a cube of edge length L = 0.600 m and mass 450 kg is suspended by a rope in an o

Question

a cube of edge length L = 0.600 m and mass 450
kg is suspended by a rope in an open tank of liquid of density
1030 kg/m^3
. Find (a) 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)
the magnitude of the total upward
force on the bottom of the cube, and
(c) the tension in the rope. (d)
Calculate the magnitude of the
buoyant force on the cube using
Archimedes’ principle. What relation
exists among all these quantities?

Explanation / Answer

Data: Mass, m = 450 kg Length, L = 0.600 m Density, = 1030 kg/m^3 Atm pressure, Patm = 1 atm = 1.01 x 10^5 Pa Solution: (a) htop = L/2          = 0.600 / 2          = 0.300
Ptop = Patm + g htop          = 1.01 x 10^5 + ( 1030 * 9.8 * 0.300 )          = 1.040 x 10^5 Pa Area of top surface, A = L^2                                        = 0.600^2                                        = 0.360 m^2 Ftop = Ptop * A          = 1.040 x 10^5 * 0.360          = 3.744 x 10^4 N Ans: Ftop = 3.744 x 10^4 N (b) hbot = 3L/2          = 1.5 * 0.600          = 0.900 m Pbot = Patm + g hbot          = 1.01 x 10^5 + ( 1030 * 9.8 * 0.90 )          = 1.100 x 10^5 Pa Fbot = Pbot * A          = 1.100 x 10^5 * 0.360          = 3.963 x 10^4 N Ans: Fbot = 3.963 x 10^4 N (c) Fbot - Ftop = 2.190x 10^3 N Tension, T = mg - ( Fbot - Top )                    = (450 * 9.8)- 2.190 x 10^3                    = 2.219 x 10^3 N Ans: Tension, T = 2.219 x 10^3 N (d) Buoyant force, Fbot - Ftop = 2.190 x 10^3 N Ans: Buoyant force, Fb = 2.190 x 10^3 N

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