Total force that a fluid exerts on a cylinder

In summary, the conversation discusses the calculation of the density of a cylinder using various dimensions and variables. The calculation includes the use of two different areas for Hg and Pr, and the total force is found to be 50.11 N. The conversation also mentions the possibility of using an integral in the calculation and the consideration of friction at the base.
  • #1
Guillem_dlc
184
15
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?
 
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  • #2
I've already got this one! Thanks
 
  • #3
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.
 
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Likes Guillem_dlc and Lnewqban

1. What is the definition of total force that a fluid exerts on a cylinder?

The total force that a fluid exerts on a cylinder is the sum of all the individual forces acting on the cylinder due to the pressure of the fluid. This force is perpendicular to the surface of the cylinder and is known as the hydrostatic force.

2. How is the total force on a cylinder calculated?

The total force on a cylinder can be calculated by multiplying the pressure of the fluid by the surface area of the cylinder. This can be expressed as F = P x A, where F is the total force, P is the pressure, and A is the surface area of the cylinder.

3. What factors affect the total force on a cylinder in a fluid?

The total force on a cylinder in a fluid is affected by the density of the fluid, the depth of the cylinder in the fluid, and the shape and orientation of the cylinder. It is also affected by the velocity of the fluid, as well as any external forces acting on the cylinder.

4. How does the total force on a cylinder change with depth in a fluid?

The total force on a cylinder in a fluid increases with depth, as the pressure of the fluid increases with depth. This is due to the weight of the fluid above the cylinder, which increases the pressure on the cylinder's surface and thus increases the total force.

5. Can the total force on a cylinder in a fluid be negative?

Yes, the total force on a cylinder in a fluid can be negative if the pressure on one side of the cylinder is greater than the pressure on the other side. This can occur if the cylinder is not completely submerged in the fluid, or if there are external forces acting on the cylinder that counteract the force of the fluid.

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