In a hydraulic lift, the radii of the pistons are 2.50 cm and 9.4 cm. A car weighing W 8.6 kN is to be lifted by the force of the large piston. (a) What force Fa must be applied to the small piston?(b) When the small piston is pushed in by 8.6 cm, how far is the car lifted?(c) Find the mechanical advantage of the lift, which is the ratio W / Fa.
Ill be shocked if you get this answer from yahoo. Or should i say the correct answer.
EDIT: It is the Area of the pistons: a) Fa Ari/Ar2 x F load pi x (r1^2) / pi x (r2^2) x 8.6 kN (2.5^2)pi / (19.6 cm^2 / 283.5 cm^2 x 8.6 kN .069 zx 8.8 kN .594 kN or 594 Newtons. EDIT: Not sure of (b)--is it distance x radius d x r, or Area r of piston x d area x dist (that is, mech. advantage/distance)? b) r1 d1 r2d2 so, d2- r1d1/r2 2., cm x 8.6 cm /9.4 cm 2.128 cm. c) W/ fa 8.6 kN/.594 kN 14.48 : 1 .
