Finding the sum and difference of two drag coefficients

  • #1

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





where:

Fd is the drag force, which is by definition the force component in the direction of the flow velocity
ρ
is the mass density of the fluid,
u
is the flow speed of the object relative to the fluid,
A
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:

Answers and Replies

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

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