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Tension of rope

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  1. Nov 5, 2015 #1
    1. The problem statement, all variables and given/known data
    1.jpg

    2. Relevant equations
    My assumption:-
    if one of the ropes have been cut, it mean the object just hang VERTICALLY with one rope with 5N load on it.Is it my assumption is correct?

    3. The attempt at a solution
    Tension of the other rope = 5N (is it correct)?? base on my assumption
     
  2. jcsd
  3. Nov 5, 2015 #2

    haruspex

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    I would assume they mean the tension at the instant after cutting the other string.
     
  4. Nov 5, 2015 #3

    phinds

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    Have you studied force vectors? I think your assumption is half correct and your conclusion is wrong. What you are right about is the hanging vertically. All of your statements about force are wrong.
     
  5. Nov 5, 2015 #4

    phinds

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    Hm ... could be. I didn't interpret it that way but I guess that could be right. stuc, my comments are not based on that assumption.
     
  6. Nov 5, 2015 #5
    ...the tension at the instant after cutting the other string

    how to calculate it? it mean my assumption is wrong

    i just check an answer..the answer is 5N...but how to get it
     
  7. Nov 5, 2015 #6

    haruspex

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    Hmm.. that does not fit with either interpretation. But I remain of the view that it's the instantaneous value they're after.
    First, can you figure out the weight of the mass?
     
  8. Nov 5, 2015 #7

    phinds

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    You have not answered my question about whether or not you have studied force vectors.
     
  9. Nov 6, 2015 #8
    there is no weight given.:H
     
  10. Nov 6, 2015 #9
    learn but always confuse when try to solve it
     
  11. Nov 6, 2015 #10

    phinds

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    What is the definition of a Newton?
     
  12. Nov 6, 2015 #11
    1N = the force needed to accelerate 1kg of mass
     
  13. Nov 6, 2015 #12
    1 ans.jpg

    Attached is the calculation..any one can verify my ans (4.33N) because the given ans in 5N.

    Update:Wrong Arrow for Fy
     
    Last edited: Nov 6, 2015
  14. Nov 6, 2015 #13
    Untitled.png
    Attached other option ans
     
  15. Nov 6, 2015 #14

    haruspex

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    As I posted, the first step is to determine the weight of the mass. No, it's not given, but there is enough information to find it. To do this, just consider the initial set up with two ropes. Write the equation for balance of forces in the vertical direction.
     
  16. Nov 6, 2015 #15
    You got 4.33 N for the upward component of force contributed by each wire. But there are two wires. So what is the weight of the object? If this weight has to be supported by only one wire, what would the tension in that wire have to be?

    Chet
     
  17. Nov 6, 2015 #16
    My step to determine the weight of the mass..is it ok? (weight of the mass = 7.071N)
    Snapshot.jpg
     
  18. Nov 6, 2015 #17
    weight of the mass = 7.071N
    Snapshot.jpg
     
  19. Nov 6, 2015 #18

    haruspex

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    No, that doesn't work. Pythagoras applies when two vectors at right angles are to be added.
    What is the component of each tension in the vertical direction?
     
  20. Nov 6, 2015 #19
    :nb)..tq for the info..will be calculate
     
  21. Nov 6, 2015 #20

    haruspex

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    Hint: you already calculated it in post #12.
     
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