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carlj0nes

  • one year ago

By using mathematical modelling wit vectors, explain how the acceleration of the object can be included to determine the value of the force P. I will attach my answer for the question without acceleration, but i need to modify it to explain how to include acceleration

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  1. carlj0nes
    • one year ago
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  2. acxbox22
    • one year ago
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    @dan815

  3. TheSmartOne
    • one year ago
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    @dan815

  4. TheSmartOne
    • one year ago
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    @Michele_Laino (when you come online later on)

  5. dan815
    • one year ago
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    okay , so do you want me to show you another way to do it? or check your work?

  6. carlj0nes
    • one year ago
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    my work is correct for non acceleration but for the same question I have been asked to By using mathematical modelling with vectors, explain how the acceleration of the object can be included to determine the value of the force P.

  7. carlj0nes
    • one year ago
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    So im assuming it is to tweak the current question to add acceleration

  8. carlj0nes
    • one year ago
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    out of interest was does Fnet refer to

  9. dan815
    • one year ago
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    |dw:1435246603669:dw|

  10. dan815
    • one year ago
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    |dw:1435246874207:dw|

  11. dan815
    • one year ago
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    that says Fapp = P

  12. carlj0nes
    • one year ago
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    Cheers @dan815 i was watching as you was doing it. I have just wrote it up neat and handed it in. Hopefully i should get feedback today.

  13. carlj0nes
    • one year ago
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    I will be asking another question to the qualified help soon which is linked with this post. posted at the bottom of this http://openstudy.com/study#/updates/557d6ee3e4b0e4e582aa72cd

  14. carlj0nes
    • one year ago
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    @dan815 I have just had feedback and apparently it isn't what he is after.|dw:1435259141823:dw| I have attached a drawing of what he wants and an equation

  15. dan815
    • one year ago
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    okay right, that is what we are doing

  16. dan815
    • one year ago
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    add all the forces

  17. dan815
    • one year ago
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    oh actuallly there was something that wasnt considered properly

  18. dan815
    • one year ago
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    once the block starts moving up, the force of friction is not acting the other way so, we have to subtract that

  19. dan815
    • one year ago
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    besides that the equations are same for all forces

  20. carlj0nes
    • one year ago
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    N=187.836 Weight is 16.62Kg therefore mg =163.042 u=0.4 P=119.772

  21. dan815
    • one year ago
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    let me check what i get

  22. dan815
    • one year ago
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    |dw:1435260170676:dw|

  23. dan815
    • one year ago
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    sub the values in see if that gives the same answer too

  24. dan815
    • one year ago
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    just make sure your signs are right for all of them

  25. carlj0nes
    • one year ago
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    Angle=0 (using zero for ease on here) =14.5 pcos0=115.95 mgsin0=40.822 uN=0.4x187.836=75.1344

  26. carlj0nes
    • one year ago
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    what do we use for v. is that the acceleration.

  27. carlj0nes
    • one year ago
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    @dan815 would you be able to check please.

  28. Michele_Laino
    • one year ago
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    in your first problem we have to apply this vector formula: \[\Large {\mathbf{N}} + {\mathbf{R}} + {\mathbf{P}} + {\mathbf{W}} = {\mathbf{0}}\] |dw:1435261769664:dw|

  29. Michele_Laino
    • one year ago
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    R is the friction force

  30. Michele_Laino
    • one year ago
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    If we rewrite that vector equation, using components, we get these subsequent scalars equations: \[\Large \left\{ \begin{gathered} - mg\sin \theta + P\cos \theta + \mu mg = 0 \hfill \\ N - mg\cos \theta - P\sin \theta = 0 \hfill \\ \end{gathered} \right.\]

  31. Michele_Laino
    • one year ago
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    which you can solve for P and N

  32. Michele_Laino
    • one year ago
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    that above is the case without acceleration

  33. Michele_Laino
    • one year ago
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    If we want to include an acceleration say a, then the new vector equation is: \[\Large {\mathbf{N}} + {\mathbf{R}} + {\mathbf{P}} + {\mathbf{W}} = m{\mathbf{a}}\]

  34. carlj0nes
    • one year ago
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    but are some of these opposing forces

  35. Michele_Laino
    • one year ago
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    So, if the vector a is directed at the top of the inclined plane, namely: |dw:1435262337778:dw|

  36. Michele_Laino
    • one year ago
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    then the new scalars equations are: \[\Large \left\{ \begin{gathered} - mg\sin \theta + P\cos \theta + \mu mg = ma \hfill \\ N - mg\cos \theta - P\sin \theta = 0 \hfill \\ \end{gathered} \right.\]

  37. Michele_Laino
    • one year ago
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    all those forces are the external forces acting on our box of mass m

  38. carlj0nes
    • one year ago
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    @Michele_Laino would you be able to check to see if I have understood what you have done. Sorry about the quality but I will re write it once I know its correct.

  39. Michele_Laino
    • one year ago
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    what is the value of \theta?

  40. carlj0nes
    • one year ago
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    angle is14.5 deg

  41. Michele_Laino
    • one year ago
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    ok! your equations are right. So substituting the value of \theta, we get: \[\Large \left\{ \begin{gathered} - 0.250mg + 0.968P + \mu mg = ma \hfill \\ N - 0.968mg - 0.250P = 0 \hfill \\ \end{gathered} \right.\] what is the value of a?

  42. carlj0nes
    • one year ago
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    I don't have a value for acceleration as it is to produce the equation that can be used. mg=163.042 p=119.772 u=0.4 and N =187.836

  43. Michele_Laino
    • one year ago
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    ok! Then we have to substitute P, \mu and mg, into the first equation. After a simple computation I get this: \[\Large ma = 140.43\]

  44. Michele_Laino
    • one year ago
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    are your forces measured in Newtons

  45. carlj0nes
    • one year ago
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    yes

  46. carlj0nes
    • one year ago
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    I also got that answer on the picture I uploaded a few minutes ago

  47. Michele_Laino
    • one year ago
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    then we have this value: \[\Large a = \frac{{140.43}}{{16.62}} \cong 8.45\;{\text{m/se}}{{\text{c}}^{\text{2}}}\]

  48. carlj0nes
    • one year ago
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    so what you are saying the acceleration speed is 8.45m/sec^2

  49. Michele_Laino
    • one year ago
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    I'm saying that the acceleration is toward the top of the inclined plane, and its magnitude is: 8.45 m/sec^2

  50. carlj0nes
    • one year ago
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    thank you for your help. Im now going to sleep as I have an early start as im helping the school take children to the science museum. I will try and write this up neat tomorrow then.

  51. Michele_Laino
    • one year ago
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    :)

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