Hydraulic Lift.( U tube) and oil

[dr3vil704]

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 [tex]\pi[/tex]r^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.

 

[stewartcs]

Give this a read...

http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html#hpress

CS

 

[Moetasim]

I would approach this problem by first identifying the key variables and equations involved. In this case, we have a hydraulic lift with a U-tube, filled with oil of a certain density, and two pistons of different radii. The car resting on the wider piston is being lifted by applying a force on the smaller piston. The key equation here is Pascal's law, which states that the pressure applied to an enclosed fluid is transmitted equally to all parts of the fluid.

To solve for the force on the smaller piston, we can use the formula P = F/A, where P is the pressure, F is the force and A is the area. We know that the pressure is constant throughout the fluid, so we can equate the pressure on the smaller piston to the pressure on the wider piston. This gives us:

P1 = P2
F1/A1 = F2/A2

We also know that the distance the smaller piston is raised is 1.2 m, so we can use this to find the difference in height between the two pistons (d2-d1). Using the formula for the volume of a cylinder, we can also relate the areas to the radii of the pistons by:

V1 = A1d1 = πr1^2d1
V2 = A2d2 = πr2^2d2

Since the volume of the oil is constant, we can equate these two equations and solve for the difference in height (d2-d1). This gives us:

d2-d1 = (πr1^2d1)/A1 - (πr2^2d2)/A2

Now we can substitute this into our original equation:

F1/A1 = F2/A2
F1/A1 = F2/A2 = (πr1^2d1)/A1 - (πr2^2d2)/A2

Solving for F2 (the force on the smaller piston), we get:

F2 = F1(A2/A1) = F1(r1^2/r2^2)(d1-d2)

Finally, we can plug in the values given in the problem to solve for F2:

F2 = (12,000 N)(.018^2/.005^2)(1.2 m) ≈ 83,520 N

Therefore, the force that must be applied