Buoyancy and Archimedes' Principle: Sample Problems and Solutions, Slides of Fluid Mechanics

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ENGR. BON RYAN ANIBAN

Free surface h x p 1 p 2

p 1 = γ h

p 2 = γ (h + x)

F 1 F 2

F 1 = γ hA

F 2 = γ (h + x)A

A

ARCHIMEDES PRINCIPLE

‘’ Any body immersed in a fluid is acted

upon by an upward fore (buoyant force)

equal to the weight of displaced fluid’’

Net upward force, Bouyant force

Fb = F 2 – F 1

= γ (h + x)A - γ hA

= γ A(h + x - h)

= γ Ax

But, Ax = volume of displaced

Bouyant Force

Fb = γ liquidVd

Fnet

By Equilibrium

Wobject

Fb = Wobject

A cube of timber 1.25 ft on each side floats in water. The specific gravity of timber is 0.60. Find the submerged depth of the cube. Solution Free surface

W

BF

1.25 ft

ΣFy = 0; BF = W γ liquidVd = γ timberVtimber (62.4 lb/ft^3 )(1.25 ft ×1.25 ft ×d) = (62.4 lb/ft^3 )(0.60)(1.25 ft)^3 d = 0.75 ft

1.25 ft

d

A block of wood 0.2 m thick is floating in sea water. The specific gravity of wood is 0.65 while that of sea water is 1.03. Find the minimum area of the block which will support a man weighing 80 kg. Solution Free surface

Wwood

BF

0.2 m

Wman

ΣFy = 0; BF = Wman + Wwood γ waterVd = (^) W man

• γ woodVwood (9810)(1.03)(0.2)(A) =^ 80(9.81)+ 9810(0.65) (0.2)(A) A = 1.053 m^2

A piece of lead (sp. Gr. 11.3) is tied to a 130 cc of cork whose specific gravity is 0.25. They float just submerged in water. What is the weight of the lead? Solution Free surface

BFC

BFL

WC

WL

ΣFy = 0; BFC + BFL = WC + WL γ wVdC +^ γ wVdL = γ CVC (9.81)(0.00013) + 9.81 = 9.81 (0.25)(0.00013) W= 1.049 N (VdL) VL= 9.466 x 10 − 6 m^3 W= 9810(11.3)(9.466 x 10 − 6 )

• γ LVL
• 9.81 (11.3) (VL)