How do I calculate the force needed for a hydraulic lift?

[jrrodri7]

Homework Statement

The small piston of a hydraulic lift has a cross sectional area of (2.8 cm^2), and the large piston connected to it has an area of (17 cm^2). What force of F must be applied to the smaller piston to maintain a load of (28000 N)?

Homework Equations

A_1 * V_1 = A_2 * V_2

The Attempt at a Solution

Since the volumes are the same, I set the A * F to be set equal to each other on both sides and solved...but I'm not seeing something in the unit conversion...Idk what to do.

 

[montoyas7940]

Maybe it would help to think of it as Force/Area. You have the right idea setting it up as a proportion.

 

[rock.freak667]

You need to convert cm^2 to m^2

Remember that 1cm=10^{-2}m square both sides of that and you will get your conversion.

 

[jrrodri7]

OH it'd be F1/A1 = F2/A2, and you I would have to convert to m^2, by X100 but squaring them? you mean make it into flat out (m)?

 

[rock.freak667]

1cm=10^{-2}m \Rightarrow 1cm^{2}=10^{-4}m^2..squaring like that to get the conversion.

 

[montoyas7940]

Why convert, rock.freak667? The units in the area will cancel. OP will be left with just Newtons.

 

[rock.freak667]

oh both are in cm^2...my bad, I read it wrong, I thought one was in m^2 and the other was in cm^2..No need to convert then.