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Equilibrium - The physics of a clothesline.

  1. Jan 3, 2013 #1
    1. The problem statement, all variables and given/known data

    A clothesline is attached to two fixed ends which are 10.0m apart. A pulley of mass 40.0kg hangs freely in the middle of the line. The sag at the centre is 0.20m. Find the tension in each half of the clothesline.

    2. Relevant equations

    3. The attempt at a solution

    not sure how to tackle this problem since I don't have the values for the angle θ that is formed at the centre. Please point me towards the right direction with this problem. Thank you.
  2. jcsd
  3. Jan 3, 2013 #2


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    Homework Helper

    Make a drawing and you will find the angle.

  4. Jan 3, 2013 #3
    ok so I made a drawing like ehild recommended and I think I figured out angle θ. This is what I got for an answer.

    tanθ = 5/0.2 = 25
    θ = tan-1*25 = 87.7

    then then formula I used to solve the problem would be 2Tcos87.7=mg

    then i rearranged the formula to read:
    T= 4884N

    Therefore the tension on each half of the clothesline is 4884N.

    please correct me if I'm wrong on this problem. Thank you.
  5. Jan 3, 2013 #4
    Your solution is ok.
  6. Jan 3, 2013 #5
    ok perfect, thank you. As part of my solution I'm asked to find the x and y components but I'm not sure how to do so, can anyone help me? they want me to use these five steps to solve the problem, the steps are as follows:
    Step 1: select the object to be studied.
    Step 2: draw a "free-body diagram" for each object chosen.
    Step 3: choose a set of x and y axes for each of the objects being analyzed, and resolve the free-body diagram into components that point along these axes. (this is the step I'm having issues with).
    Step 4: set up the equations in such a way that the sum of the x-components of the forces is zero, and the sum of the y-components is also equal to zero.
    Step 5: solve the equations for the unknown quantities you are looking for.

    Again, thanks for your help.
  7. Jan 3, 2013 #6
    How did you get the equation 2Tcos87.7=mg before? Describe this in steps for them.
  8. Jan 3, 2013 #7
    ok so that's basically all I have to do? put it into words...?
  9. Jan 3, 2013 #8
    Yeah man!!
  10. Jan 3, 2013 #9
    Alright then, funny how sometimes I can miss the simplest things, thanks a lot.
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