a hydraulic lift has a narrow cylinder of diameter 4cm and wide cylinder of diameter 30cm .Calculate the force that must be applied to the liquid in the small cylinder to lift a 1500kg car​

Respuesta :

The force that must be applied is 261.3 N

Explanation:

According to Pascal's principle, the pressure is transmitted equally in all parts of the fluid: so, the pressure on the narrow cylinder must be equal to the pressure on the wide cylinder.

So we can write:

[tex]p_1 =p_2\\\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]

where:

[tex]F_1[/tex] is the force applied on the small cylinder

[tex]A_1 = \pi r_1^2[/tex] is the area of the first cylinder, with [tex]r_1 = 4 cm[/tex] being the radius

[tex]F_2[/tex] is the force on the wide cylinder, which is the weight of the car, so:

[tex]F_2 = mg=(1500 kg)(9.8 m/s^2)=14700 N[/tex]

where m = 1500 kg is the mass of the car

[tex]A_2 = \pi r_2^2[/tex] is the area of the wide cylinder, with [tex]r_2=30 cm[/tex] being its radius

Substituting and solving for [tex]F_1[/tex], we find the force on the narrow cylinder:

[tex]F_1 = \frac{F_2 A_1}{A_2}=\frac{(14700)(\pi (4)^2)}{\pi (30)^2}=261.3 N[/tex]

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