In a hydraulic lift the radii of the pistons are 3.20 cm and 16.0 cm. An SUV of mass 1850kg 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 15.0 cm, how far is the car lifted? (c) Find the mechanical advantage of the lift?
The ratio of the radii of the pistons is 5, so ratio of area is 5^2 25, 25 is the mechanical advantage. So force on small piston 1850kg/25 74kg Car is lifted 15cm/25 0.6cm
Pressure 1 Pressure 2 F/A1 F/A2 1850 kg / ( pi x 1/ 4 x 0.16^2 ) F / ( pi x 1/ 4 x 0.032^2 ) Force needed is 74 kg using F1x pi x d x L F2x pi x d x L2 1850 x pi x 0.16 x L 74 x pi x 0.032 x .15m L 1.2 x 10 ^ -3 m
