A hydraulic lift is used to jack a 780kg car 15cm off the floor. The diameter of

Question

A hydraulic lift is used to jack a 780kg car 15cm off the floor. The diameter of the output piston is 19cm , and the input force is 280N . a. What is the area of the input piston? b. What is the work done in lifting the car 15cm ? c. If the input piston moves 15cm in each stroke, how high does the car move up for each stroke? d. How many strokes are required to jack the car up 15cm ?

Explanation / Answer

(a) Pressure in equals pressure out: P1 = P2 (P = F / A) = 280/(Pie*(19/2)^2 *10^-4) =9875.54 pa F1 / A1 = F2 / A2 --> A1 = (F1 * A2) / F2 (F2 = m*g) A1 = (280*(Pie*(19/2)^2 *10^-4)) / (780*9.81) =10.37 cm^2 (b) Work = force out × distance moved: W = F2 * y = (y = 15cm = 0.15m) W = (780*9.81)*0.15 = 1147.77 J (c) Volume moved by piston 1 equals volume moved by piston 2: V1 = V2 (V = A * h) A1 * h1 = A2 * h2 --> h2 = (A1 * h1) / A2 = ... (h1 = 15cm = 0.15m) h2 = 0.55 cm (d) Number of strokes required equals total height divided by incremental height: n = y / h2 = 15/0.55 = 23 approx :)

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