Integrating pressure over area to get friction force

In summary, the conversation discusses experiments involving the calculation of resistance force on a cylindrical body (cable) being pulled through water-saturated sand. The formula used is based on stress components and is put into MATLAB, where it breaks down. Two members of the team use a different approach, calculating the pressure/stress on each point of the 2D-surface and summing it to get the force. The speaker disagrees with this approach and suggests integrating over the area to get the force immediately. They also mention subtracting densities of sand and water for a weighted average, but this is not necessary as pressure increases linearly with depth. The speaker is unsure about the correct way to calculate the friction force and is willing to provide more information if
  • #1
Ortix
64
0
I'm doing some experiments where I need to calculate the resistance force on a cylindrical body (cable) when it's being pulled through water saturated sand

We derived formula from a theory which was originally based on a square body by using stress components. This way we know the pressure at any given depth on the cable.

UCTTdRC.png

This depicts a square body under sand saturated in water.
(It's in dutch but I'm sure we all can figure out what zand and kabel mean ;) )

The formula we derived (which is a bit too big to put here) was put into MATLAB and this is where it breaks down. 2 other guys from my team calculate the resistance force like so (quick sample code):

Code:
for theta = 0:0.1:360
    sigma_point(i) = sigma(theta); // sigma is the formula as a function of theta. Theta is converted to rads
    i = i+1;
end

force = sum(sigma_point)/length(sigma_point)*area_of_cable*coeff_fric

What they do is they calculate the pressure/stress on each point of the 2D-surface with steps of 0.1 degrees. They sum that and divide that by the amount of steps they have taken. Finally they multiply it by the area of the cable and the coefficient of friction which is tan(phi), also known as the angle of friction.

But I'm not buying this. It just doesn't make sense to me. Every fiber in my body is screaming that we should integrate over the area to get the force immediately. I tried it but I'm getting forces in ranges of kilo Newtons which doesn't make sense.

What I did was (i can't get formulas to work):

  1. integral from 0 to 2pi of sigma(theta)*radius*length*d_theta
  2. multiplied the above result with the coefficient of friction.
I hope I was clear enough because I'm confusing myself here as well.
My question is basically whether the first MATLAB approach is correct and if not what would be a correct way to calculate the friction force.

I'll gladly supply more information/explanation if necessary
 
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  • #2
Why do you subtract densities of sand and water? I would expect a weighted average here.

Ortix said:
Every fiber in my body is screaming that we should integrate over the area to get the force immediately.
That's what they are doing, numerically. It is not necessary, however. Pressure increases linearly with depth. The average pressure is just the pressure at the middle of the cable - assuming it does not move upwards or downwards through the sand too fast.
 

FAQ: Integrating pressure over area to get friction force

1. What is the concept of integrating pressure over area to get friction force?

The concept is based on the idea that when an object is in contact with a surface, the pressure exerted by the object on the surface is distributed over its entire area. By integrating this pressure over the area of contact, the resulting force is equivalent to the friction force acting on the object.

2. How is this concept used in real-world applications?

This concept is used in various engineering and scientific fields, such as in designing structures to withstand frictional forces, calculating the braking distance of vehicles, and determining the stability of objects on inclines.

3. Is the friction force always equal to the integrated pressure over area?

No, the friction force can vary depending on various factors such as the surface materials, roughness, and the normal force between the two surfaces. The integrated pressure over area is just one way of calculating the friction force in idealized situations.

4. Can this concept be applied to non-uniform pressure distributions?

Yes, this concept can be extended to non-uniform pressure distributions by dividing the area of contact into smaller sections and integrating the pressure over each section separately.

5. What are the limitations of using this concept to calculate friction force?

This concept assumes that the pressure is evenly distributed over the entire area of contact, which may not always be the case in real-world situations. Other factors such as surface roughness, adhesion, and lubrication can also affect the friction force and may not be accurately represented by this concept.

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