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Hanging Meter Stick

  1. Apr 12, 2004 #1
    A meter stick (L = 1 m) with mass M = 0.409 kg is supported by two strings, one at the 0 cm mark and the other at the 90 cm mark.

    a) What is the tension in the string at 0 cm?
    b) What is the tension in the string at 90 cm?
    c) Now the string at 0 cm is cut. What is the tension in the remaining string immediately thereafter?

    I got parts a and b figured out, but I can't get c...Can I please get some help...The attachment is a picture of what the problem looks like.
     

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  2. jcsd
  3. Apr 12, 2004 #2

    chroot

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    Part c is the easiest. The meterstick just becomes a generic weight of 0.409 kg hanging from a string. The tension in the string is just 0.409 g.

    - Warren
     
  4. Apr 12, 2004 #3
    No, that's answers not correct...I thought that it would work out like that too, and the tension would be equal to the total weight of the meter stick, but this answer is not right...
     
  5. Apr 12, 2004 #4
    This is what the help for the problem says, but I still don't understand it...

    HELP: Draw a FBD to find all the forces and torques. A convenient axis for computing the torques is the position of the remaining string.
    HELP: Unlike the previous parts, neither the net force nor the net torque vanishes. The net force leads to an equation for the acceleration of the CM. The net torque leads to an equation for the angular acceleration. The acceleration and angular acceleration are related to each other (HOW?). Put all this together to find the tension.
     
  6. Apr 12, 2004 #5

    chroot

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    Sorry, you're right -- I missed the word "immediately" in the question. I'm afraid I don't have time to figure this one out right now. :sad:

    - Warre
     
  7. Apr 12, 2004 #6
    10*0.409g/9 maybe? (i.e 4.454N for g = 9.8)

    Ahh, I made the same mistake. :redface:
     
  8. Apr 12, 2004 #7
    no problem, can someone just help me out when you get the chance, i can't get this to work out in any way...I don't know where I'm going wrong...
     
  9. Apr 12, 2004 #8

    HallsofIvy

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    By the way, chroot, I think we should be careful, when responding to people's homework questions, to distinguish between "mass" and "weight". I would certainly never say a "weight of 0.409 kg " or refer to tension in terms of grams.
     
  10. Apr 12, 2004 #9
    Actually I think chroot was saying the tension would be 0.409kg times g, i.e mg.
     
  11. Apr 12, 2004 #10

    Doc Al

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    When the left string is cut, the stick pivots about the 90cm mark. Take torques about that point:
    (1) mgx = I α (Where "x" is the distance from the pivot to the center of mass of the stick; I is the rotational inertia of the stick about the pivot.)
    Note that the acceleration of the c.m. is related to α by a = αx. Thus you can solve for "a".

    Now consider the forces acting on the stick:
    (2) mg - T = ma
    This will give you T, the tension in the remaining string.
     
  12. Apr 12, 2004 #11

    krab

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    I seem to be in a minortiy here, but it appears obvious to me that immediately after cutting string1, string2 is unaffected. So answer in c. is same as answer in b.
     
  13. Apr 12, 2004 #12

    Doc Al

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    Interesting point. I guess it depends on what is meant by immediately. It will take some time for the tension in the string to readjust to the new demand on it.
     
  14. Apr 12, 2004 #13

    chroot

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    What I meant was "a generic weight," i.e. a piece of lead, of mass 0.409 kg.

    - Warren
     
  15. Apr 12, 2004 #14

    Thanks for the help everyone, I figured out the answer to the problem...
     
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