Question
In the figure below, a cube of edge length L = 0.550 m and mass 440 kg is suspended by a rope in an open tank of liquid of density 1030 kg/m3.
(a) Find Ftop, the magnitude of the total downward force on the top of the cube from the liquid and the atmosphere, assuming atmospheric pressure is 1.00 atm.
(b) Find Fbottom, the magnitude of the total upward force on the bottom of the cube.
(c) Find T, the tension in the rope.
(d) Calculate Fbu, the magnitude of the buoyant force on the cube using Archimedes' principle.
Explanation / Answer
Data: Mass, m = 440 kg Length, L = 0.550 m Density, = 1030 kg/m^3 Atm pressure, Patm = 1 atm = 1.01 x 10^5 Pa Solution: (a) htop = L/2 = 0.550 / 2 = Ptop = Patm + g htop = 1.01 x 10^5 + ( 1030 * 9.8 * 0.275 ) = 1.038 x 10^5 Pa Area of top surface, A = L^2 = 0.550^2 = 0.3025 m^2 Ftop = Ptop * A = 1.038 x 10^5 * 0.3025 = 3.14 x 10^4 N Ans: Ftop = 3.14 x 10^4 N (b) hbot = 3L/2 = 1.5 * 0.550 = 0.825 m Pbot = Patm + g hbot = 1.01 x 10^5 + ( 1030 * 9.8 * 0.825 ) = 1.09 x 10^5 Pa Fbot = Pbot * A = 1.09 x 10^5 * 0.3025 = 3.31 x 10^4 N Ans: Fbot = 3.31 x 10^4 N (c) Fbot - Ftop = 1.7 x 10^3 N Tension, T = mg - ( Fbot - Top ) = (440 * 9.8)- 1.7 x 10^3 = 2.612 x 10^3 N Ans: Tension, T = 2.612 x 10^3 N (d) Buoyant force, Fbot - Ftop = 1.7 x 10^3 N Ans: Buoyant force, Fb = 1.7 x 10^3 N