Question

A hydraulic car jack needs to be designed so it can lift a 2903.57 lb car assuming that a person can exert a force of 24.41 lbs. If the piston the person is pushing on had a radius of 3.26 cm, what should the diameter of the piston be that is used to raise the car?

Answer 1

Diameter of the piston would be 0.71 m (71.1 cm)

Step-by-step explanation:

From the principle of pressure;

`(F_(1) )/(A_(1) )` =
`(F_(2) )/(A_(2) )`

Let
`F_(1)` = 2903.57 lb,
`F_(2)` = 24.41 lbs,
`r_(2)` = 3.26 cm = 0.0326 m.

`A_(2)` =
`\pi r^(2)`

=
`(22)/(7)` x
`(0.0326)^(2)`

= 0.00334
`m^(2)`

So that:

`(2903.57)/(A_(1) )` =
`(24.41)/(0.00334)`

`A_(1)` =
`(2903.57*0.00334)/(24.41)`

= 0.3973

`A_(1)` = 0.4
`m^(2)`

The radius of the piston can be determined by:

`A_(1)` =
`\pi r^(2)`

0.3973 =
`(22)/(7)` x
`r^(2)`

`r^(2)` =
`(0.3973*7)/(22)`

= 0.1264

r =
`√(0.1264)`

= 0.3555

r = 0.36 m

Diameter of the piston = 2 x r

= 2 x 0.3555

= 0.711

Diameter of the piston would be 0.71 m (71.1 cm).