How Do You Calculate the Force Affecting Pressure in a Viscosity Experiment?

AI Thread Summary
In a viscosity experiment, the calculation of pressure difference (p) involves understanding the relationship between force and area, specifically the cross-sectional areas of the tube and valve. The participant struggles with determining the force needed to calculate pressure, questioning whether gravity could be considered as the force in this context. They acknowledge a confusion in their understanding of pressure as force per unit area. The discussion highlights the importance of accurately applying fluid mechanics principles to derive the necessary equations for viscosity measurement. Clarifying these concepts is essential for successfully conducting the experiment.
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1. Homework Statement





where:
l = the length of the tube in cm
r = the radius of the tube in cm
p = the difference in pressure of the two ends of the tube in dynes per cm2
c = the coefficient of Viscosity in poises (dyne-seconds per cm2)
v = volume in cm3 per second
pi = 3.14159... (i couldn't get the symbol to appear)

I am doing an experiment in which i measure viscosity for different liquids. However, I'm having problems in obtaining value "p". I know that pressure is force times area. In this case, area should be the cross sectional area for the tube at one end, and the cross sectional area of the valve opening at the other end (if I'm not mistaken). However, what will be the force?




2. Homework Equations





where:
l = the length of the tube in cm
r = the radius of the tube in cm
p = the difference in pressure of the two ends of the tube in dynes per cm2
c = the coefficient of Viscosity in poises (dyne-seconds per cm2)
v = volume in cm3 per second
pi = 3.14159... (i couldn't get the symbol to appear)



3. The Attempt at a Solution

I know that pressure is force times area. In this case, area should be the cross sectional area for the tube at one end, and the cross sectional area of the valve opening at the other end (if I'm not mistaken). However, what will be the force?
 
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Isn't pressure defined as force divided by area?
 
oooo, typo. your right, i meant force PER UNIT area. but still, any ideas on my problem?
 
\frac{r^{4}(pi)p}{8cl}
 
No, I was horrible at fluid mechanics. I was just making sure you didn't write it down wrong on your paper and then get confused. Aside from that, I'm useless.
 
Would the force be gravity?
 
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