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melomori [17]
3 years ago
11

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?
Physics
1 answer:
marishachu [46]3 years ago
4 0

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).

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