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A hydraulic lift has two pistons: a small one of cross-sectional area 5.00 cm2 and a large one of cross-sectional area 300 cm2. What minimum force must be applied to the small piston in order for the large piston to lift a truck of mass 4500 kg

Respuesta :

Answer:

735.75 N

Explanation:

We are given;

Area of small lift; A1 = 5 cm²

Area of large lift; A2 = 300 cm²

Mass required to lift large lift;m = 4500 kg

Since Force = mg, then, Force required to lift large lift;F2 = 4500 × 9.81 = 44145 N

By Pascal's law,

F1/A1 = F2/A2

We want to find minimum force that must be applied to the small piston;F1, so let's make F1 the subject.

Thus;

F1 = (F2 × A1)/A2

Plugging in the relevant values, we have;

F1 = (44145 × 5)/300

F1 = 735.75 N

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