# Fluid dynamics

1. Jun 6, 2009

### naspri4

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

Imagine a hydraulically operated dentist's chair having a mass of 190.0 kg and in it is sitting a 52.8 kg patient. The large piston beneath the chair has a diameter of 5.00 cm, whereas the small piston, moved by a pedal on which the dentist steps, has a diameter of 1.00 cm. What is the pressure in the interconnecting fluid when the dentist operates the chair?

2. Relevant equations

F=PA

3. The attempt at a solution

so there was another question right below this that asked for how much force he must exert on the smaller piston, so using F=PA i found P to equal 121.24 then i used that number with the area of the smaller piston and found the F of the smaller pistion to be 95.17N. So i assume the pressure must be right, but i dont get what the first question is asking for because although it is labeled in Mpa the answer is still wrong.

(finding pressure)
190 + 52.8 = 242.8

242.8 * 9.8 = 2389.44

2389.44= (P)(A=(pie)(2.5squared))

2389.44=(P)(19.625)

P=121.24