Hi. I need help with this problem. My strategy so far is to somehow use P =F/A, but I can't figure out what to do with the given diameters. I tried finding the pistons' areas using 2pi(r)(h) for area of a cylinder, and then somehow solving for pressure, but I don't have h =/ Help, please!
Problem:
Imagine a hydraulically operated dentist's chair having a mass of 205.0 kg and in it is sitting a 57.8 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 2.00 cm.
What is the pressure in the interconnecting fluid when the dentist operates the chair? (MPa)
How much force (N) must he exert on the small piston?
Some calculations:
205 kg = 2009 N
57.8 kg = 566.44 N
2 cm = .02 m; 5 cm = .05 m
Problem:
Imagine a hydraulically operated dentist's chair having a mass of 205.0 kg and in it is sitting a 57.8 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 2.00 cm.
What is the pressure in the interconnecting fluid when the dentist operates the chair? (MPa)
How much force (N) must he exert on the small piston?
Some calculations:
205 kg = 2009 N
57.8 kg = 566.44 N
2 cm = .02 m; 5 cm = .05 m
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(205 + 57.8) x 9.8 = 2,575.44N. This is the force on the oil.
Now you DO need the large piston area, = (.025^2 x pi) = 0.00196375m^2.
(2,575.44/0.00196375) = 1.3115MPa.
2,575.44/ (5^2/2^2) = 412.1N.
Now you DO need the large piston area, = (.025^2 x pi) = 0.00196375m^2.
(2,575.44/0.00196375) = 1.3115MPa.
2,575.44/ (5^2/2^2) = 412.1N.
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You are using the formula for the lateral area of a cylinder; the one you want is for the area of a circular end of the cylinder.
And you don't even need that!
Per your calculations, Fmax = 2009 + 566.44 = 2575.44 N
Since the pistons are both circular, their cross sectional areas are in proportion to the square of their diameters:
a/A = 2.00/5.00 = 0.40
Fs = Fmax*a/A = 2575.44*0.40 = 1030.2 N
And you don't even need that!
Per your calculations, Fmax = 2009 + 566.44 = 2575.44 N
Since the pistons are both circular, their cross sectional areas are in proportion to the square of their diameters:
a/A = 2.00/5.00 = 0.40
Fs = Fmax*a/A = 2575.44*0.40 = 1030.2 N
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u see....Pascal stated a law which goes some what like...: when u apply pressure on a liquid , the pressure gets distributed uniformly throughout.....
now ..i guess i found the wrong part of ur approach....u are right with ur formula P= F/A
but there's something wrong in the calculation of the area.... i mean....the area would be just for the base of the piston (or say the cylinder) i.e. (pie radius^2)
now ..i guess i found the wrong part of ur approach....u are right with ur formula P= F/A
but there's something wrong in the calculation of the area.... i mean....the area would be just for the base of the piston (or say the cylinder) i.e. (pie radius^2)
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keywords: Pressure,Force,Area,Problem,Pressure/Force/Area Problem