USAPhO 2014 F=ma exam #19, (tension, static equilibrium)

[Agrasin]

Unfortunately there's no link to the test. This is from the test administered a month ago.

Homework Statement

A choppah is flying horizontally at constant speed. A perfectly flexible uniform cable is suspended beneath the choppah; air friction on the cable is NOT negligible.

Which of the following diagrams best shows the shape of the cable as the choppah flies through the air to the right?

I'll try to help you visualize the cable shapes depicted in the problem:

1) the cable is straight and vertical
2) the cable is straight and slanted, like a y = x graph
3) the cable is curved, like a y=x^2 curve, Q1 (but not exactly)
4) the cable is curved the opposite way, like y = √x , Q1, (but not exactly, you get the idea)
5) the cable looks like y = x^3 , Q1 & Q3, (not exactly though, more squashed into the y-axis)

Homework Equations

This is a visualization problem, I believe. Handy to remember that the force of gravity, the tension at each point, and the force of air resistance each play a role.

The Attempt at a Solution

I put 3 as the answer. I thought that is the mass was negligible, it would be 2, a straight slanted line. However, gravity acts at the center of mass, pulling down the slant in the middle.

Now I realized I hadn't taken air resistance into account.

Your solution? Thanks.

 

[haruspex]

I put 3 as the answer.

Sounds right.

I thought that is the mass was negligible, it would be 2, a straight slanted line.

Why slanted? Consider a short section at the bottom. You only have two forces on it. So what must the slope of the cable be there?

However, gravity acts at the center of mass, pulling down the slant in the middle.

Now I realized I hadn't taken air resistance into account.

Without air resistance, why would the answer not be (1)?
If you want to get a clearer idea of the exact shape, here's a way to think about the problem:
At equilibrium, every part of the cable has the same airspeed. Now, this won't be quite true, but suppose the air resistance is proportional to the length of cable, regardless of its orientation in the airstream. What does that tell you (magnitude and direction) about the force of air resistance on a short section length dl? What is the force of gravity on such a section? What is the net of those two forces? Does that give you a way of representing the problem in a static set-up?

 

[Agrasin]

How do you refute this argument I heard for case 2? (straight slanted cable)Assume that it is a massless string connecting a number of point masses. Consider the bottom point mass. It gets 3 forces: gravity mg, wind force going left horizontal F, and tension T1 towards right-up direction with angle (with horizon) a. So mg=T1sina, F=T1cosa. So tan(a)=(mg)/F
No consider the second from bottom point mass. It gets 4 forces: gravity mg, wind force F, downward tension T1 and upward tension T2. The angle with horizon of T1 is a and that of T2 is b.
Vertical: T2sinb = mg+T1sina=2mg
Horizontal: T2cosb = F+T1cosa=2F. So tan(b)=(mg)/F
So, a=b. That means the rope shape is straight.

 

[haruspex]

How do you refute this argument I heard for case 2? (straight slanted cable)


Assume that it is a massless string connecting a number of point masses. Consider the bottom point mass. It gets 3 forces: gravity mg, wind force going left horizontal F, and tension T1 towards right-up direction with angle (with horizon) a. So mg=T1sina, F=T1cosa. So tan(a)=(mg)/F
No consider the second from bottom point mass. It gets 4 forces: gravity mg, wind force F, downward tension T1 and upward tension T2. The angle with horizon of T1 is a and that of T2 is b.
Vertical: T2sinb = mg+T1sina=2mg
Horizontal: T2cosb = F+T1cosa=2F. So tan(b)=(mg)/F
So, a=b. That means the rope shape is straight.

Yes, sorry, I was too quick to agree that (3) sounded right. In fact, that last part of my previous post leads to (2) as the answer.
At each short segment dl of the cable we have, apart from tension, the same horizontal force F.dl and the same vertical force g.dl. These combine to produce the same constant force at a constant angle. Therefore the set-up is equivalent to gravity, increased a little and acting at an angle.
This is effectively the same as your argument

 

[Agrasin]

Thanks for that. Although I'm not happy I got it wrong.

I've heard your argument that the force of gravity and horizontal force is the same at each point, but I don't think that's true.

At each point as you go higher, the segment dL feels greater weight downward (mg) and same air resistance F. So as you go down, the angle changes and the cable looks like number 3.

I'm not disagreeing with you. Many people have told me it's case 2. But what's wrong with my thought process?

 

[haruspex]

At each point as you go higher, the segment dL feels greater weight downward (mg)

No, it feels greater tension. The weight of that segment is ρg.dL.

 

[Agrasin]

Yes exactly, so the tension vector for each point gets greater the higher up the point is. The air resistance vector is constant and to the left. So the resultant vector changes for each point on the cable, resulting in Case 3, right?

What's wrong with that logic?

 

[haruspex]

Yes exactly, so the tension vector for each point gets greater the higher up the point is. The air resistance vector is constant and to the left. So the resultant vector changes for each point on the cable, resulting in Case 3, right?

What's wrong with that logic?

You don't a priori know what the consequence of tension will be on the shape since it will in turn depend on the shape. Ignore tension to begin with. What you then have is the identical force, same magnitude and direction, on every segment length dl. This is exactly the same as for a cable suspended with no horizontal movement. All that's different is the strength and direction of 'gravity'. So the straight line solution must still be valid. You can now introduce tension, always in the direction opposing the combined gravity and drag force, with a magnitude proportional to distance from the free end.

 

[greendog77]

Yes exactly, so the tension vector for each point gets greater the higher up the point is. The air resistance vector is constant and to the left. So the resultant vector changes for each point on the cable, resulting in Case 3, right?

What's wrong with that logic?

The magnitude of tension does get greater the higher up the point is, but there are two tension forces: one upwards, and one downwards. The tension forces cancel out, such that the net tension force is up along the rope, of constant magnitude. I can work this out more rigorously for you if you would like. Very interesting problem, btw.

 

[lu.william.m]

Hi,

Can you consider the downdraft of the air from the helicopter blades? A real helicopter's downdraft is about 70-115mph while the record speed goes at 250 mph. Also, downward airspeed is lesser at the bottom - http://www.copters.com/aero/hovering.html.

Original question here: http://imgur.com/0XkhXvk

I am quite convinced of its being curved because of decreasing downdraft velocity.

 

[greendog77]

no, you are not supposed to take into account the downdraft of air.

 

[lu.william.m]

why not? it's a helicopter, and the blades have to spin if its going to stay moving horizontal?

 

[greendog77]

Typically physics problems like this will not factor in such a vague condition. Usually we ignore parts of the problem that are not focused on by the text. In this case, we ignore the helicopter to focus on the rope. Also, several conditions can offset the theory.

1. The helicopter is lightweight, and downdraft air is negligible
2. The rope is long, and downdraft air is negligible after a small fraction of the rope is passed