In a hydraulic lift the radii of the pistons are 1.75cm and 10.5cm. A car weighing 18.5kN is to be lifted by the force of the large piston.
a.) what force must be applied to the small piston?
b.) when the small piston is pushed 1.00cm, how far is the car lifted?
c.) find the mechanical advantage of the lift (this is the ratio of the large force to the small force)
The Attempt at a Solution
a.) F= (r/R)^2 (F)
= (1.75 cm/10.5 cm)^2 (18.5 kN)
= .514 or 514 N
b.) Win(f)= Wout(F)= fh=FH=H= (f/F)h= (pi)r^2/(piR^2=(r/R)^2
= (1m/R)^2 = (100cm/10.5cm)^2
c.) MA= F/f= 18.5kN/1.75cm
I have no clue if I am doing this right.. can anyone help?