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?

Respuesta :

Answer:

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

Explanation:

From the principle of pressure;

[tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]

Let [tex]F_{1}[/tex] = 2903.57 lb, [tex]F_{2}[/tex] = 24.41 lbs, [tex]r_{2}[/tex] = 3.26 cm = 0.0326 m.

[tex]A_{2}[/tex] = [tex]\pi r^{2}[/tex]

    = [tex]\frac{22}{7}[/tex] x [tex](0.0326)^{2}[/tex]

    = 0.00334 [tex]m^{2}[/tex]

So that:

[tex]\frac{2903.57}{A_{1} }[/tex] = [tex]\frac{24.41}{0.00334}[/tex]

[tex]A_{1}[/tex] = [tex]\frac{2903.57*0.00334}{24.41}[/tex]

    = 0.3973

[tex]A_{1}[/tex] = 0.4 [tex]m^{2}[/tex]

The radius of the piston can be determined by:

[tex]A_{1}[/tex] = [tex]\pi r^{2}[/tex]

0.3973 = [tex]\frac{22}{7}[/tex] x [tex]r^{2}[/tex]

[tex]r^{2}[/tex] = [tex]\frac{0.3973*7}{22}[/tex]

   = 0.1264

r = [tex]\sqrt{0.1264}[/tex]

 = 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).