5
$\begingroup$

For a physics/ engineering contest, I want to use a large balloon as an anchor point for a pulley. This would allow me to raise and drop masses.

However, in testing, when I pull on the pulley the balloon just lowers and when I stop pulling the string the balloon rises with the item in tow.

My theory for why this didnt work has to do with the momentum of the balloon vs the device being lifted

Is the balloon not large enough? Should i use multiple? Is this even possible?

Edit

Here is an illustration and a better description:Balloon Anchor

If I would like to use a balloon to hold up a pulley to allow items to be lifted into the sky.

In a real world test, when I pull the lifting rope (brown line). The balloon lowers. When I stop pulling, the balloon then lifts the payload back to the original position.

Edit 2

Some more observations that may help the discussion:

  • There are guidelines that tether the balloon in place it would continue to rise if able.
  • No matter how softly I pull the pulley line the balloon lowers.
  • As noted in a comment below, I estimate the balloon is capable of lifting 4 to 6 lbs total and the payload is only 1/4 to 1/2 lbs
  • The friction of the pulley is very small it is free spinning and the lines in this case are small strings

Also, while all I have is my gut reaction from seeing and feeling the forces; I would like to direct the attention back towards the momentum and/or inertia of the balloon. It seems "easier" to move the balloon. It does return to its equilibrium, but seems easier to affect with smaller forces. Perhaps due to its small mass?

Improve this question
$\endgroup$
1
  • $\begingroup$ can you post an image or a diagram of your machine so that it help us to figure out the problem $\endgroup$
    – Deiknymi
    Mar 6, 2013 at 15:50

5 Answers 5

Reset to default
3
$\begingroup$

I don't think this can work, even with your clarification that the balloon is tethered to the ground. Let's identify all the forces. The balloon has a buoyancy, pulling it upwards. I will call that $F_b$ (b for balloon or buoyancy). The balloon pulls the pulley up. I will call the tension in the rope tethering the pulley to the ground $F_s$ (for stake). That pulls the pulley down. Gravity pulls the package down($F_p$ for payload). Then you pull on the rope with an applied force $F_a$. That force pulls up on the payload, and down on the pulley twice.

There are two cases, equlibrium and non-eqilibrium:

  1. in equilibrium, there is no net force on the payload, so $F_p = F_a$. There is also no net force on the pulley, so $2F_a + F_s = F_b$. Combining and rearranging terms gives you $2F_p = F_b - F_s$. In this case, imagine that you're holding the rope with enough tension to keep everything from moving.
  2. Now you pull on the rope a little harder, so $F_a$ increases, and now $F_a > F_p$. Let's calculate the net force on the pulley. It is $2F_a + F_s - F_b$. Substituting in the above equation gives us $2F_a + F_b - 2F_p - F_b = 2(F_a - F_p) > 0$. So the pulley moves down.

That result is independent of the buoyancy of your balloon. Normally these things can work because the pulley is constrained not to move. I don't see how to do that here: any increase in $F_b$ will be offset by a matching increase in $F_s$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ the balloon is staked to the ground with guide lines $\endgroup$
    – Bill Bonar
    Mar 6, 2013 at 18:50
  • $\begingroup$ Thank you for the thought out response. The one concept I don't see represented in your forces is once the balloon lowers at all the tethering line is lax and is no longer pulling with F(s) $\endgroup$
    – Bill Bonar
    Mar 7, 2013 at 18:25
  • $\begingroup$ I am unsatisfied with that as well. That seems to be a big hole in my argument, and I can't figure it out. I would appreciate it if anyone has any ideas. $\endgroup$
    – Colin McFaul
    Mar 8, 2013 at 3:41
2
$\begingroup$

For a simple pulley of the weight of the payload and the force being applied on the other end of the rope. When you are trying to pull the weight upward, you are applying additional downward force on the balloon, so your total downward force is 2x the weight of the payload. The upward force of the balloon is less than that, so the balloon descends until you've stopped pulling, then the balloon only has to lift the mass of the payload, so it goes up again.

The problem you will face is that in order to not descend when you pull on the pulley, the balloon will need to be capable of lifting 2x the weight of the payload. So, it will ascend with the payload when you aren't pulling!

