- #1
sc0tt
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How much force is necessary to lift a ring, diameter 20mm, made of fine wire, and placed on the surface of water at 20 degrees celsius?
All the equations from the chapter:
Shearing stress = F/A = Viscosity*(velocity parallel/h)
Kinematic Viscosity = viscosity/density
Change in pressure = 4(surface tension/diameter)
Capillarity height = 4Tcos(theta)/(g*diameter*density)
Bulk modulus = -V*(change in pressure/change in V)
compressibility = 1/Bulk modulus
PV=RT
PV^n=constant
I'm learning fluid dynamics on my own for possible research so I don't have any professor or guidance, just a book. This makes solving problems much more difficult. With this problem specifically, I'm lost because I feel like the mass of the ring must be known as part of the problem, but no other information is given. This leads me to believe it has something to do with surface tension, compressibility, or shearing stress (this equation contains force).
My initial reaction was to solve the first equation for force and substitute the second equality in for shear stress. This leaves:
F=A*(viscosity*velocity parallel/h)
but there is no velocity parallel or required height.
The other equations don't seem to be of any help.
The answer the book gives is 9.15*10^-3 N
Thanks for any help,
Scott
All the equations from the chapter:
Shearing stress = F/A = Viscosity*(velocity parallel/h)
Kinematic Viscosity = viscosity/density
Change in pressure = 4(surface tension/diameter)
Capillarity height = 4Tcos(theta)/(g*diameter*density)
Bulk modulus = -V*(change in pressure/change in V)
compressibility = 1/Bulk modulus
PV=RT
PV^n=constant
I'm learning fluid dynamics on my own for possible research so I don't have any professor or guidance, just a book. This makes solving problems much more difficult. With this problem specifically, I'm lost because I feel like the mass of the ring must be known as part of the problem, but no other information is given. This leads me to believe it has something to do with surface tension, compressibility, or shearing stress (this equation contains force).
My initial reaction was to solve the first equation for force and substitute the second equality in for shear stress. This leaves:
F=A*(viscosity*velocity parallel/h)
but there is no velocity parallel or required height.
The other equations don't seem to be of any help.
The answer the book gives is 9.15*10^-3 N
Thanks for any help,
Scott