h, F1 F = P,A = h,pgA FB = F2 - F, F2 F2 = P2A = h2pgA %3D Fluid of density p Figure 14.20 Pressure due to the weight of a fluid increases with depth because p = hpg . This change in pressure and associated upward force on the bottom of the cylinder are greater than the downward force on the top of the cylinder. The differences in the force results in the buoyant force FB. (Horizontal forces cancel.)

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Chapter14: Fluid Mechanics
Section: Chapter Questions
Problem 50P: The tip of a nail exerts tremendous pressure when hit by a hammer because it exerts a large force...
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Referring to Figure 14.20, prove that the buoyant force on the cylinder is equal to the weight of the fluid displaced (Archimedes’ principle). You may assume that the buoyant force is F2 − F1 and that the ends of the cylinder have equal areas A . Note that the volume of the cylinder (and that of the fluid it displaces) equals (h2 − h1)A .

h,
F1
F = P,A = h,pgA
FB = F2 - F,
F2
F2 = P2A = h2pgA
%3D
Fluid of
density p
Figure 14.20 Pressure due to the weight of a fluid increases
with depth because p = hpg . This change in pressure and
associated upward force on the bottom of the cylinder are
greater than the downward force on the top of the cylinder. The
differences in the force results in the buoyant force FB.
(Horizontal forces cancel.)
Transcribed Image Text:h, F1 F = P,A = h,pgA FB = F2 - F, F2 F2 = P2A = h2pgA %3D Fluid of density p Figure 14.20 Pressure due to the weight of a fluid increases with depth because p = hpg . This change in pressure and associated upward force on the bottom of the cylinder are greater than the downward force on the top of the cylinder. The differences in the force results in the buoyant force FB. (Horizontal forces cancel.)
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