Lifting a car using a hydraulic lift

[baddin]

1. The hydraulic lift at a repair shop is filled with oil. The car rests on a 25-cm-diameter piston. To lift the car, compressed air is pushed down on the 6-cm-diameter piston.
a) What air-pressure force will support a 1300kg car level with the compressed air piston?
b) How much must the air-pressure-force be increased to lift the car by 2m.



2. Homework Equations :
p = p + ρ*g*h
ρ=900kg/m^2

3. The Attempt at a Solution
I did a) by doing p + ρ*g*0 + F1/A1 = p + ρ*g*0 + F2/A2
Then used F1 = F2 * (A1/A2) to find that F1 is 730N, which is correct according to the solutions.

Then I attempted b) and used a similar approach.
I tried:
p + F1/A1 + ρ*g*(0) = p + ρ*g*h + F2/A2*
So I rearranged to get F1 = A1*ρ*h*g + (F2)*(A1/A2)
Finding A1, A2, and F2 easily, and using h =2m and ρ=900kg/m^3 and g = 9.8m/s^2, I keep getting F1 = 784N. So the increase in force should be 784N - 730N = 54N, but my book tells me the answer is 920N.

 

[ehild]

b) How much must the air-pressure-force be increased to lift the car by 2m.


Then I attempted b) and used a similar approach.
I tried:
p + F1/A1 + ρ*g*(0) = p + ρ*g*h + F2/A2*
So I rearranged to get F1 = A1*ρ*h*g + (F2)*(A1/A2)
Finding A1, A2, and F2 easily, and using h =2m and ρ=900kg/m^3 and g = 9.8m/s^2, I keep getting F1 = 784N. So the increase in force should be 784N - 730N = 54N, but my book tells me the answer is 920N.

The amount of oil does not change during lifting the car. If the level increases in one arm, it should decrease in the other one. So the oil level in the compressed arm is below zero.

ehild.

 

[baddin]

Thank you so much! I used the volume's of liquid displaced to calculate how much the smaller piston would go down and found that it would move down by 34.72m, while as the other larger piston would move up by 2m.
Then I used the following formula to find F1 = A1*(ρ*g*h) + F2*(A1/A2)
And got 1650N
Then 1650-730 =920N =D Thank you very much!

 

[ehild]

You are welcome

ehild