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[CHECK] tension problem a little help

  1. May 1, 2008 #1
    [CHECK] tension problem!! a little help

    A 2.0 kg ball is attached to a vertical post with two strings, one 2.0 m long and the other 1.0 m long as shown in the figure. If the ball is set whirling in a horizontal circle, what is the minimum speed necessary for the lower string to be taught? If the ball has a constant speed of 6 ms-1, find the tension on each string.


    |\
    | \
    | \ -2.0m
    | \
    | \
    |----0 -2.0kg
    |
    | ^1m


    For the first part, figured that the 2m string has two components of tension Tsin60 and Tcos60.
    Tsin60=mg=19.6
    Tcos60=mv^2/r=2v^2
    tan60=19.6/2v^2
    v=5.66 m/s

    is that right? I calculated the angle that the 2m string made w/ the x-axis and went from there...

    Now..for the second part..
    i figured that the tension in the 1m string is the centripital force which is:
    T=mv^2/r=2*36/1= 72N
    Now for the second string..what would the tension be?
    i'm guessing i have to use the Tsin60 and the Tcos60
    Tsin60=19.6N
    Tcos60=72N
    T=sqrt(19.6^2+72^2)
    T=74.6N

    I hope this is right, i did all the work, i'm just looking for confirmation please
     
  2. jcsd
  3. May 1, 2008 #2

    tiny-tim

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    Hi warmfire540! :smile:

    How far apart are the ends of the two strings? :confused:
     
  4. May 1, 2008 #3
    They are 1.73m apart using pythagorean's theorem
     
  5. May 2, 2008 #4

    tiny-tim

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    Hi warmfire540! :smile:

    I asked because your diagram doesn't look like that!
    That's fine … except you forgot to take the square-root. :rolleyes:
    Nooo … the centripetal acceleration is matched by the horizontal components of all the forces … in this case, the tensions in both strings.

    Try again! :smile:
     
  6. May 2, 2008 #5
    Hmm, i got v^2=19.6/2tan60
    v^2=5.66
    v=2.38m/s



    So, i don't understand then what the centripital force would be, how does the 2m string interfere with this? are my tensions wrong for part 2? Do i have part 1 right at least?

    THANKS!!
    -warmfire540
     
  7. May 2, 2008 #6

    tiny-tim

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    Hi warmfire540! :smile:

    Yes, part 1 is right now! :smile:

    For part 2, do the vertical components first … that gives you the tension in the top string, doesn't it?

    Then do the horizontal components. :smile:
     
  8. May 2, 2008 #7
    Well..the vertical component for the 2m string is:
    Tsin60

    for the 1m string
    None

    The horizontal component for the 2m string is:
    Tcos60

    for the 1m string
    mv^2/r

    Now what? how would i figure out the tensions in each string?
     
  9. May 3, 2008 #8

    tiny-tim

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    unknowns

    Hi warmfire540! :smile:

    You really don't get unknowns, do you?

    For problems like this, draw a clear diagram.

    Label all the known lengths, forces, etc with numbers.

    Label all the unknown ones with letters.

    It's the only way to do it without making mistakes.

    In this case, you've already labelled the tension in the top string T.

    So label the tension in the bottom string U (for want of a better letter).

    Then instead of:
    you should write:
    The horizontal component for the 1m string
    U

    You now have two equations … on the left-hand of each, are the total components forces in each direction … on the right-hand side, the acceleration in that direction! :smile:

    Don't try to do these things in your head, or make assumptions … in this case, you somehow assumed the 1m string was supplying the centripetal acceleration, and because you hadn't given it a letter, you couldn't see a way out of it!​
     
  10. May 3, 2008 #9
    So, given v=6m/s how would i find the two tensions on the strings
    I suppose now i don't understand ALL the forces that are acting on both strings
    I have drawings down but still don't get it
    does the 2m string affect the 1m string? i figured only the angle on the 2m string was affected?

    i can't figure out the tensions on both strings..
     
  11. May 3, 2008 #10

    tiny-tim

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    Hi warmfire540! :smile:
    ooh … that's not a sensible physics question … physics is equations … you can only ask "do these two forces appear in the same equation?"

    As I said:
    So … just do the equations in that order …

    first, what is T (you don't need v for that)? :smile:
     
  12. May 3, 2008 #11

    Well T is the tension..
    for the 2m string Tsin60+tcos60=Tension
    for the 1m string
    There is no vertical component, only horizontal
    so T=mv^2/r right?
    Tension=2*36/1
    Tension=72N for the smaller string
    UNLESS some force in the 2m string affects this 72N
    that's where i'm confused..is there a force making 72N greater or smaller?
     
  13. May 3, 2008 #12

    tiny-tim

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    No! :frown:

    T is the tension in the top string.

    To find it, just take vertical components of force. :smile:
     
  14. May 4, 2008 #13

    Blah! confused..
    Well, one string at time..the T (T) in the 2m string...:
    T(2m)=Tcos60i+Tsin60j = 0.5Ti+0.87Tj
    Obviosly I need to find T

    for the 1m string:
    U=mv^2/r = 72Ni

    The sum of the x-componenets, and the sum of y-componenets must be zero since there is no net force right?

    I don't know..i'm just confusing myself trying to find the tensions of both strings
     
  15. May 4, 2008 #14

    tiny-tim

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    Hi warmfire540! :smile:
    Bad idea. :frown:

    I agree that forces are vectors, and that what you have written is completely correct.

    But in practice, it's usually a disaster to try to deal with more than one direction at a time.

    So forget these vectors … just use one component at a time! :smile:

    In this case, as you say:
    So Tsin60º = mg.

    Note that the reason we have done the y-compoonents first is that U and the acceleration are both horizontal … which makes the vertical equation particularly easy!

    ok … that gives you T on a plate.

    Now do the x components … that's more of a dog's dinner, with T U and the acceleration all taking part. But you know T by now, so that'll be ok. :smile:
     
  16. May 4, 2008 #15

    okay okay
    so this is because the the ball is not changing it's direction vertically! right..the net force must be zero
    So
    Tsin60=mg
    T=mg/sin60
    T=22.6N
    ^^That is the tension in the 2m string..final.. velocity won't effect this...(right?)


    Well T=22.6, U=centripital force?
     
  17. May 4, 2008 #16
    It may be hard to intuitively imagine that the top string AND the bottom string are providing the centripetal force, but you have basically proven it to yourself now. Look at the tension T you found, the total force is 22.6N but the force in the y or vertical direction is only mg, so where is the rest of that string's force going?

    --Bob
     
  18. May 5, 2008 #17

    tiny-tim

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    Bingo! :biggrin: You're getting the hang of this! :smile:
    S'right! Independent of velocity. :smile:
    Nooo … don't take short-cuts …

    (btw, it's "centripetal", with an e … the "pet" part means "seeking", as in "petition")
    You need an equation with T and U on the left-hand-side, and acceleration on the right-hand-side.

    Don't you? :biggrin:
     
  19. May 5, 2008 #18


    hmmm..

    mv^2=Tcos60+U
    this is the centripetal force equals the x component of the 2m string and the tension on the 1m string (U)..
     
  20. May 6, 2008 #19

    tiny-tim

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    That's it! :smile:

    (dont forget the "/r")
     
  21. May 6, 2008 #20
    okay! yayy :biggrin:

    so mv^2/r=Tcos60+U
    72=22.6cos60+u 22.6cos60 = 11.3
    6.4=u
    the tension in the 1m string is 6.4N
    is that all? that's it?!
    problem solved?
     
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