Question

A hydraulic system is used to lift a 2100 kg (20,600 N) vehicle in a auto repair shop which is sitting on a piston with a cross-sectional area of 0.56 cm2. What is the minimum force that must be applied to the attached piston of area 0.035 cm2 in order to lift the vehicle?

Answer 1

1287.5 N

Step-by-step explanation:

We are given that

Mass of vehicle ,m=2100 kg

Weight of vehicle ,F=20,600 N

Area,A=
`0.56 cm^2=0.56* 10^(-4) m^2`

`1 cm^2=10^(-4) m^2`

We have to find the minimum force that must be applied to the attached piston of area
`0.035 cm^2` in order to lift the vehicle.

`A'=0.035cm^2=0.035* 10^(-4) m^2`

Apply the pascal's law

`(F)/(A)=(F')/(A')`

`(20600)/(0.56* 10^(-4))=(F')/(0.035* 10^(-4))`

`F'=(20600)/(0.56* 10^(-4))* 0.035* 10^(-4)=1287.5 N`