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Finding the tension

  1. Dec 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Garfield (a cat) and John (his human) did not go to the Olympics (not feline up to
    it). However, they got in trouble. John sits on a frictionless skateboard on a
    horizontal parking lot. A light strong rope tied to a solid post passes horizontally
    to the skate board, passes around a pulley on the skateboard, returns horizontally
    to pass above the post, around a second pulley and vertically down into a yawning
    chasm. Garfield hangs on tightly to the bottom end of the rope. Garfield has a
    mass of 20 kg (too much lasagna), John and the skateboard have a combined mass
    of 60 kg. Calculate the tension in the rope before John hits the post or Garfield
    bottoms out.


    3. The attempt at a solution

    My FBD diagrams

    http://s1176.beta.photobucket.com/user/LolaGoesLala/media/c_zps40e98e7e.png.html

    Ok so this is what i came up with to find the tensin between the cat and the by and skateboard

    m
    Fnet = Fg - T
    ma = md - T

    M
    Fnet = Fg + Fn + T
    Fnet = T
    Ma = T

    Combining
    a = (mg/m+M)

    To find the tension
    Ma = T
    M(mg/m+M) = T
    T = (M x m/M +m)g
    T = ( 60000g x 20000g/60000g + 20000g)(9.8m/s^2)
    T = 147000N

    so that is the tension between john/skateboard and the cat
    however how do i account for the nail and the tension between john/skateboard and the nail ?
    Here is a rough sketch of what i think it looks like

    http://s1176.beta.photobucket.com/user/LolaGoesLala/media/f_zpsafd5a9e3.png.html
     
  2. jcsd
  3. Dec 30, 2012 #2

    gneill

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    Staff: Mentor

    Is there a mechanical advantage involved? How many rope segments are pulling on each mass? How might this affect your FBD?

    Are the accelerations of the two objects the same? If not, how are they related?
     
  4. Dec 30, 2012 #3
    Wait are those questions that you are asking me? or? Because all its asking for is the tension before john/skateboard hit the post
     
  5. Dec 30, 2012 #4

    gneill

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    Staff: Mentor

    You need to address the questions in order to solve the problem correctly; their answers affect the equations you write to determine the tension.
     
    Last edited: Dec 30, 2012
  6. Dec 30, 2012 #5
    I looked at all the prospects and then found the equations
     
  7. Dec 30, 2012 #6

    gneill

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    Staff: Mentor

    I would suggest that you reconsider the net force acting on John and the related rates of the motions of John and Garfield. The pulley arrangement makes a difference.
     
  8. Dec 30, 2012 #7
    So are you trying to say that for john and the skateboard i would need to include the tenion twice since there is another piece of tension connected to the nail/post.
    So like this?

    M
    Fnet = Fg + Fn + T + T
    Fnet = T+ T
    Ma = T^2
     
  9. Dec 30, 2012 #8

    gneill

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    Staff: Mentor

    That's the idea, yes.
    Well, you should keep X and Y components separate (Fg and Fn act in a different direction than the T's), but the sum of the forces acting in the X-direction will be T + T. That's 2T (not T squared).

    Then consider how the accelerations of the two masses are related; the pulley arrangement will impose a relationship between them --- they are not equal.
     
  10. Dec 30, 2012 #9
    if i do that it does not work, plz see my process

    m
    Fnet = Fg - T
    ma = md - T

    M
    Fnet = Fg + Fn + T + T
    Fnet = T + T
    Ma = 2T

    Combining
    a = (mg + T/m+M)

    To find the tension
    Ma = 2T
    M(mg+T/m+M) = 2T
    Mmg + MT = 2T (m+M)
    Mmg + MT = 2Tm + 2TM
    Mmg = 2Tm + TM
    Mmg = T(Tm +M)

    Like this is confusing me?
     
  11. Dec 30, 2012 #10

    gneill

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    Staff: Mentor

    What is the 'd' in 'md'? Shouldn't that be mg?
    Okay, BUT!!! The a's for the two masses are NOT THE SAME. They are related, but they are not the same value. The pulley arrangement makes them unequal. Check: If the cat descends by some distance 'y', by how much distance 'x' will the skateboard move?

    Write an equation relating the two accelerations (And label them differently. Maybe a1 and a2).
    I don't see how you can write the above if they don't share the same acceleration. You need to sort out the relationship between the two accelerations before you can meaningfully combine the equations.
     
    Last edited: Dec 30, 2012
  12. Dec 30, 2012 #11
     
  13. Dec 30, 2012 #12

    gneill

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    Staff: Mentor

     
  14. Jan 4, 2013 #13
     
  15. Jan 4, 2013 #14
  16. Jan 4, 2013 #15

    gneill

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    Staff: Mentor

    The rope has only one tension for its entire length, since the pulleys are massless and frictionless.

    Take a close look at the forces acting on the skateboard pulley. Can you assign a tension to each of the "connections" there? In other words, how does your T2 relate to T?

    Also, you still haven't stated the relationship that exists between the motion of the cat and the motion of the skateboard; how far does the skateboard move when the cat moves by some amount?
     
  17. Jan 4, 2013 #16
    Well i have two ways i can do this. Since you said that the "rope has only one tension for its entire length, since the pulleys are massless and frictionless." I am going to assign the T2 to the little rope that connects the skateboard and the pulley. Well if the cat moves down 2m. the skateboard will move ahead 1m right? its a 1:2 ratio
     
  18. Jan 4, 2013 #17

    gneill

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    Staff: Mentor

    Yes, but what is the relationship between T2 and T? Isolate that pulley and draw the FBD for it.
    Correct. That should tell you the ratio of a2 to a1 also (and v2 to v1 if it's of interest).
     
  19. Jan 4, 2013 #18
    well for the T2 and T1

    the ratio i came up with was T2 = 2T1

    like this:
    http://s1176.beta.photobucket.com/user/LolaGoesLala/media/dfdsdsd_zps2ce0a7cd.png.html

    so since T2 = 2T1 and a1 = 2a2

    (1) m1a1 = m1g - T
    (2) m2a2 = T2

    Times the first equation by two

    2m1a1 = 2m1g - 2T
    m2a2 = 2T

    2m1(2a2) = 2m1g
    m2a2

    4m1a2 + m2a2 = 2m1g
    a2 = 2m1g / (4m1 +m2)

    when solving
    a2 = 2(20kg)(9.8m/s^2) / 4(20kg)+(60kg)
    a2 = 2.8 m/s^2

    since a1 = 2a2
    a1 = 5.6 m/s^2

    and using the first equation
    m1a1 = m1g - T
    T = m1g - m1a1
    T = (20kg)(9.8m/s^20 - (20kg)(5.6m/s^2)
    T = 84 N
     
  20. Jan 4, 2013 #19

    gneill

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    Staff: Mentor

    That looks good! Well done.
     
  21. Jan 4, 2013 #20
    THANK YOU :D
    it would have not been done without your help. :)
     
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