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451_PartUniversity Physics Solution

451_PartUniversity Physics Solution - Fluid Mechanics 14.61...

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Fluid Mechanics 14-15 14.61. I DENTIFY : Apply Newton°s 2nd law to the car. The buoyancy force is given by Archimedes° principle. (a) S ET U P : The free-body diagram for the floating car is given in Figure 14.61. sub ( V is the volume that is submerged.) E XECUTE : y y F ma = 0 B mg = w sub 0 V g mg ρ = Figure 14.61 3 3 sub w / (900 kg)/(1000 kg/m ) 0.900 m V m ρ = = = 3 3 sub obj / (0.900 m )/(3.0 m ) 0.30 30% V V = = = E VALUATE : The average density of the car is 3 3 (900 kg)/(3.0 m ) 300 kg/m . = car water / 0.30; ρ ρ = this equals sub obj / . V V (b) S ET U P : 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. E XECUTE : When the car is full submerged sub , V V = the volume of the car and 3 3 2 4 water (1000 kg/m )(3.0 m )(9.80 m/s ) 2.94 10 N B Vg ρ = = = × 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: 4 2 3 water (2.94 10 N 8820 N)/(9.80 m/s ) 2.10 10 kg m = × = × And 3 3 3 water water water / (2.10 10 kg)/(1000 kg/m ) 2.10 m V m ρ = = × = The fraction this is of the total interior volume is 3 3 (2.10 m )/(3.00 m ) 0.70 70% = = E VALUATE : The average density of the car plus the water inside it is 3 3 (900 kg 2100 kg)/(3.0 m ) 1000 kg/m , + = so car water ρ ρ = when the car starts to sink. 14.62. I DENTIFY : For a floating object, the buoyant force equals the weight of the object. fluid submerged B V g ρ = . S ET U P : Water has density 3 1.00 g/cm ρ = . E XECUTE : (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. E VALUATE : 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. I DENTIFY : For a floating object the buoyant force equals the weight of the object. The buoyant force when the wood sinks is water tot B V g ρ = , where tot V is the volume of the wood plus the volume of the lead. / m V ρ = . S ET U P : The density of lead is 3 3 11.3 10 kg/m × . E XECUTE : 3 wood (0.600 m)(0.250 m)(0.080 m) 0.0120 m V = = . 3 3 wood wood wood (600 kg/m )(0.0120 m ) 7.20 kg m V ρ = = = . wood lead ( ) B m m g = + . Using water tot B V g ρ = and tot wood lead V V V = + gives water wood lead wood lead ( ) ( ) V V g m m g ρ + = + . lead lead lead m V ρ = then gives water wood water lead wood lead lead V V m V ρ ρ ρ + = + . 3 3 4 3 water wood wood lead 3 3 3 lead water (1000 kg/m )(0.0120 m ) 7.20 kg 4.66 10 m 11.3 10 kg/m 1000 kg/m V m V ρ ρ ρ = = = × × . lead lead lead 5.27 kg m V ρ = = . E VALUATE : 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 ( ) V g m m g ρ = + 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. I DENTIFY : The fraction f of the volume that floats above the fluid is fluid 1 , f ρ ρ = where ρ is the average density of the hydrometer (see Problem 14.29). This gives
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