Question

A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a diameter of 0.15 m, approximately how much mass can be lifted by the jack

Answer 1

1400.38N

Step-by-step explanation:

Step one

Given data

P1= 250N

D1= 0.02m

A1= πD1^2/4

substitute

`A1= 3.142*0.02^2/4\\\\A1=3.142*10^-4`

D2= 0.15m

A1= πD2^2/4

`A2= 3.142*0.15^2/4\\\\A2=1.76*10^-3`

Required

The load P2

Step two:

Applying the hydraulic expression for a non-compressible fluid

we know that

Pressure= force/are

P1/A1=P2/A2

`250/3.142*10^-4= P2/1.76*10^-3`

cross multiply we have

P2= 1.76*10^-3*250/3.142*10^-4

P2=0.44/3.142*10^-4

P2=1400.38N