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

Answer:

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

Explanation:

From the principle of pressure;

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

Let $F_{1}$ = 2903.57 lb, $F_{2}$ = 24.41 lbs, $r_{2}$ = 3.26 cm = 0.0326 m.

$A_{2}$ = $\pi r^{2}$

    = $\frac{22}{7}$ x $(0.0326)^{2}$

    = 0.00334 $m^{2}$

So that:

$\frac{2903.57}{A_{1} }$ = $\frac{24.41}{0.00334}$

$A_{1}$ = $\frac{2903.57*0.00334}{24.41}$

    = 0.3973

$A_{1}$ = 0.4 $m^{2}$

The radius of the piston can be determined by:

$A_{1}$ = $\pi r^{2}$

0.3973 = $\frac{22}{7}$ x $r^{2}$

$r^{2}$ = $\frac{0.3973*7}{22}$

   = 0.1264

r = $\sqrt{0.1264}$

 = 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).