# Force and Continuity?

1. Mar 30, 2008

### jrrodri7

1. The problem statement, all variables and given/known data

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)?

2. Relevant equations

A_1 * V_1 = A_2 * V_2

3. 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.

2. Mar 30, 2008

### montoyas7940

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

3. Mar 31, 2008

### 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.

4. Mar 31, 2008

### jrrodri7

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

5. Mar 31, 2008

### rock.freak667

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

6. Mar 31, 2008

### montoyas7940

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

7. Apr 1, 2008

### 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.