Dentist Chair

  • Thread starter BuBbLeS01
  • Start date
  • #1
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Homework Statement


A dentist's chair of mass 207kg is supported by a hydraulic lift having a large piston of cross-sectional area 1403.0cm^2. The dentist has a foot pedal attached to a small piston of cross-sectional area 78.0cm^2. What force must be applied to the small piston to raise the chair.


Homework Equations


P=F/A
Fp/Ap = Fc/Ac


The Attempt at a Solution


Fp = Fc/Ac * Ap
Fp = 207kg/1403 * 78 = 112.78

Okay so I have already gotten the answer 112.78N but I don't understand a few things about this question...
- Why does Fp/Ap = Fc/Ac??
- Why is 207kg relate to 1403cm^2?
- What is the subscripts p and c???
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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Suggestion in your calculations when you said...
BuBbLeS01 said:
Fp = 207kg/1403 * 78 = 112.78
F=mg...so the force should be the weight..so you have to multiply by 9.81Nkg^-1 or whatever value you take as g and convert the cm^2 to m^2


"Why does Fp/Ap = Fc/Ac??"

Does this not mean that [itex]P_p=P_c[/itex] which just basically says that pressure is transmitted equally throughout a fluid...i.e. the pressure at one end is equal to the pressure at the other.

the subscripts p and c would be piston and chair respectively
 

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