Launching Balloon: Calculating Force F(t)

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In summary, the conversation discusses the use of Newton's Second Law to calculate the force, F(t), acting on a vertical length of rope that is being pulled up at a constant velocity, v. There is a question about defining the system and whether it should include the entire length of the rope or just the vertical section. It is suggested to use a drum to eliminate transitional acceleration and simplify the problem. Ultimately, it is decided that F(t) = m(t)g when considering the entire rope as the system.
  • #1
PeterPumpkin
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The following came from a discussion about launching a balloon.

Suppose you have a heavy coiled rope of total length, L, and constant linear density, MU. You take hold of one end of the rope and pull it vertically up with a force, F(t), so that the tip of the rope moves at a constant velocity, v.

What is F(t), assuming the rope is so long that some of it remains coiled on the ground?

In drawing our free-body diagram, two questions arise:
1) Can you define our system to be just the vertical length of rope (of length, y(t)) --- see figures 1 & 2?
2) If so, how do we calculate the force that the coiled portion exerts on the vertical section, F(coil on rope)? There must be a force, otherwise the coiled portion wouldn’t unwind.

Thanks in advance.
 

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  • #2
Is this a homework question? Do you have any work to show us?

As a hint, though, it may be useful to delete the velocity from the free body diagrams and consider the difference between the two static scenarios...
 
  • #3
No. It is not a homework question!
 
  • #4
You'd know the rate of mass flow transitioning from not moving to moving upwards at the fixed velocity. There's a period of time where the speed of the unwinding rope is faster than the upward velocity, because it also has a horizontal component.

The unknown is the amount of time it takes for each section of the coiled rope to transition into upwards movement, specifically, the relationship between acceleration and time of the rope as it unwinds.

As time goes on, the mass of the rope moving vertically increases, while the mass of the rope sections in transition remains near constant, so the limit of this is simply the weight of the suspended rope as it moves at constant speed.

You could simplify this by having the rope on a drum unwinding at constant speed, eliminating the transitional acceleration of the rope.
 
  • #5
What I was wondering about was that the system is not fixed. IE Can we define our system to be the vertical length of rope even though the length of rope increases?
 
  • #6
PeterPumpkin said:
What I was wondering about was that the system is not fixed. IE Can we define our system to be the vertical length of rope even though the length of rope increases?
Using a drum that the rope unwinds from approximates this, as the rope no longer experiences any linear acceleration as it spools off the drum, and if the drum is friction free, it's not adding any load to the system. The effective diamter decreases as layers of rope are peeled off, but this can be ignored to simplify the problem.
 
  • #7
Does this mean we could apply Newton's second law to the vertical section of the rope ("our system") as:

F(t) - m(t)g - Force(coil on rope) = ma

where we take Force(coil on rope) = 0

and a = 0 (as rope rises at constant vertical velocity)
 
  • #8
PeterPumpkin said:
Does this mean we could apply Newton's second law to the vertical section of the rope ("our system") as:
F(t) - m(t)g - Force(coil on rope) = ma
where we take Force(coil on rope) = 0
and a = 0 (as rope rises at constant vertical velocity)
Yes, in which case, F(t) = m(t)g.
 
  • #9
OK. What if we took the WHOLE rope as our system. Afterall, we are free to define our system as we wish.

For the sake of simplicity, go back to the original posting where the rope was coiled on the table.

Then the rope is subject to a Normal force of
N(t)=(L-y(t))*MU*g
where MU = the length density of the heavy rope and
L-y(t) is the length of the coiled segment.

Surely Newton’s Second Law must apply to our system.
m*a = F(t) - y(t)*MU*g + N(t)

Can we do this?
 

1. What is the purpose of calculating force in a balloon launch?

The purpose of calculating force in a balloon launch is to understand the amount of force that is being exerted on the balloon as it is launched into the air. This can help determine the efficiency and effectiveness of the launch, as well as potential risks or issues that may arise.

2. How is force calculated in a balloon launch?

Force is calculated in a balloon launch by using the formula F(t) = m * a, where m is the mass of the balloon and a is the acceleration. The mass can be determined by weighing the balloon and any additional objects attached to it. The acceleration can be calculated by measuring the change in velocity of the balloon over a specific time period.

3. What factors can affect the force in a balloon launch?

Several factors can affect the force in a balloon launch, including the mass and weight of the balloon, the amount and type of gas or air inside the balloon, wind or air resistance, and any additional objects attached to the balloon. The launch location and weather conditions can also have an impact on the force exerted on the balloon.

4. Why is it important to calculate force in a balloon launch?

Calculating force in a balloon launch is important because it can help determine the success and safety of the launch. Too much force could cause the balloon to burst or lose control, while too little force may result in a failed launch or inability to reach the desired altitude. Additionally, understanding the force can also aid in making improvements or adjustments for future launches.

5. Are there any potential risks associated with calculating force in a balloon launch?

While there are some risks associated with calculating force in a balloon launch, such as potential errors in measurement or equipment malfunction, these can be minimized by using proper techniques and equipment. It is important to carefully follow safety guidelines and procedures to ensure a successful and safe launch.

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