Question

g A hydraulic press has a safety feature which consists of a hydraulic cylinder with a piston at one end and a safety valve at the other. The cylinder has a radius of 0.0200 m and the safety valve is simply a 0.00750-m radius circular opening at one end, sealed with a disk. The disk is held in place by a spring with a spring constant of 950 N/m that has been compressed 0.0085 m from its natural length. Determine the magnitude of the minimum force that must be exerted on the piston in order to open the safety valve.

Answer 1

58.32 N

Step-by-step explanation:

Area of a circle =
`\pi`
`r^(2)`

where r is the radius of the circle.

The cylinder has a radius of 0.02 m, its area is;

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

=
`(22)/(7)` x
`(0.02)^(2)`

=
`(22)/(7)` x 0.0004

= 1.2571 x
`10^(-3)`

Area of the cylinder is 0.0013
`m^(2)`.

The safety valve has a radius of 0.0075 m, its area is;

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

=
`(22)/(7)` x
`(0.0075)^(2)`

=
`(22)/(7)` x 5.625 x
`10^(-5)`

= 1.7679 x
`10^(-4)`

Area of the valve is 0.00018
`m^(2)`.

From Hooke's law, the force on the safety valve can be determined by;

F = ke

`F_(2)` = 950 x 0.0085

= 8.075 N

Minimum force,
`F_(1)`, required can be determined by;

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

`(F_(1) )/(0.0013)` =
`(8.075)/(0.00018)`

`F_(1)` =
`(0.0013 *8.075)/(0.00018)`

= 58.32

The minimum force that must be exerted on the piston is 58.32 N.