a. In a hydraulic lift, the radii of the pistons are 2.63 cm and 10.8 cm. A car

Question

a. In a hydraulic lift, the radii of the pistons are 2.63 cm and 10.8 cm. A car weighing W = 10.8 kN is to be lifted by the force of the large piston. What force Fa must be applied to the small piston? b. When the small piston is pushed in by 12.1 cm, how far is the car lifted? c. Find the mechanical advantage of the lift, which is the ratio W/Fa.

Explanation / Answer

help: BY P = F(a)/A(a) = F(b)/A(b) [here a & b representing small & large piston] =>F(a) = F(b) x A(a)/A(b) =>F(a) = 10.8 x 1000 x [{pi x (2.63 x 10^-2)^2}/{pi x (10.40 x 10^-2)^2}] =>F(a) = 10800 x [6.92/108.16] =>F(a) = 690.98 N (b) By V = A(a) x L(a) = A(b) x L(b) =>L(b) = A(a) x L(a)/A(b) =>L(b) = [6.92/108.16] x 9.8 =>L(b) = 0.6269 cm =>L(b) = 6.27mm (c) M.A. = W/F(a) = 10800/690.98 = 15.63

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