A hydraulic system is used to lift a 2100 kg (20,600 N) vehicle in a auto repair shop which is sitting on a piston with a cross-sectional area of 0.56 cm2. What is the minimum force that must be applied to the attached piston of area 0.035 cm2 in order to lift the vehicle?

Answer :

lublana

Answer:

1287.5 N

Explanation:

We are given that

Mass of vehicle ,m=2100 kg

Weight of vehicle ,F=20,600 N

Area,A=[tex]0.56 cm^2=0.56\times 10^{-4} m^2[/tex]

[tex]1 cm^2=10^{-4} m^2[/tex]

We have to find the minimum force that must be applied to the attached piston of area [tex]0.035 cm^2[/tex] in order to lift the vehicle.

[tex]A'=0.035cm^2=0.035\times 10^{-4} m^2[/tex]

Apply the pascal's law

[tex]\frac{F}{A}=\frac{F'}{A'}[/tex]

[tex]\frac{20600}{0.56\times 10^{-4}}=\frac{F'}{0.035\times 10^{-4}}[/tex]

[tex]F'=\frac{20600}{0.56\times 10^{-4}}\times 0.035\times 10^{-4}=1287.5 N[/tex]

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