How to calculate tension in a cable

[twinplums]

If an aircraft is flying straight and level and drops a load but it remains attatched to the aircraft by a cable (hung load), how do I calculate the tension in the cable. I know the aircraft speed, the weight og the load and the length of the cable.

Thanks

 

[berkeman]

If an aircraft is flying straight and level and drops a load but it remains attatched to the aircraft by a cable (hung load), how do I calculate the tension in the cable. I know the aircraft speed, the weight og the load and the length of the cable.

Thanks

Welcome to the PF.

Do you also have the angle that the cable makes with the vertical? That would help a lot in simplifying the calculation.

 

[twinplums]

Thanks,

I do not know an angle as it is only theoretical,

but I would like to know how to calculate 90deg to the vertical i.e. directly behind the aircraft and

a 30deg below the horizontal

 

[haruspex]

It can never get to horizontal. The tension would be unlimited.
At θ to the vertical, we have
Vertical load = L
Horizontal load (drag) = H
Tension = T = sqrt(H^2+L^2)
tan(θ) = H/L
cos(θ) = L/T
So T = L.sec(θ).
At 30 degrees, T = 2L/√3

 

[Philip Wood]

Won't the string be vertical? The load on board the plane has the same horizontal velocity component as the plane, and will retain this velocity component as it is lowered. [I'm appealing to Newton's first law (the law of inertia).]

Edit: I was being silly. Air resistance will, of course, be significant.

 

[haruspex]

Won't the string be vertical? The load on board the plane has the same horizontal velocity component as the plane, and will retain this velocity component as it is lowered. [I'm appealing to Newton's first law (the law of inertia).]

I would expect drag to be significant, so the string will trail.
In fact, drag on the string may be important, so it won't even be straight. But if you only care about the max tension, the angle at the top of the string should give the right answer.

 

[K^2]

Yes, typically, you get a curve, because the drag on the cable adds up, but the tension is constant throughout. Won't be quite catenary, but should be close. If weight is heavy and provides significant drag, however, straight line will give reasonable approximation.

 

[twinplums]

Thank you for your replies,

The aircraft speed is 390 km per h, the load is 90Kg and the cable is 1.5 meters.

 

[Philip Wood]

In the unlikely event that air resistance is much smaller than the weight of the load (883 N), then the cable will hang almost vertically, and the tension in it will be approximately 883 N.

If air resistance isn't negligible (compared with 883 N), we need to know how large it is. It can't be estimated knowing the speed of the plane, unless we know the size and shape of the load. Even then, it would be no easy matter.

 

[K^2]

Yes, shape and size of the load can be used to estimate drag. Altitude will also matter. Cable is short enough for the curving to be insignificant, so it can be estimated as a straight line.

 

[AlephZero]

It can never get to horizontal. The tension would be unlimited.

Not necessarily. The object beinig towed might be generating a lift or buoyancy force, so its weight is zero or even negative. This applies to an aircraft towing a sailplane (glider) for example.

 

[haruspex]

Yes, typically, you get a curve, because the drag on the cable adds up, but the tension is constant throughout. Won't be quite catenary, but should be close. If weight is heavy and provides significant drag, however, straight line will give reasonable approximation.

Depends how the air resistance depends on the angle of the section of cable.
If the air resistance is constant per unit length of cable (i.e. not affected by the angle) then it would be a catenary. If, at the opposite extreme, it only depends on the vertical length of each section of cable then it would be a parabola. So I'd guess something between the two.

 

[haruspex]

Thank you for your replies,

The aircraft speed is 390 km per h, the load is 90Kg and the cable is 1.5 meters.

Hmm.. at that speed and short cable length you could get a nasty surprise.
As it drops, it will take a short time for the air resistance to retard it relative to the plane. If the cable becomes taut before that process completes it will then swing upwards and may strike the underbelly. It's all going to depend on the drag coefficient of the object. How dense is it?

 

[K^2]

Depends how the air resistance depends on the angle of the section of cable.
If the air resistance is constant per unit length of cable (i.e. not affected by the angle) then it would be a catenary. If, at the opposite extreme, it only depends on the vertical length of each section of cable then it would be a parabola. So I'd guess something between the two.

That's basically what I was thinking. Though, I didn't think to establish the second limiting case. Good call on the parabola. Yes, it'd be somewhere in between these two. It's probably worth pointing out that a short segment of a catenary is well-approximated by a parabolic segment.