Tension in landing cable on an aircraft carrier

  • #1

Homework Statement


[/B]
Keep in mind this is a Top Gun-themed homework assignment.
Cougar comes in for a shaky landing. His 20422 kg airplane traveling at 85 m/s strikes the deck at 3.5 degrees below the horizontal. Cougar's plane snags the landing cable stretched across the deck. The landing cable is a cable which feeds out from beneath the deck of the carrier from two points 15 m apart from each other. The cable is hydraulically controlled so that it maintains a constant tension. The airplane travels forward 60 meters before stopping. Consider the landing cable to be the only thing that stops the aircraft, the wheel brakes are not applied.
Calculate the magnitude of tension in the landing cable.

Homework Equations


[/B]
W = 1/2mv^2 (in this case)
W = ∫Fxdx

The Attempt at a Solution


[/B]
Since there's no direct way to find tension, I started with W = .5(20422)(85cos3.5)^2, solving for work done by the cable on the plane, which gave me -73499523.13 J. This answer was another part of the question and has been marked as correct.

In order to solve for tension I tried setting 73499523.13 J equal to ∫Tsin(arctan(x/7.5))dx from 0 to 60 meters, or 73499523.13 = T(7.5sqrt(.0177777777(60)^2 + 1)-7.5sqrt(1)).

This gave me T = 73499523.13/(7.5sqrt(65) - 7.5), or 1387649.215 Newtons.

Using my calculator to integrate the tension function and I arrived at the same answer. 1387649.215 N is not the correct answer, so I would like help on this problem, as the rest of my classmates are stumped as well.

I apologize ahead of time for formatting errors and for the lack of illustrations provided in the homework itself.
 
Last edited:

Answers and Replies

  • #2
SteamKing
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Homework Statement


[/B]
Keep in mind this is a Top Gun-themed homework assignment.
Cougar comes in for a shaky landing. His 20422 kg airplane traveling at 85 m/s strikes the deck at 3.5 degrees below the horizontal. Cougars plane snags the landing cable stretched across the deck. The landing cable is a cable which feeds out from beneath the deck of the carrier from two points 15 m apart from each other. The cable is hydraulically controlled so that it maintains a constant tension. The airplane travels forward 60 meters before stopping. Consider the landing cable to be the only thing that stops the aircraft, the wheel brakes are not applied.
Calculate the magnitude of tension in the landing cable.

Homework Equations


[/B]
W = 1/2mv^2 (in this case)
W = ∫Fxdx

The Attempt at a Solution


[/B]
Since there's no direct way to find tension, I started with W = .5(20422)(85cos3.5)^2, solving for work done by the cable on the plane, which gave me -73499523.13 J. This answer was another part of the question and has been marked as correct. In order to solve for tension I tried setting 73499523.13 J equal to ∫Tsin(arctan(x/7.5))dx from 0 to 60 meters, or 73499523.13 = T(7.5sqrt(.0177777777(60)^2 + 1)-7.5sqrt(1)). This gave me T = 73499523.13/(7.5sqrt(65) - 7.5), or 1387649.215 Newtons. Using my calculator to integrate the tension function and I arrived at the same answer. 1387649.215 N is not the correct answer, so I would like help on this problem, as the rest of my classmates are stumped as well.

I apologize ahead of time for formatting errors and for the lack of illustrations provided in the homework itself.
You should draw a sketch of the plane and the arrester cable (that's what it's actually called) looking down on the deck of the carrier. You've assumed in your calculations that there is only one tension you are dealing with. The arrester cable system is actually arranged more like this:
maxresdefault.jpg

Note: Ignore the numbers on the graphic. They are not for this problem.

In the graphic above, the plane has snagged the arrester cable and is moving to the right.

Also, it's a lot easier to follow a string of calculations if they aren't buried in a paragraph of text.
 
  • #3
I'll reformat my calculations after this, and I did draw a diagram but I don't have anything to take a picture with at the moment.
I did not know about the pistons as they aren't mentioned anywhere in the problem, but I think I may have only solved for half of the tension. I just wasn't sure if I was correct up until the end.
 
  • #4
SteamKing
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I'll reformat my calculations after this, and I did draw a diagram but I don't have anything to take a picture with at the moment.
I did not know about the pistons as they aren't mentioned anywhere in the problem, but I think I may have only solved for half of the tension. I just wasn't sure if I was correct up until the end.
You don't have to worry how the tension in the arrester cable is maintained, only that it remains constant while slowing the aircraft.

This image was the best one I could find online which showed how the arresting system worked for purposes of illustrating the current problem.
 
  • #5
haruspex
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I think I may have only solved for half of the tension
Yes, it looks like you forgot there are two instances of the tension, as far as the aircraft is concerned.
You don't need any calculus to solve this. Just think about the work done on the cable.
 

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