Finding the sum and difference of two drag coefficients

In summary: I'm guessing that LC1 and LC2 are the "lift load cells" that are mentioned in the text.In summary, the problem involves determining the sum and difference of the coefficients of drag between two individual items that make a torque AUT. The problem provides information such as the diameter of each bottle, wind velocity, forces of the lift load cells, and the force of the drag load cell. The solution involves using the drag force equation, where the drag force is equal to the force component in the direction of the flow velocity. The problem also involves using torque calculations by using the radius, force, and angle of the load cell. The key to solving the problem is understanding the signs of the two torques and realizing that the lift
  • #1
Lunia Oriol
3
0

Homework Statement


I have to determine the sum and difference of the two coefficients of drag between two individual items that make a torque AUT. I'm given the diameter of each bottle (6.35 cm), wind velocity (14.3 m/s), the forces of the lift load cells (+0.0741 N, -0.0741 N), and the force of the drag load cell (0.846 N). Here is an image for clarification.

https://www.dropbox.com/s/k362escpst0m3f0/Photo Jan 12, 22 05 05.jpg?dl=0

Homework Equations



c2a53b7e1a1a64041f3dfca19baf22b2ec0d5c0e


where:

Fd is the drag force, which is by definition the force component in the direction of the flow velocity
ρ
a1d651c28959a0f15127c097ff4488b123d9e708
is the mass density of the fluid,
u
880f91e25cd451d89d1f6d0d06852b56a7b74a32
is the flow speed of the object relative to the fluid,
A
c6aaf5ce10d6add44b973e28fb3d95f37abf3721
is the reference area.

The Attempt at a Solution


I’m pretty sure i found the sum of the drag coefficients since I know the force of the load cell, the velcity, cross area, and density of air (my drag coefficient sum was 2.18), but I don’t know if I’m supposed to now used the force of the lift load cell since the torque is the radius times the perpendicular force. Thanks in advance.

Edited for clarity.
 
Last edited:
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  • #2
Lunia Oriol said:
the diameter between the two bottles
Diameter of the bottles?
Lunia Oriol said:
the lift load cells
Lift? Isn't the table horizontal? Some of the text is truncated in your image.
Lunia Oriol said:
if I’m supposed to now used the force of the lift load cell
Yes, of course. What equation can you write?
 
  • #3
haruspex said:
Diameter of the bottles?

Lift? Isn't the table horizontal? Some of the text is truncated in your image.

Yes, of course. What equation can you write?


I re-uploaded an image that displays the full problem. I determined that there were two load cells that measured lift because the two drag forces pointing to the right are normal to the lift load cells.

This is where me and my group got lost. I know that torque is a measurement of the radius times the force times sin(theta). If I were to use the force of one load cell (+0.0741), the distance from the middle of the bottle to the sting, and the angle of the force (sin90), I could certainly find the torque. I also know that torque equals the radius times (force2 - force1). But we couldn't go on from there. Initially we thought that because the lift load cells had essentially the same magnitude of force, the difference of the drag coefficients would be zero because we would be plugging in the same values to find the coefficients of drag, but it didn't seem plausible.
 
  • #4
Lunia Oriol said:
two load cells that measured lift
I still do not understand why you call it lift. That implies vertical force, no? The top diagram is the view from above, so all the forces are in the horizontal plane.
Lunia Oriol said:
the lift load cells had essentially the same magnitude of force
But what are the signs of the two torques? Do they both act the same way about the sting?
 
  • #5
Lunia Oriol said:
This is where me and my group got lost. I know that torque is a measurement of the radius times the force times sin(theta). If I were to use the force of one load cell (+0.0741), the distance from the middle of the bottle to the sting, and the angle of the force (sin90), I could certainly find the torque. I also know that torque equals the radius times (force2 - force1). But we couldn't go on from there. Initially we thought that because the lift load cells had essentially the same magnitude of force, the difference of the drag coefficients would be zero because we would be plugging in the same values to find the coefficients of drag, but it didn't seem plausible.
haruspex said:
I still do not understand why you call it lift. That implies vertical force, no? The top diagram is the view from above, so all the forces are in the horizontal plane.

But what are the signs of the two torques? Do they both act the same way about the sting?

No, they act in opposite directions. If it's not lift, then what's the force LC1 and LC2 are measuring?
 
  • #6
Lunia Oriol said:
No, they act in opposite directions.
How do you get that? The two forces are of opposite sign and are on opposite sides of the axis.
Lunia Oriol said:
If it's not lift, then what's the force LC1 and LC2 are measuring?
The three FLC forces are presumably supplied by apparatus to hold the platform steady. The values would be measured by strain gauges.
 

1. How do you find the sum of two drag coefficients?

The sum of two drag coefficients can be found by adding the two coefficients together. This is because drag coefficients are a measure of the resistance an object experiences when moving through a fluid, and when two objects are moving together, their individual drag coefficients combine to create a new overall drag coefficient.

2. What is the difference between two drag coefficients?

The difference between two drag coefficients is the variation in the amount of resistance each coefficient represents. This can be due to differences in the shape, size, or surface characteristics of the objects experiencing drag.

3. Can you find the sum or difference of more than two drag coefficients?

Yes, you can find the sum or difference of any number of drag coefficients as long as they are all acting on the same object or system. Simply add or subtract each coefficient to find the overall effect of drag on the object.

4. How do you measure drag coefficients?

Drag coefficients can be measured through experiments or simulations. In an experiment, an object is placed in a controlled flow of fluid and the resistance it experiences is measured. In a simulation, mathematical equations and computer models are used to calculate the drag coefficient based on the object's properties and the fluid's properties.

5. Can drag coefficients be negative?

Yes, drag coefficients can be negative. This typically occurs when an object is experiencing lift, which is an upward force that counteracts the force of drag. In this case, the negative drag coefficient represents the force of lift acting on the object.

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