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Unformatted text preview: 1-34 Review Problems 1-85A hydraulic lift is used to lift a weight. The diameter of the piston on which the weight to be placed is to be determined. F2F1D210 cm25 kgWeight 2500 kg Assumptions1 The cylinders of the lift are vertical. 2 There are no leaks. 3 Atmospheric pressure act on both sides, and thus it can be disregarded. Analysis Noting that pressure is force per unit area, the pressure on the smaller piston is determined from kPa23.31kN/m23.31m/skg1000kN14/m)10.()m/skg)(9.8125(4/2222211111==⋅===ππDgmAFPFrom Pascal’s principle, the pressure on the greater piston is equal to that in the smaller piston. Then, the needed diameter is determined from m1.0=→⋅=→===2222222222221m/skg1000kN14/)m/skg)(9.812500(kN/m23.314/DDDgmAFPPππDiscussion Note that large weights can be raised by little effort in hydraulic lift by making use of Pascal’s principle. 1-86A vertical piston-cylinder device contains a gas. Some weights are to be placed on the piston to increase the gas pressure. The local atmospheric pressure and the mass of the weights that will double the pressure of the gas are to be determined. WEIGTHS GAS AssumptionsFriction between the piston and the cylinder is negligible. AnalysisThe gas pressure in the piston-cylinder device initially depends on the local atmospheric pressure and the weight of the piston. Balancing the vertical forces yield kPa95.7==⋅−=−=kN/m66.95m/skg1000kN1)/4m12.()m/skg)(9.815(kPa1002222pistonatmπAgmPPThe force balance when the weights are placed is used to determine the mass of the weights kg115.3=→⋅++=++=weights222weightsweightspistonatmm/skg1000kN1)/4m12.()m/s)(9.81kg5(kPa95.66kPa200)(mmAgmmPPπA large mass is needed to double the pressure. 1-35 1-87An airplane is flying over a city. The local atmospheric pressure in that city is to be determined. AssumptionsThe gravitational acceleration does not change with altitude. PropertiesThe densities of air and mercury are given to be 1.15 kg/m3and 13,600 kg/m3.AnalysisThe local atmospheric pressure is determined from kPa91.8==⋅+=+=kN/m84.91m/skg1000kN1m))(3000m/s)(9.81kg/m(1.15kPa582223planeatmghPPρThe atmospheric pressure may be expressed in mmHg as mmHg688===m1mm1000kPa1Pa1000)m/s)(9.81kg/m(13,600kPa8.9123atmHggPhρ1-88The gravitational acceleration changes with altitude. Accounting for this variation, the weights of a body at different locations are to be determined. Analysis The weight of an 80-kg man at various locations is obtained by substituting the altitude z (values in m) into the relation ⋅×−==−226m/skg1N1)m/s103.32kg)(9.807(80zmgWSea level: (z = 0 m): W = 80×(9.807-3.32x10-6×0) = 80×9.807 = 784.6 NDenver: (z = 1610 m): W = 80×(9.807-3.32x10-6×1610) = 80×9.802 = 784.2 NMt. Ev.: (z = 8848 m): W = 80×(9.807-3.32x10-6×8848) = 80×...
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This note was uploaded on 03/19/2008 for the course MAE 301 taught by Professor Hassan during the Spring '08 term at N.C. State.
- Spring '08