Line tension attached to Parasail?

In summary: This means that the wind speed would need to increase or the size of the parachute would need to decrease. In summary, the force on the rope when parasailing with a 3600lb boat, a 180lb pilot, and a 28 foot parachute in 15mph winds traveling at 25mph is approximately 4278.75 lbs. To lift the boat out of the water, the force on the rope would need to be greater than 3600 lbs, which could be achieved by increasing the wind speed or decreasing the size of the parachute.
  • #1
mav7
1
0
Not sure what to expect here.. but figured I'd at least ask!

Background: I heard of examples where a parasail can actually lift a boat out of the water. Also, I've seen instances where increased wind speed as caused the rope to break, which is rated for 8000lbs.

I'm wondering how you can estimate the force that will be applied to the rope based on certain variables. If I left anything out, let me know.

Let's say we have a 3600lb boat (weight includes passengers, fuel and gear) traveling 25mph, with a parasail attached. The pilot is 180lbs. There is a wind speed of 15mph. The boat is traveling into the wind. I'm assuming we will need some sort of variable for drag created by the parachute, not sure what this would be. But let's say the parachute has a 28 foot diameter if that helps.

Q1) What will the tension/force be on the rope, in lbs?

Q2) What conditions would need to be met for the boat to be lifted out of the water?

Thanks in advance...
 
Physics news on Phys.org
  • #2
To answer your first question, the tension/force on the rope will depend on the design of the parasail and the weight of the boat. In general, however, it can be estimated by using the equation: Force (in lbs) = Weight of Boat (in lbs) + Weight of Pilot (in lbs) + Drag force (in lbs)The drag force is dependent on the size of the parachute, the speed of the boat, and the wind speed. To calculate this, you will need to use a formula such as the one from NASA: Drag Force (in lbs) = 0.5 * Air Density (in kg/m3) * Velocity^2 * Parachute Area (in m2)So, for your example, the total force on the rope would be: Force (in lbs) = 3600 + 180 + (0.5 * 1.225 * (25^2) * 0.785) Force (in lbs) = 4278.75 lbsTo answer your second question, for the boat to be lifted out of the water, the force on the rope needs to be greater than the weight of the boat. So, in your example, the force on the rope needs to be greater than 3600 lbs.
 
  • #3




Hi there, thank you for your question about line tension attached to parasail. It's great that you are curious about the forces involved in this activity. I can provide some insights into this topic.

Firstly, let's discuss the concept of line tension. Line tension is the force applied to a rope or line when it is pulled taut. In the case of a parasail, the line tension is the force that is keeping the parasail attached to the boat and allowing it to stay in the air. This tension is affected by various factors such as the weight of the boat, the speed of the boat, and the wind speed and direction.

To answer your first question, estimating the force on the rope can be a complex calculation as it depends on multiple variables. However, we can use some basic principles to get an approximate answer. The formula for force is mass times acceleration (F=ma). In this case, the mass would be the combined weight of the boat, passengers, and pilot. The acceleration would be the speed of the boat, which is 25mph. So, the force on the rope would be approximately 9000lbs (3600lbs + 180lbs = 3780kg x 25mph = 9000lbs). This is a rough estimate as it does not take into account the drag created by the parachute, which would decrease the force on the rope.

Moving on to your second question, the conditions that would need to be met for the boat to be lifted out of the water are related to the force on the rope. In order for the boat to be lifted, the force on the rope would need to be greater than the weight of the boat. This would require a strong wind speed and a properly designed and rated parasail. Additionally, the angle of the rope would also play a role. If the rope is pulled at a steep angle, the force on the rope would be greater and could potentially lift the boat.

I hope this helps to answer your questions. It's important to note that these calculations are simplified and do not take into account all the variables and complexities involved in parasailing. It is always important to follow safety guidelines and use properly rated equipment when engaging in any activity. If you have any further questions, please don't hesitate to ask. Thank you!
 

1. What is line tension attached to parasail?

Line tension attached to parasail refers to the force exerted by the parasail's lines on the boat or other object it is attached to. This tension is what allows the parasail to stay airborne and be controlled by the operator.

2. How is line tension measured?

Line tension can be measured using a line tension gauge, which is a specialized instrument that attaches to the parasail's lines and measures the amount of force being exerted. Some modern parasail systems also have built-in sensors that can measure line tension.

3. What factors affect line tension in parasailing?

There are several factors that can affect line tension in parasailing, including wind speed, boat speed, the size of the parasail, and the angle of the lines. Higher wind and boat speeds, as well as larger parasails and steeper line angles, will result in higher line tension.

4. Can line tension be adjusted during a parasailing ride?

Yes, line tension can be adjusted during a parasailing ride by either adjusting the boat speed or the angle of the lines. These adjustments can be made by the operator to control the height and speed of the parasail.

5. Is there a maximum safe amount of line tension for parasailing?

Yes, there is a maximum safe amount of line tension for parasailing. The exact amount will vary depending on the specific equipment and conditions, but it is important for operators to monitor line tension and make adjustments as needed to ensure the safety of the riders.

Similar threads

Replies
16
Views
340
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
5K
  • Mechanics
Replies
18
Views
4K
  • Other Physics Topics
Replies
26
Views
5K
  • Introductory Physics Homework Help
Replies
4
Views
10K
  • Introductory Physics Homework Help
Replies
10
Views
3K
  • General Engineering
Replies
7
Views
2K
Back
Top