As described in the manual, consider a 10 cc syringe is connected to a 50 cc syr
ID: 2202137 • Letter: A
Question
As described in the manual, consider a 10 cc syringe is connected to a 50 cc syringe. Information of the syringes are as follows: (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. 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? 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?Explanation / Answer
First, the cross- sectional areas of both syringes. 10cc piston has an area of 169.72 mm^2. 50cc has an area of 615.75 mm^2. Ratio of areas, = 615.75/ 169.72, = 3.628:1. Mass of 10cc piston = 16g. You then apply 50g., = 66g. 66g x 3.628 = 239.448g. But the 50cc piston has a mass of 60.6g, so subtract that. = 178.848g force on the probe. Now, do the piston masses balance? 16g x 3.628 = 58.048g. The question says 60.6g. It's a trifle unbalanced.? If it balanced exactly (maybe it does, if you use a different figure for pi) then: FA = FB/3.628. The relationship would be linear.
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