Question

A hydraulic lift is designed to lift cars up to 2000 kg in mass.If the lift under the car is 1.0 m by 1.2m and the area of the input piston is 10 cm by 10 cm, how much force must be exerted on the input piston to lift a 2000-kg car? A) 136 N B 24x 10^4 N C) 160 N D) 196 N E) 17N

Answer 1

C) 160 N

Step-by-step explanation:

`m` = mass of the car = 2000 kg

Weight of the car is given as

`W = mg \\W = (2000) (9.8)\\W = 19600 N`

`F_(i)` = Force exerted on the input piston

`A_(i)` = Area of input piston = 10 cm x 10 cm = 0.1 m x 0.1 m = 0.01 m²

`A_(lift)` = Area of lift = 1 m x 1.2 m = 1.2 m²

Using Pascal's law

`(F_(i))/(A_(i)) = (W)/(A_(lift))\\(F_(i))/(0.01) = (19600)/(1.2)\\F_(i) = 160 N`