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Unformatted text preview: Problem Set 4: Problem 5.
Problem: It takes a time to raise a car of mass M and the supporting hydraulic car-lift platform of mass 1 M to a height H. The efficiency of the motor driving the lift is 80%. Let g denote gravitational 6 acceleration and note that 1 hp = 550 ftlb/sec. (a) Find the average output power, Pp , delivered by the hydraulic pump to lift the car and platform. (b) Find the average electric power, Pe , required. (c) Compute Pp and Pe in hp for M g = 3000 lb, H = 6.4 ft, and = 17 sec. Solution: (a) The average output power is Pp = F V = (Wcar + WLif t ) V = (Mcar + MLif t ) gV where W and M denote weight and mass, respectively. We are given Mcar = M and MLif t = Hence, 7 Pp = M gV 6 Finally, V = H/ , wherefore 7 M gH Pp = 6 (b) Now, Pe = Pp / and = 4 , which tells us that 5 Pe = (c) For the given values, we have Pp = 7 (3000 lb)(6.4 ft) ft lb = 1318 = 2.4 hp 6 17 sec sec Pp = 5 Pp = 3.0 hp 4 7 M gH 35 M gH = 6 24 1 6M. ...
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This note was uploaded on 10/12/2009 for the course AME 301 at USC.