Calculating Car Lift Height with a Hydraulic Lift

[BeiW]

Homework Statement

A hydraulic lift is used to jack a 970 kg car .12m off the floor. The diameter of the output piston is .18 m, and the input force is 270 N.

If the input piston moves .13 m in each stroke, how high does the car move up for each stroke?

2-3. Relevant equations and attempt at solving
F_in=F_out
That becomes pghA=pghA
p = density
g=gravity
h=height
A=area

So I get 270=p(9.8)(.13)(A) which is the input force.
I could set that equal to the output force:
270=p(9.8)(.13)(A₁)=(p)(9.8)(h₂)(.02545)
.02545 is the area of a circle when I plug in .09 for radius.

The p's would cancel, and I'm left with the variables A₁ and h₂.

I know I can use the system of equations to solve this, but I want to know if I set up the equations correctly. And also, what do I do with the .12 m (that the car is lifted off the ground)??

 

[hy23]

i believe the .12m is given so that you can find the initial height of the input piston since they are not in equilibrium

 

[BeiW]

i believe the .12m is given so that you can find the initial height of the input piston since they are not in equilibrium

In that case, would I add the .12m to the .13 for h₁?

 

[hy23]

.12m refers to the car or height of the output piston, the input piston moves by .13m, the distance the car moves is related by the force/piston area ratios of the input and output

 

[rwisz]

Your equations and approach seem to be on the right track. To account for the .12 m lift of the car, you would need to add that to the final height h2. So your final equation would be:

270 = p(9.8)(.13)(A1) = p(9.8)(h2 + .12)(.02545)

From here, you can solve for h2 by rearranging the equation and plugging in the values for p, g, and A1. This will give you the height that the car moves up for each stroke of the hydraulic lift.