A cubical object is of side L = 1.28ft and weight W = 2360 lb in vacuum. It is s
ID: 2014396 • Letter: A
Question
A cubical object is of side L = 1.28ft and weight W = 2360 lb in vacuum. It is suspended in an open tank of liquid of density p = 2.12 slugs/ft^3 at a depth L/2 to its top by a vertical thin wire that is normal to one face. A) Find the total downward force extended by the liquid and the standard atmosphere on the top of the object. B) Find the total upward force on the bottom of the object. C) Find the tension in the wire. D) Calculate the buoyant force on the object using Archimedes' principle.Explanation / Answer
Let L = 1.28 feet
L = 1.28(0.3048m)
L = .390144m
Area = A = L2 = .152m2
Mass of object = m = 2360lb
m = 2360(0.4535kg)
m = 1070.26kg
Density of liquid = = 2.12slugs/ft3
= 2.12(515.37 kg/m3)
= 1092 kg/m3
a) Pressure at depth htop = L/2 is:
Ptop = Patm + ghtop
Ptop = 1.013 x 105Pa + (1092)(9.8)(.1951)
Ptop = 1.034 x 105Pa
Force on the top = Ptop x Area
1.034 x 105Pa * .152m2
Ftop = .1572 x 105N
b) Pressure at depth hbot = 3L/2 is:
Pbot = Patm + ghbot
Pbot = 1.013 x 105Pa + (1092)(9.8)(.5857)
Pbot = 1.0757 x 105 Pa
Force on the bottom:
Fbot = Pbot x Area
Fbot = 1.0757 x 105 Pa * .152
Fbot = .1635 x 105
c) From Archimede's Principle:
Fbot - Ftop = g(hbot - htop)*A
= gL3
= (1092)(9.8)(.390144)3
= 635.5 N
Tension
T = mg - (Fbot - Ftop)
T = (1070.26)(9.8) - (635.5N)
T = 9853.04 N
d) Buoyancy Force:
Fbuo = Fbot - Ftop
Fbuo = 635.5N
Hope this helps!