Question

A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. What force must be applied to the smaller piston to obtain a force of 1600 N at the larger piston

Answer 1

The force is
`F_1 = 400.8 \ N`

Step-by-step explanation:

From the question we are told that

The first diameter is
`d_1 = 4.0 \ cm = 0.04 \ m`

The second diameter is
`d_2 = 8.0 \ cm = 0.08 \ m`

Generally the first area is

`A_1 = \pi * (d^2_1 )/(4)`

=>
`A_1 = 3.142 * (0.04^2)/(4)`

=>
`A_1 = 0.00126 \ m^2`

The second area is

`A_2 = \pi * (d^2_2 )/(4)`

`A_2 = 3.142 * (0.08^2)/(4)`

`A_2 = 0.00503 \ m^2`

For a hydraulic press the pressure at both end must be equal .

Generally pressure is mathematically represented as

`P = (F)/(A)`

=>

`(F_1)/(A_1 ) = (F_2)/(A_2 )`

=>
`F_1 = (1600)/(0.00503) * 0.00126`

=>
`F_1 = 400.8 \ N`