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Tension on a rope

  1. Dec 28, 2013 #1
    1. The problem statement, all variables and given/known data

    My boat is tethered to a rope that can withstand a maximum of 8000 pounds of tension. The wind will exert, at maximum, 3000 pounds of force on the boat. How do I calculate the tension on the rope based on the angles that the rope is at? does it make a difference if the boat remains stationary while the wind blows, or if the wind causes it to move a bit? Thanks for the help (:

    Ty = Tsin(theta)
    Tx = Tcos(theta)

    2. Relevant equations


    T = m*a ?

    I also found this equation: F = (mv^2) / r for tension caused by a moving object, I'm not sure if that applies though....


    The wind only acts in the x-direction, and in the y-direction the force from the water will balance any other y-forces out.

    3. The attempt at a solution


    It seems to me like the tension on the rope will be 3000lbf no matter what the angle. The angle would change the amount of force exerted in either direction, x or y, but the NET force would remain the same. HOWEVER, http://www.engineeringtoolbox.com/rope-angle-tension-increase-d_1507.html describes an increase in tension as the angle increases, with the least tension being when the side of the boat and the rope are perpendicular. If this is true, what causes this? why? what equations give the numbers of tension increase they've described? Thanks! (:
     
  2. jcsd
  3. Dec 28, 2013 #2

    Simon Bridge

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    The resultant force due to the action of the wind on the boat may depend on the boat's orientation to the wind.

    The resulting tension in the rope also depends on what else is acting on the boat (and the rope) - bobbing up and down may make a difference, and so would contact with a dock or other obstacle that could create a pivot.

    The engineering toolbox example in the link you provided looks like it is for the situation where a beam is supported horizontally by the rope. The vertical double-line represents a wall, the horizontal line is a beam.
    The rope has to create a moment about the pivot where the beam joins the wall which balances the moment due to the weight of the beam.

    Calculating the tension in the rope involves drawing free-body diagrams.
    If there is only the rope and the wind acting on the boat, and the boat is stationary, then the rope will not make an angle to the wind.
    If the rope makes an angle to the wind, and the boat is stationary, then there is another force on the boat.
    If the rope makes an angle to the wind, and there is no other force on the boat, then the boat is not stationary - it will accelerate.
     
  4. Dec 29, 2013 #3

    haruspex

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    Are we to assume the rope is horizontal, or is the angle the rope "is at" its angle to the horizontal?
     
  5. Dec 29, 2013 #4

    Simon Bridge

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    ... oh and the horizontal thing ...
     
  6. Apr 13, 2016 #5
    I'm looking for the same answer but, my question is, if a boat weighs 1000 lbs and is being pulled by another boat at 10 mph what is the tension on the roap in fresh water the water is calm and it's not windy.
    I don't think the angles matter but if it did then let's say it was pull paralell to the water.
    If anyone can help me with the formula for this it would be great.
     
  7. Apr 13, 2016 #6

    Simon Bridge

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    Angles do matter so thanks... note: what is the net force needed to pull something at constant velocity?
    The formula you want is Newton's second law: ##\sum \vec F = m\vec a##
    You need two free body diagrams.
     
  8. Apr 13, 2016 #7

    haruspex

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    That's a very different question. Post it in a new thread, filling in the template. In particular, quote the problem exactly as given to you and show some attempt of your own. If the problem came with a diagram, please upload that if you can.

    As your question stands, there is no answer. The weight of the boat is a vertical force, while the tension in the rope is horizontal, so there is no evident relationship between the two forces. The force the rope needs to overcome is the drag from the water.
     
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