A hydraulic lift is used to jack a 970-kg car 12 cm off the floor. The diameter

Question

A hydraulic lift is used to jack a 970-kg car 12 cm off the floor. The diameter of the output piston is 18 cm, and the input force is 250N. a) What is the area of the input piston? b)What is the work done in lifting the car 12 cm? c)If the input piston moves 13 cm in each stroke, how high does the car move up for each stroke? d)How many strokes are required to jack the car up 12 cm? e)Show that energy is conserved. ( Please show work, thank you!!!)

Explanation / Answer

1. The output force = m*g = 870kg*9.8m/s^2 = 8530N; the input force = 250N..so the ratio of forces is 8530N/250N = 34.1.. a)S o the area of the input = area of the output/34.1= (pi*9^2)/34.1 = 7.46cm^2 b)W = m*g*y = 870kg*9.8m/s^2*0.11m = 938J c)The volume of fluid moved is constant so A*h)input = A*h)output.. If h input = 13cm, then the h output = Ain/Aout*hin = hin/34.1 = 13cm/34.1 = 0.381cm d) Now 11cm = n*0.381cm => n = 11/0.381 = 28.8 = 29 strokes e)W input = 29*0.381x10^-2*250N = 27.5J 2.The vol of water removed = 2800m^2*8.00m = 22400m^3...Now mass = denisty *Vol = 1000kg/m^3*22400m^3 = 2.24x10^7kg 3.We have F/A)plunger = F/A) needle..so F needle = F/A)plunger*A needle = 2.2N/(pi*0.55cm)^2*pi*( 0.0125cm)^2 = 1.14x10^-3N b) The pressure in the vein is 18mmHg ..converting to Pa we get 18mm*1.013x10^5Pa/760mm = 2400N/m^2...To achieve this pressure in the needle we need F/A = 2400Pa.. So F = A*2400 = pi*(0.125x10^-3)^2*2400 = 1.18x10^-4N From the problem we know the ratio of plunger to needle = 2.2/1.14x10^-3 = F/1.18x10^-4 Therefore F = 2.2*1.18x10^-4/1.14x10^-3 = 0.227N 4. We know that rho*V)rock = 9.88 and density of water*V = 9.88-5.68 So equating V we get 9.88/rho = (9.88-5.68)/1000.. so rho = 9.88*1000/(9.88-5.68) = 2350kg/m^3

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