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.

(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?

Any Help PLZ?

Answer 1

I don't get the big picture of what is going on in your lab manual problem. That is why I have not responded to your previous post. Perhaps someone else will.

Answer 2

it basically two tubes connected with a line, one to 10 cc has a weight put on top of it the second one 50cc has a force probe put on top of it. the weight changes force in first and causes the force and pressure in second to change. it basically using Pascal's Principal.

Answer 3

The pressure between them is the same.
Pressure on either pistion is force/area
and of course area is PI(d^2/4)

so force will be inversely prop to diameter squared.

If you need more help, post back.

Answer 4

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

or about 1.7N

check that.