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Force vs. time graph (Impulse)

  1. Oct 16, 2010 #1

    JJBladester

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    1. The problem statement, all variables and given/known data

    A 4-lb collar which can slide on a frictionless vertical rod is acted upon by a force P which varies in magnitude as shown. Knowing that the collar is initially at rest, determine its velocity at (a) t = 2s (b) t = 3s.

    Answers:
    (a) 29.0 ft/s
    (b) 77.3 ft/s

    impulse%20momentum%20vertical%20rod.jpg

    2. Relevant equations

    [tex]mv_1+\sum Imp_{1\rightarrow 2}=mv_2[/tex]

    3. The attempt at a solution

    [itex]mv_1+[/itex] Area under F vs. t graph [itex]=mv_2[/itex]

    [tex]v_2=\frac{\left (\frac{1}{2} \right )(2)(10)}{\frac{4}{32.17}}=80.43ft/s[/tex]

    I did not get the book's answer of 29.0 ft/s for part (a). Why?
     
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  3. Oct 16, 2010 #2

    PhanthomJay

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    I don't get either answer. Don't forget that the net force acting on the collar must include the weight of the collar acting down.
     
  4. Oct 16, 2010 #3

    JJBladester

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    Right, forgot about the weight force and its associated impulse (Wt).

    [itex]mv_1+[/itex] Area under F vs. t graph [itex]+Wt = mv_2[/itex]

    [tex]v_2=\frac{0+\left (\frac{1}{2} \right )(2)(10) + (4)(2)}{(4/32.17)}=144.8ft/s[/tex]

    Closer, I believe... But not the book's answer.
     
  5. Oct 16, 2010 #4

    PhanthomJay

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    But this is still incorrect...the applied force acts up and the weight acts down...
     
  6. Oct 16, 2010 #5

    JJBladester

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    [itex]mv_1+[/itex] Area under F vs. t graph [itex]-Wt = mv_2[/itex]

    [tex]v_2=\frac{0+\left (\frac{1}{2} \right )(2)(10) - (4)(2)}{(4/32.17)}=16.09ft/s
    [/tex]

    Closer...
     
  7. Oct 16, 2010 #6
    mmm closer....

    stop doing formulas and think.
    If the collar weight was 100 lb, what would be the speed after 2s or 3s, or 10s, 20s or any seconds ?
     
  8. Oct 16, 2010 #7
    I'm not familiar with the units used in this question and I had to read between the lines to use the 32.17 factor.Non SI units are a pain.Anyway my answers agree with the books answers.
     
  9. Oct 16, 2010 #8

    JJBladester

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    The speed (velocity) at any time would be vi-gt, if we weren't including an external force P. There is an external force present, so there's more to it than that.
     
  10. Oct 16, 2010 #9

    PhanthomJay

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    And that's what I get also!
     
  11. Oct 16, 2010 #10
    The collar will not start to lift/move until F becomes equal to 4lb.Think graph and shifting your x axis upwards.
     
  12. Oct 16, 2010 #11
    Now you are sitting on a chair I suppose. If a force of 4 lb was applied to you upward, would you be moving at vi-gt ???
     
  13. Oct 16, 2010 #12

    JJBladester

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    I weigh 170lb, so it would take 170lb of force to counter my weight, then any additional force would start me upward.
     
  14. Oct 16, 2010 #13
    The collar behaves no differently than you.
     
  15. Oct 16, 2010 #14
    For your graph think of the relevant resultant force on the collar.This doesn't vary from 0 to 10 in 2 seconds and then levels off,it varies from 0 to six(10-4) in 1.2 seconds and then levels off.
     
  16. Oct 16, 2010 #15

    JJBladester

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    I think I'm starting to understand this. It takes the pulling force P 4lb to even get the collar started.

    From the graph in the problem, we have y=5x from t=0 sec to t=2 secs. So, in order for the pulling force P to even move the collar, we need to know what the time of the equation is when pulling force = 4lb. It turns out to be 4/5, or 1.2.

    [tex]\frac{\frac{1}{2}(1.2)(6)+(0.8)(10)-(4)(2)}{(4/32.17)}=29 ft/s[/tex]

    The first term in the numerator is P acting from t=0 to t=1.2 seconds to counteract the collar's weight. The 2nd term is what happens from 1.2 seconds to 2 seconds (after the weight of the collar is overcome), and the third term in the numerator is to subtract the weight*time of the collar. The denominator is the mass of the collar.
     
  17. Oct 16, 2010 #16
    you have the right answer for part a but I am not sure how you got there.In the first 0.8 second the collar does not move.In the next 1.2 seconds it does move being subjected to a resultant force which varies from 0 to 6 in 1.2 seconds.The force then remains constant at 6 for a further 1 sec.Look at the graph you presented in post 1 and resketch it by drawing a line parallel to the x axis and meeting the y axis at 4.The change of momentum is then equal to the area under the graph going down to the horizontal line you have just drawn.
     
  18. Oct 16, 2010 #17

    JJBladester

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    Now that's a good explanation. Whew, long tired day... Think I have the flu... But to all who responded for guidance, THANK YOU. Full solution posted here (with the book's answers being obtained):

    impulse%20momentum%20collar%20graph.png
     
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