its top a distance d above the surface as shown in the figure. [5 pts] . a) What is mass density po of the cube? [5 pts] b) What is the total pressure acting on the bottom surface of the cube under water? [5 pts] c) What is the total pressure acting on the top surface of the cube above the water? [10/pts] d) If a mass m is now placed on top of the cube, what is the smallest value of m such that the cube will be fully submerged under water?

Explanation / Answer

total volume of block = L^3

volume immersed Vimmerse = (L-d)*L^2

buoyancy force Fb = rho_w*Vimmeresed*g ( upward)

gravitational force Fg = -M*g   ( down ward)

M = mas of the block = rho_0*V*g = rho_0*L^3*g

the block is in equilibrium

Fnet = 0

Fb + Fg = 0

rho_w*Vimmersed*g - rho_0*V*g = 0

rho_w*L^2*(L-d) - rho_0*L^3 = 0

rho_w*(L-d) - rho_0*L = 0

rho_0 = rho_w*(L-d)/L

rho_0 = 1025*(0.415-0.324)/0.415

rho_d = 225 kg/m^3

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(b)

Pbottom = P0 + rho_w*g*(L-d)

P = 10^5+ (1025*9.8*(0.415-0.324))

P = 100914.095Pa

============================

(c)

Ptop = Po = 10^5 Pa

=====================

(d)

volume immeresed = V = L^3

Fb = rho_w*Vimmersed*g

total weight Fg = m*g + M*g

M = rho_0*V*g

Fb = Fg

rho_w*L^3*g = mg + rho_0*L^3*g

m = (rho_w - rho_0)*L^3

m = (1025-225)*0.415^3

m = 57.2 kg

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