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Using Newton's Laws Question involving tension

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data

    An old rope can now safely suspend 120 kg. When the rope is tied to a beam, it hangs down with a vertical length of 12.0 m. Calculate the minimum time required for an 85 kg person starting from rest to climb the entire length of the rope without breaking it.

    2. Relevant equations

    F = ma
    Five key equations

    3. The attempt at a solution

    a = f/m
    a = 1176 N [down] / 85 kg
    a = 13.835 m/s2

    using d = vi(t) + 1/2at2
    t = sqrt(1.73)
    t = 1.3 s

    But the correct answer is 2.4 s. What did I do wrong? Can someone please help?
     
  2. jcsd
  3. Mar 30, 2013 #2

    TSny

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    Hello and welcome to PF!

    Did you take into account all of the forces acting on the person? A free-body diagram for the person might help.
     
  4. Mar 30, 2013 #3

    Simon Bridge

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    Draw the free body diagram for the person and apply ##\sum F=ma##.
    Do all the algebra before putting the numbers in.
     
  5. Mar 30, 2013 #4
    As the others are saying, you need to add up ALL accelerations and forces.

    [tex]\sum F=ma[/tex] What forces will be acting on the rope when the person begins to accelerate while climbing the rope?
     
  6. Mar 31, 2013 #5
    But I included all the forces and drew a FBD. The only forces are the force of tension and the force of gravity. I included the force of gravity when I calculated the maximum force (1176 N) that can be exerted on the rope.

    F = 120 (-9.8 m/s^2 [up])
    F = 1176 N [down]
     
  7. Mar 31, 2013 #6

    TSny

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    Newton's second law is [itex]\sum[/itex]F = ma. When you apply this formula to the person, the left side must include all forces acting on the person Did you draw a FBD for the person? How many forces act on the person? What is the numerical value of each of those forces? What are the directions of those forces?
     
  8. Mar 31, 2013 #7

    TSny

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    What does the above force represent? Is it a force on the person? Why is it upward?

    Same questions for this force.
     
  9. Mar 31, 2013 #8
    F = -1176 N [up]
    = 1176 N [down] represents the maximum force of tension that be applied to the rope before it snaps. The force that act on the person as he is climbing up is the force of tension he climbs up the rope and the force of gravity is pulling him down. (1175 N Down because gravity is downwards) There are no other forces in the vertical direction nor in the horizontal direction.
     
  10. Mar 31, 2013 #9

    TSny

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    OK. The person exerts a force down on the rope and Newton's third law says the rope exerts an equal but opposite force on the person. So, one of the forces acting on the person is the upward force from the rope. As you said, the other force on the person is the force of gravity acting down on the person. Good. But, how did you get 1175 N for the force of gravity?
     
  11. Mar 31, 2013 #10
    I got 1175 N as the force of gravity because I used F = ma and plugged in 120 kg as the mass and 9.8 m/s^2 [down] as the acceleration. I used the acceleration due to gravity to find the force of gravity acting on the object.
     
  12. Mar 31, 2013 #11

    TSny

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    For the force of gravity on the person, you should use the mass of the person (85 kg).
     
  13. Mar 31, 2013 #12
    So I should do 1176 - 85(9.8) = 343
    then 343/85 = 4.03
    then if I plug that into the key equation of acc I get 2.4 s which is the correct answer. But Im not sure why? Why does subtracting 1176 - 85(9.8) work out to eventually get the right answer. If they are both forces of gravity shouldnt adding them work, but adding them up doesnt work out?
     
  14. Mar 31, 2013 #13

    TSny

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    They are not both forces of gravity. Only one force of gravity acts on the person. The other force that acts on the person is the force which the rope exerts upward on the person.

    Note that the two forces act in opposite directions. So, when you add them as vectors to get the total force on the person, the magnitude of the net force is obtained by subtracting the magnitudes of the two forces.
     

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    Last edited: Mar 31, 2013
  15. Mar 31, 2013 #14
    But to get 1176 N [down] didn't we use acceleration due to gravity to solve it, so then how can it be the force of tension? Isn't 1176 N [down] simply the force of gravity acting on an object that has 120 kg mass?
     
  16. Mar 31, 2013 #15

    TSny

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    If the rope is hanging by itself and then you start adding weights to the bottom, you find that the rope breaks when you hang more than 120 kg. That means that the rope can only withstand a tension that is numerically equal to the weight of a 120 kg mass. But of course that doesn't mean that the tension force is the same type of force as the force of gravity.

    Instead of hanging weights on the end of the rope, we could have just pulled on the end of the rope with a spring scale and noted that the rope breaks when the spring scale reading exceeds 1176 N. Now the 1176 N has nothing to do with a force of gravity. It's the maximum force the spring scale (or anything else) can exert on the rope before the rope breaks.

    So, think of the 1176 N as the maximum force the climbing person can exert on the rope before the rope breaks. If the person pulls down on the rope with 1176 N then the rope pulls upward on the person with 1176 N. This force that the rope exerts upward on the person is not a force of gravity. It's a force of interaction between the molecules in the person's hands and the molecules in the rope.
     
    Last edited: Mar 31, 2013
  17. Mar 31, 2013 #16
    Ok I think I get it now! So 1176 [up] (because of the third law) - 833 (force of gravity acting downwards on the person) = the total force. Alright thank you so much for your help today!
     
  18. Mar 31, 2013 #17

    TSny

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    Yes, you have it now. Nice work.
     
  19. Mar 31, 2013 #18
    We seem to be doing the same homework!

    http://www.lakeheadschools.ca/scvi_staff/childs/Gr11_physics_web/downloadable_content/unit3/textpdf3/phys11_3_5.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
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