Hydraulic Lift.( U tube) and oil

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SUMMARY

The discussion centers on calculating the force required to lift a 12,000 N car using a hydraulic lift system consisting of a U-tube filled with oil. The wider arm has a radius of 0.18 m, while the narrower arm has a radius of 0.05 m. The formula A1 × d1 = A2 × d2 is utilized to relate the areas and distances of the two arms, where A represents area and d represents vertical distance. The density of the oil, at 800 kg/m³, is also a critical factor in determining the force needed on the smaller piston after the car has been raised 1.2 m.

PREREQUISITES
  • Understanding of hydraulic systems and Pascal's principle
  • Knowledge of area calculations for circular surfaces (A = πr²)
  • Familiarity with basic physics concepts such as force, mass, and gravity
  • Ability to perform unit conversions (e.g., cm to m)
NEXT STEPS
  • Calculate the areas of the pistons using the formula A = πr²
  • Apply Pascal's principle to determine the relationship between forces and distances in hydraulic systems
  • Explore the impact of fluid density on force calculations in hydraulic lifts
  • Investigate real-world applications of hydraulic systems in automotive and industrial settings
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Students studying physics, engineers working with hydraulic systems, and anyone interested in the mechanics of lifting heavy objects using fluid dynamics.

dr3vil704
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Homework Statement


A 12,000 N car is raised using a hydraulic lift, which consist of a U - tube with arm of unequal areas, filled with oil with a density of 800 kg/m^3 and capped at both ends with tight-fitting pistons. The wider arm of the U tube had a radius of 18. cm and the narrower arm has a radius of 5 cm. The car rest on the piston on the wider arm of the U-tube, The piston are initially at the same level. What is the force that must applied to the smaller piston in order to lift the car after it has been raised 1.2 m. Neglect the weight of the pistons

Ok, so I converted the radius to Meter, just for future reference. 5 cm = .005 m, 18 cm = .018 m.

I'm thinking of the Hyrdralift lift formula. A1 X d1 = A2 X d2. A= area, and D= vertical distance The sub 1 represent the smaller arm, and sub 2 represent the wider arm. and since it a circular surface, I should use \pir^2 for the Area.

So I could do d1=(A2/A1) X d2...
But then, I have no idea what to do with the density of the oil and use that distance to find the force (Mass X Gravity) of the smaller arm.
 
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