Fluids: Concept about Fluid Pressure in a relation with Force and Area

In summary: For a): Was the density of blood provided?No.For b):The problem seems to refer to the outward force that tends to radially expand the given length of an artery or vein.The wet surface of that length is calculated by imaginarily making a longitudinal cut of that cylindrical segment of that blood vessel and unfolding it.It then becomes a rectangle of 2.00 cm long and (π x 0.15) cm wide (which is the perimeter of the internal/wet surface of the vessel segment).Consider that the problem is not asking about the total pressure, but the delta pressure between foot and head.The wet surface is the surface of
  • #1
Nova_Chr0n0
16
3
Homework Statement
(a) Calculate the difference in blood pressure between
the feet and top of the head for a person who is 1.65 m tall. (b)
Consider a cylindrical segment of a blood vessel 2.00 cm long and
1.50 mm in diameter. What additional outward force would such a
vessel need to withstand in the person’s feet compared to a similar
vessel in her head?
Relevant Equations
P = F/A
I've already got the correct answer in letter (a), which is 17140.2 Pascals. My question will be focusing about the letter b of the question and here is my solution:

(b)

FORMULA:
P = F/A
F = P*A

My understanding about this problem is I have to use the pressure that I got in letter (a) to calculate the force needed by the blood vessel to withstand 17140.2 Pascals

Given:
P = 17140.2 Pa
Diameter = 1.5 x 10^-3 m (converted to meter)
height = 0.02 m (converted to meter)

Solution:
>> Getting the area

Area = πr^2
Area = π[(1.5 x 10^-3)/2]^2
Area = 1.767 x 10^-6 m^2

>> Solving for the Force
F = 17140.2(1.767 x 10^-6)
F = 0.030 N

When I tried to look for the answer, I found 1.61 N as the final force. I think I've got the area calculation wrong. Isn't the area needed to solve the problem is the one that the force makes contact with? In the cylindrical segment stated in the problem, I've assumed that the force would be in contact with its base area, which is just the circle. Is my understanding about the area needed completely wrong? Please enlighten me. Thanks!
 
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  • #2
Nova_Chr0n0 said:
I've already got the correct answer in letter (a), which is 17140.2 Pascals.
The final answer to part (a) should be expressed with a suitable number of significant figures.
(But the unrounded value should be used in subsequent calculations, as you have correctly done.)

Also, note that the unit 'pascal' is written with a lower-case 'p', though the symbol is 'Pa'.

Nova_Chr0n0 said:
>> Getting the area
Area = πr^2
You are asked about the 'outward force'. So the first question to ask yourself is: on what surface(s) is the pressure producing an outwards force?
 
  • #3
Steve4Physics said:
The final answer to part (a) should be expressed with a suitable number of significant figures.
(But the unrounded value should be used in subsequent calculations, as you have correctly done.)

Also, note that the unit 'pascal' is written with a lower-case 'p', though the symbol is 'Pa'.You are asked about the 'outward force'. So the first question to ask yourself is: on what surface(s) is the pressure producing an outwards force?
Is it the whole surface of the cylindrical vessel and not only its base area?
 
  • #4
It is a segment of a vessel. There are no base walls, only lateral surface.
 
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  • #5
Nova_Chr0n0 said:
What additional outward force would such a vessel need to withstand
To a physicist the question makes no sense. Force is a vector; it acts in a single direction, not 'outwards', which is multiple directions. The net force on the vessel would be extremely small, being only the weight of fluid in it, and then only if the vessel is horizontal.
An appropriate question to ask would be the extra tension in the wall of the vessel.

Edit: The equation P=F/A only applies to flat surfaces. More generally, it is ##\vec F=\int P \cdot\vec{dA}##. For the cylindrical segment, ##\int \vec{dA}=0##.
 
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  • #6
Nova_Chr0n0 said:
>> Solving for the Force
F = 17140.2(1.767 x 10^-6)
F = 0.030 N

When I tried to look for the answer, I found 1.61 N as the final force. I think I've got the area calculation wrong.
For a):
Was the density of blood provided?

For b):
The problem seems to refer to the outward force that tends to radially expand the given length of an artery or vein.

The wet surface of that length is calculated by imaginarily making a longitudinal cut of that cylindrical segment of that blood vessel and unfolding it.

It then becomes a rectangle of 2.00 cm long and (π x 0.15) cm wide (which is the perimeter of the internal/wet surface of the vessel segment).

Consider that the problem is not asking about the total pressure, but the delta pressure between foot and head.

Unfolding-a-cylinder-to-a-plane.png
 
  • #7
The problem is not asking about the pressure but the force. Which does not make sense, as already mentioned by @haruspex. It does not matter if it is the difference or not. The net force on the walls is zero. As we expect, the blood vessels do not accelerate through the body. There is a tension in the walls, though.
 

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