Improve this answer
$\endgroup$
1
  • $\begingroup$ I can almost guarantee that the balloon is capable of lifting at minimum 2x the weight of at least some of the devices I tested. For example the balloon, was able to float one device that was nearly 1 pound + egg (our contest limit) but the same behavior was present when attempting to lift the smallest devices less than 4oz + egg Something else must be at work here. If i had to guess the balloon is pulling with 4-6 lbs of lifting power. $\endgroup$
    – Bill Bonar
    Mar 6, 2013 at 18:48
1
$\begingroup$

When you try to pull the rope across the pulley a component of the applied force act's downward as shown in figure due to which the downward force due to the mass and fcos(θ) overcome's the force applied by the balloon and it start's coming down. But when you stop applying the force again the force applied by the balloon increases and it start's rising up and the mass also start's rising up as the length of the string is shorter as compared to the previous one and the balloon start's to rise till al the forces are balancedenter image description here

To stop this according to me you can use a pulley with lock mechanism which allows you to decide when the pulley is free to rotate and when it is not controlling it it'e mechanism with another rope, and also you should use a pulley with teeth on the side which is in contact wit the rope and a bigger balloon with a buoyant force greater than the force applied by the payload . What will happen by this set-up when you will be holding the balloon the balloon will apply a force which will be grater than the weight of the payload which can cause it rise up here the work of the locking mechanism and the surface with the teeth comes when you wish to rise the payload just release the pulley and it will start rolling allowing the payload to rise up and when you wish to stop the rise just lock the pulley and the teethed pulley will prevent the slipping of the rope over it which can cause it to rise up. And you should also have another grove at the side of the pulley fitted with a rope so that if you want to bring the payload down you can just pull the rope to make the pulley turn in the opposite direction causing it to descend. Hope it helped you, if you have any problem in understanding my idea let me know

Improve this answer
$\endgroup$
4
  • $\begingroup$ This is a correct place to start in understanding the lateral forces, though you should visualize what happens as the balloon move to be halfway between the load and the person/thing pulling on the rope: disaster of the sort that riggers have nightmares over. $\endgroup$
    – dmckee --- ex-moderator kitten
    Mar 7, 2013 at 19:40
  • $\begingroup$ Sorry but I didn't got what you said if you can explain please $\endgroup$
    – Deiknymi
    Mar 7, 2013 at 19:44
  • $\begingroup$ As soon as you start to pull on the rope the balloon will move sideways. As it nears the center the load will be pulled up and across, at which point it may shift. And, if it does, which way will it move? Directly toward the person doing the pulling. $\endgroup$
    – dmckee --- ex-moderator kitten
    Mar 7, 2013 at 19:48
  • $\begingroup$ Yes I know it will move toward's the rigger's I just only neglected that part while answering the question because he only wanted a way to control it's up and down motion $\endgroup$
    – Deiknymi
    Mar 7, 2013 at 19:53
0
$\begingroup$

After skimming through the other answers, it seems no-one is thinking about friction. I hardly believe the frictional forces in this system would be negligible. Still, I'm going to do so as well. Because in any case I can only see this working if the balloon has a high buoyancy while it is prevented from ascending to its maximal height by a ceiling.

Colin McFaul puts his finger on the vital issue: if the balloon is just allowed to ascend to its maximal height, of course pulling on it will cause it to descend. It was at equilibrium! All forces cancel each other out perfectly in equilibrium. So putting an extra downward force on it (pulling the rope) will make it move towards a new equilibrium position at a lower height.

This doesn't happen if you hold the balloon back, don't let it go to the maximal height its buoyancy allows for. In other words: use a ceiling. Then, if the ceiling is sufficiently low (or conversely if the buoyancy if sufficiently strong) pulling the rope will add an additional downward force, but the balloon still has some extra buoyancy in it (which it couldn't convert into extra height due to the ceiling) so it will stay where it is for a while longer, until it cannot maintain equilibrium at the same position.

So that's what I would try: use a ceiling. Make it sufficiently low and get as much buoyancy from the balloon(s) as possible.

Improve this answer
$\endgroup$
0
$\begingroup$

Thinking out of the box i have brought an answer to this that works on paper. By using the forces listed in the original question (Balloon lift @ 6lbs and payload @ 8oz) i have included the following forces as being necessary to engineer a solution. First the weight+gravity for the pulley (I list at 4oz), secondary pulley system (.5oz friction force i will elaborate later), payload rope (I list at 2oz) and pull force (2lbs per tug). The setup for this relies on taking the guide lines off the lift balloon completely and instead attaching 2 sets of 3 guide lines to two additional pulleys which are anchored about 2 feet below the payload balloon when at rest. These two additional pulleys are kept afloat by two mini balloons each anchored to either side of each pulley keeping one just above the other in air. To complete the configuration the main pulley is attached directly to the bottom of the payload balloon. the main rope being used to raise the payload is fed through the two offset pulleys below the payload balloon and then fed through the main pulley attached to the payload balloon, in essence creating two different pulley systems for this without adding to the weight of the payload/main pulley. The idea is to not anchor the balloon, but use the balloon's lift as a dynamic force in addition to the pull force applied during each tug on the main rope. As each tug is applied the pulley systems will reduce the force required to pull the payload with the secondary pulley system guiding the line so that the system will not drift with each tug. After each tug the payload balloon is allowed to rise pulling the payload up in addition to the force applied during each tug. By not anchoring the balloon we allow the balloon to do most of the pulling.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ This seems like a how-to and not an explanation. $\endgroup$
    – Kyle Kanos
    Nov 9, 2015 at 11:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.