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Force and Continuity?

  • Thread starter jrrodri7
  • Start date
31
0
1. 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)?


2. Homework 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.
 

Answers and Replies

360
21
Maybe it would help to think of it as Force/Area. You have the right idea setting it up as a proportion.
 
rock.freak667
Homework Helper
6,230
31
You need to convert cm^2 to m^2

Remember that [itex]1cm=10^{-2}m[/itex] square both sides of that and you will get your conversion.
 
31
0
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)?
 
rock.freak667
Homework Helper
6,230
31
[itex]1cm=10^{-2}m \Rightarrow 1cm^{2}=10^{-4}m^2[/itex]..squaring like that to get the conversion.
 
360
21
Why convert, rock.freak667? The units in the area will cancel. OP will be left with just newtons.
 
rock.freak667
Homework Helper
6,230
31
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.
 

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