p0378

p0378 - is gravitational acceleration So the volume of the...

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434 CHAPTER 3. EFFECTS OF GRAVITY ON PRESSURE 3.78 Chapter 3, Problem 78 Problem: In air, an irregularly-shaped metal object weighs 5 tons. When it is sub- merged in a tank filled with water at 50 o F, a force of 4 tons is required to lift it from the bottom of the tank. What is the volume of the object? What is its density? Solution: If the object’s weight is W and the force to lift it when submerged is F ,then W = F + F buoy = F + ρ gV where V is the object’s volume, ρ is the density of water and g
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Unformatted text preview: is gravitational acceleration. So, the volume of the irregularly-shaped metal object is V = W − F ρ g We are given W = 10000 lbs and F = 8000 lbs. Also for 50 o F, the density of water is 1.94 slug/ft 3 . Thus, V = 10000 lb − 8000 lb p 1 . 94 slug / ft 3 Q p 32 . 174 ft / sec 2 Q = 32 . 0 ft 3 Also, denoting the object’s density by ρ c , W = ρ c gV = ⇒ ρ c = W gV = 10000 lb p 32 . 174 ft / sec 2 Q p 32 . 0 ft 3 Q = 9 . 7 slug ft 3...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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