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

Answer:

58.32 N

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}$

  = $\frac{22}{7}$ x $(0.02)^{2}$

  = $\frac{22}{7}$ 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}$

    = $\frac{22}{7}$ x $(0.0075)^{2}$

    = $\frac{22}{7}$ 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;

$\frac{F_{1} }{A_{1} }$ = $\frac{F_{2} }{A_{2} }$

$\frac{F_{1} }{0.0013}$ = $\frac{8.075}{0.00018}$

$F_{1}$ = $\frac{0.0013 *8.075}{0.00018}$

    = 58.32

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