Read the activity 2-1:Force and Pressure in the lab manual (pages 167-168). As described in the manual, consider a 10 cc syringe is connected to a 50 cc syringe. Information of the syringes are as follows:

Question

(mass of 10 cc piston) = 16 g
(mass of 50 cc piston) = 60.6 g
(diameter of 10 cc piston) = 14.7 mm
(diameter of 50 cc piston) = 28.0 mm.

(a) What is the pressure exerted by the 10 cc piston onto the air underneath?
kPa

(b) If a 50 g mass is placed on the 10 cc piston, what is the pressure exerted by the 10 cc piston onto the air underneath?
kPa

(c) If a force probe is attached to the 50 cc piston, what do you expect will be the force measurement while a 50 g mass is placed on the 10 cc piston?
N

(d) Now, the 50 g mass is removed from the 10 cc piston, and another force probe is attached to the 10 cc piston. Now both 10 cc piston and 50 cc piston have force probes attached to them. Consider that the 10 cc piston is pressed down with a force FA, and let's call FB to be the force measured by the force probe on the 50 cc piston. If you write FA as a function of FB, what is the slope of the function?


totally stumped help anyone?

Answer 1

ok so. I solved first part with F/A i got .9239 kPa.

second part i solved to i just added the forces. I got 3.811 kPa.

The last two i am having difficulty with....

Answer 2

So since i know F/A=F/A

I'm just confused on how it affects one another....

Answer 3

i tried 646.8/169.71=F2/615.752 and tried solving for F but i am getting 2346.759 and it just not right..

Answer 4

should not F2=F1*A2/A so F2=646.8*(615.752/169.71)

F2=2346.759

Answer 5

The important thing to remember is that the pressure is uniform throughout the fluid (air?), so that the ratio of piston weight (plus applied force) to piston area is the same on both sides (i.e., both syringes) . You seem to have grasped that already.

Answer 6

Lets take the small pisiont with its 50 g nass

.05g/PI(.0147)^2/4 = Forcereadin/PI(.027)^2/4

or

forcereading=.05g*(.027/.0147)^2