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Unformatted text preview: Bautista, Aldo – Homework 11 – Due: Nov 15 2005, 4:00 am – Inst: Maxim Tsoi 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The small piston of a hydraulic lift has a crosssectional area of 9 . 5 cm 2 and the large piston has an area of 60 cm 2 , as in the figure below. F 9 . 5 cm 2 area 60 cm 2 What force F must be applied to the small piston to raise a load of 25 kN? Correct answer: 3958 . 33 N. Explanation: Let : A 1 = 9 . 5 cm 2 , A 2 = 60 cm 2 , W = 25 kN , and F = F . According to Pascal’s law, the pressure ex erted on A 1 must be equal to the one exerted on A 2 . The pressure P 1 = F A 1 must be equal to the pressure P 2 = W A 2 due to the load. F A 1 = W A 2 , so F = A 1 A 2 W = (9 . 5 cm 2 ) (60 cm 2 ) (25000 N) = 3958 . 33 N . 002 (part 1 of 2) 10 points A block of volume 0 . 58 m 3 floats with a frac tion 0 . 68 of its volume submerged in a liquid of density 1320 kg / m 3 , as shown in the figure below. The acceleration of gravity is 9 . 8 m / s 2 . V B liquid Find the magnitude of the buoyant force on the block. Correct answer: 5101 . 96 N. Explanation: Let : V L = volume of displace liquid V B = 0 . 58 m 3 , ρ L = 1320 kg / m 3 , and V L V B = 0 . 68 . Basic Concepts: Archimedes’ Principle: Buoyant Force on Object=Weight of Fluid Displaced by Object. Solution: From Archimedes’ Principle, we have ρ L V L g = ρ B V B g , so V L V B = ρ B ρ L = 0 . 68 F Buoyant,Part 1 = Weight of displaced liquid = ρ L µ ρ B ρ L ¶ V B g = (1320 kg / m 3 )(0 . 68) × (0 . 58 m 3 )(9 . 8 m / s 2 ) = 5101 . 96 N . 003 (part 2 of 2) 10 points A block of density ρ B 2 and volume V B 2 is placed on top of the first block. Bautista, Aldo – Homework 11 – Due: Nov 15 2005, 4:00 am – Inst: Maxim Tsoi 2 V B 2 V B 1 liquid If the volume V B 2 is given, find V B 2 such that the two blocks are just submerged, as shown in figure above. 1. V B 2 = µ ρ L ρ B 2 ρ B 1 ρ L ¶ V B 1 2. V B 2 = µ ρ L ρ B 1 ρ L ρ B 2 ¶ V B 1 3. V B 2 = µ ρ L ρ B 1 ρ B 2 ρ L ¶ V B 1 correct 4. V B 2 = µ ρ L + ρ B 2 ρ L + ρ B 1 ¶ V B 1 5. V B 2 = µ ρ L ρ B 2 ρ L ρ B 1 ¶ V B 1 6. V B 2 = µ ρ L + ρ B 1 ρ L + ρ B 2 ¶ V B 1 Explanation: In equilibrium X F y = 0, which means: F Buoyant,Part 2 = m B 1 g + m B 2 g = ρ B 1 V B 1 g + ρ B 2 V B 2 g , where ρ B 1 is the density of the first block. But we also know that: F Buoyant,Part 2 = Weight of displaced liquid = ρ L ( V B 1 + V B 2 ) g . Therefore ρ L ( V B 1 + V B 2 ) g = ρ B 1 V B 1 g + ρ B 2 V B 2 g . (1) Solving for V B 2 : V B 2 = µ ρ L ρ B 1 ρ B 2 ρ L ¶ V B 1 . (2) We can find ρ B 1 using the equilibrium condi tion for Part 1: Weight of block 1 = F Buoyant,Part 1 ρ B 1 V B 1 g = ρ L X V B 1 g , which gives us ρ B 1 = X ρ L . Substituting this into Eq. 2: V B 2 = µ ρ L X ρ L ρ B 2 ρ L ¶ V...
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
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