451_PartUniversity Physics Solution

451_PartUniversity Physics Solution - Fluid Mechanics...

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Fluid Mechanics 14-15 14.61. IDENTIFY: Apply Newton&s 2nd law to the car. The buoyancy force is given by Archimedes& principle. (a) SET UP: The free-body diagram for the floating car is given in Figure 14.61. sub ( V is the volume that is submerged.) EXECUTE: y y Fm a = 0 Bm g = ws u b 0 Vgm g ρ = Figure 14.61 33 sub w / (900 kg)/(1000 kg/m ) 0.900 m Vm == = sub obj / (0.900 m )/(3.0 m ) 0.30 30% VV = EVALUATE: The average density of the car is (900 kg)/(3.0 m ) 300 kg/m . = car water /0 . 3 0 ; ρρ = this equals sub obj /. (b) SET UP: When the car starts to sink it is fully submerged and the buoyant force is equal to the weight of the car plus the water that is inside it. EXECUTE: When the car is full submerged sub , = the volume of the car and 2 4 water (1000 kg/m )(3.0 m )(9.80 m/s ) 2.94 10 N BV g = × The weight of the car is 2 (900 kg)(9.80 m/s ) 8820 N. mg Thus the weight of the water in the car when it sinks is the buoyant force minus the weight of the car itself: 42 3 water (2.94 10 N 8820 N)/(9.80 m/s ) 2.10 10 kg m = × And 3 water water water / (2.10 10 kg)/(1000 kg/m ) 2.10 m × = The fraction this is of the total interior volume is (2.10 m )/(3.00 m ) 0.70 70% EVALUATE: The average density of the car plus the water inside it is (900 kg 2100 kg)/(3.0 m ) 1000 kg/m , += so car water = when the car starts to sink. 14.62. IDENTIFY: For a floating object, the buoyant force equals the weight of the object. fluid submerged B Vg = . SET UP: Water has density 3 1.00 g/cm = . EXECUTE: (a) The volume displaced must be that which has the same weight and mass as the ice, 3 3 9.70 gm 9.70 cm 1.00 gm/cm = . (b) No; when melted, the cube produces the same volume of water as was displaced by the floating cube, and the water level does not change. (c) 3 3 9.70 gm 9.24 cm 1.05 gm/cm = (d) The melted water takes up more volume than the salt water displaced, and so 3 0.46 cm flows over. EVALUATE: The volume of water from the melted cube is less than the volume of the ice cube, but the cube floats with only part of its volume submerged. 14.63. IDENTIFY: For a floating object the buoyant force equals the weight of the object. The buoyant force when the wood sinks is water tot B = , where tot V is the volume of the wood plus the volume of the lead. / mV = . SET UP: The density of lead is 11.3 10 kg/m × . EXECUTE: 3 wood (0.600 m)(0.250 m)(0.080 m) 0.0120 m V . wood wood wood (600 kg/m )(0.0120 m ) 7.20 kg = . wood lead () B mm g =+ . Using water tot B = and tot wood lead V gives water wood lead wood lead ( ) g g +=+ . lead lead lead = then gives water wood water lead wood lead lead m V + . 43 water wood wood lead 3 lead water (1000 kg/m )(0.0120 m ) 7.20 kg 4.66 10 m 1000 kg/m V −− = × −× . lead lead lead 5.27 kg . EVALUATE: The volume of the lead is only 3.9% of the volume of the wood. If the contribution of the volume of the lead to B F is neglected, the calculation is simplified: water wood wood lead Vgm mg and lead 4.8 kg m = . The result of this calculation is in error by about 9%.
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14-16 Chapter 14 14.64. IDENTIFY: The fraction f of the volume that floats above the fluid is fluid 1, f ρ =− where ρ is the average density of the hydrometer (see Problem 14.29). This gives fluid 1 .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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451_PartUniversity Physics Solution - Fluid Mechanics...

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