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Linear Impulse and Momentum

  1. Mar 5, 2015 #1
    A stop block, s, prevents a crate from sliding down a [itex] \theta = 33 \hspace{2 mm} degrees [/itex] incline. A tensile force [itex] F = (F_{o}t) N [/itex] acts on the crate parallel to the incline, where [itex] F_{o} = 325 \frac{N}{s} [/itex]. If the coefficients of static and kinetic friction between the crate and the incline are [itex] \mu_{s} = 0.325 [/itex] and [itex] \mu_{k} = 0.225 [/itex], respectively, and the crate has mass of [itex] m = 50.8 kg [/itex], how long will it take until the crate reaches a velocity of [itex] v = 2.44 \frac{m}{s} [/itex] as it moves up the incline.

    upload_2015-3-5_13-0-14.png

    2. Relevant equations


    3. The attempt at a solution
    I know when [itex] f < f_{max} = \mu_{s}mgcos(\theta) [/itex], the system isn't moving, and the net force is:
    [itex] F - mgsin(\theta) - f = 0 [/itex].
    [itex] f = F-mgsin(\theta) < \mu_{s}mgcos(\theta) [/itex]
    When [itex] t = t_{1} [/itex], [itex] f = \mu_{s}mgcos(\theta) [/itex] so I have:
    [itex] 325t_{1} - mgsin(\theta) = \mu_{s}mgcos(\theta) [/itex]
    [itex] t_{1} = \frac{mg(sin(\theta)+\mu_{s}cos(\theta))}{325} [/itex]
    Now plugging in my values:
    [itex] t_{1} = \frac{(50.8)(9.81)(sin(33)+(0.325)cos(33))}{325} [/itex]
    [itex] t_{1} = 1.25 s [/itex]
    So, after 1.25 seconds, the block moves up along the incline and the net force is:
    [itex] F - mgsin(\theta) - \mu_{k}mgcos(\theta) [/itex]
    The impulse of the net force is:
    [itex] \int_{t}^{t_{1}} (F-mgsin(\theta)-\mu_{k}cos(\theta)) = m(v-0) [/itex]
    [itex] \int_{t}^{t_{1}}(325t - mgsin(\theta)-\mu_{k}mgcos(\theta)) = mv [/itex]
    [itex] 162.5(t^{2}-t_{1}^{2}) - mg(sin(\theta)-\mu_{k}cos(\theta))(t-t_{1}) = mv [/itex]
    [itex] 162.5(t^{2}-(1.25)^{2})-177.4(t-1.25) = 123.95 [/itex]
    [itex] 162.5t^{2} - 177.4t - 126.76 = 0 [/itex]
    When I solve this quadratic, I get [itex] t = 1.58 s [/itex]. However, this answer is not correct and I am not quite sure where I went wrong. Any help is appreciated. Thanks.
     
  2. jcsd
  3. Mar 5, 2015 #2

    TSny

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    Does your answer represent the total time since t = 0? [Edit: Yes, it does. Your work looks good, but I haven't checked your numbers. Let me try it.]
     
  4. Mar 5, 2015 #3
    Yes, I'm really not sure where I went wrong on this one...I'm thinking maybe I entered a value incorrectly but I can't see where. I was thinking that having another set of eyes look over the problem would really help :)
     
  5. Mar 5, 2015 #4

    TSny

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    When you factored out -mg in the third to last line, did you mess up a sign?
     
  6. Mar 5, 2015 #5
    I believe that I did, I had:
    [itex] -mgsin(\theta)-\mu_{k}mgcos(\theta) [/itex]
    [itex] -mg(sin(\theta)-\mu_{k}cos(\theta) [/itex]
    instead of
    [itex] -mg(sin(\theta)+\mu_{k}cos(\theta)) [/itex] Thank you !
     
  7. Mar 5, 2015 #6

    TSny

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    Yes. I hope it works out now.
     
  8. Mar 5, 2015 #7
    I'm trying it now...
     
  9. Mar 5, 2015 #8
    so after making the correction, I ended up with
    [itex] 162.5t^{2} - 365.458t - 126.7645 = 0 [/itex]
    Solving that quadratic I got:
    [itex] t = 2.55 s [/itex]
    This answer is incorrect, the correct answer according to Pearson is [itex] t = 2.01 s [/itex] :/
     
  10. Mar 5, 2015 #9

    TSny

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    I get a different number in place of your -126.7645.
     
  11. Mar 5, 2015 #10
    is it 78.96 or something close to that?
     
  12. Mar 5, 2015 #11

    TSny

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    Yes.
     
  13. Mar 5, 2015 #12
    Ok, I see where I went wrong now, thank you very much!
     
  14. Mar 5, 2015 #13

    TSny

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    OK. Good.
     
  15. Mar 5, 2015 #14
    and using that number I do get 2.01 seconds after using the quadratic equation :)
     
  16. Mar 5, 2016 #15
    Why do you have the time you found in the upper limit?
     
  17. Mar 5, 2016 #16
    Why do you have the time you found in the upper limit?
     
  18. Mar 5, 2016 #17

    TSny

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    You are right, t1 should be the lower limit. When _N3WTON_ evaluated the integral, t1 was used as the lower limit. So, it's OK.
     
  19. Mar 6, 2016 #18
    Okay, thank you for that, was just making sure, I noticed that the numbers made sense.
     
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