Total force that a fluid exerts on a cylinder

AI Thread Summary
The discussion focuses on calculating the total force exerted by a fluid on a cylinder, with a calculated cylinder density of 5479 kg/m³. The user presents a detailed formula involving parameters such as angular velocity, radius, and fluid viscosities to derive a force of 50.11 N. There is confusion regarding the distinction between total force and torque, with suggestions that the total force should account for buoyancy. Additionally, the impact of neglecting friction at the base is debated, as it affects the velocity gradient. The conversation highlights the complexities of fluid dynamics calculations related to cylindrical objects.
Guillem_dlc
Messages
188
Reaction score
17
Homework Statement
A cylinder of diameter ##d=12,0\, \textrm{cm}## and height ##L=1,1\, \textrm{m}## is immersed floating at the interface between mercury (##\rho_{hg}=13580,0\, \textrm{kg}/\textrm{m}^3## and ##\mu_{hg}=0,0015\, \textrm{Pa}\cdot \textrm{s}##) and liquid paraffin (##\rho_{pr}=850,0\, \textrm{kg}/\textrm{m}^3## and ##\mu_{pr}=0,2\, \textrm{Pa}\cdot \textrm{s}##) within a glass tube of diameter ##D=12,2\, \textrm{cm}##. The cylinder is at ##c=0,2\, \textrm{cm}## from the bottom of the tube, the part immersed in mercury has a length of ##b=40,0\, \textrm{cm}## and the part immersed in liquid paraffin has a length of ##70,0\, \textrm{cm}##, as shown in the figure.

The cylinder is rotated by ##100,0 \, \textrm{rpm}##. Neglecting the friction at the base of the cylinder and the tube, determine the total force, in absolute value, that the fluid exerts on the cylinder, at ##\textrm{N}##.
Relevant Equations
##F=\tau A##
Figure:
508922CF-69E6-4502-9C76-4AA5FE2E244D.jpeg


I have calculated the density of the cylinder: ##5479,0\, \textrm{kg}/\textrm{m}^3##.

Attempt at a Solution:
$$d=0,12,\,\, L=1,1,\,\, D=0,122,\,\, e=0,002,\,\, c=0,02,\,\, b=0,4,\,\, a=0,7$$
$$\omega =100\, \textrm{rpm}=10,472\, \textrm{rad}/\textrm{s}\quad e=0,122-0,12=0,002$$
We know that: ##F=\tau A=\mu \dfrac{\omega r}{e}\cdot A\rightarrow##
We have two ##\mu##'s and two different areas:
  • Hg ##\rightarrow A=\pi r^2+2\pi r\cdot b##
  • Pr ##\rightarrow A=\pi r^2+2\pi ra##
$$\rightarrow F=\dfrac{\omega r}{e}(\pi r^2+2\pi rb+\pi r^2+2\pi ra)(\mu_{Hg}+\mu_{Pr})=$$
$$=\dfrac{\omega r}{e}(2\pi r^2+2\pi r(b+a))(\mu_{Hg}+\mu_{Pr})=\dfrac{\omega 2\pi r^2}{e}(1+b+a)(\mu_{Hg}+\mu_{Pr})$$
$$=50,11\, \textrm{N}$$
Here I don't know when I should use the integral and when I shouldn't. Would you do it like this?
 
Physics news on Phys.org
I've already got this one! Thanks
 
It says to neglect the friction at the base. If you don't neglect it, you have to consider that the velocity gradient varies across it.
Doesn't seem right that it asks for the total force. The total force would be the buoyancy. What you have calculated appears to be a torque. Maybe it’s the translation.
 
  • Like
Likes Guillem_dlc and Lnewqban
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top