The small piston of a hydraulic lift has a radius of Rs = 8.0 cm, and its large piston has a radius of Rl = 40 cm. The lift raises a load of 15,000 N. The force that must be applied to the small piston is:
Hint:
Fl * Al = Fs * As → Fl * π * Rl^2 = Fs * π * Rs^2
Fs = Fl * Rs^2 / Rl^2
a) 65 N;
b) 600 N;
c) 120 N;
d) 24,000 N;