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milliex51

HEEEELP!! How do I calculate the acceleration of the cart shown, given the following assumption: A frictional resistance of magnitude 2.0 N is acting on the cart.

  • 2 years ago
  • 2 years ago

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  1. milliex51
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    |dw:1324432388146:dw|

    • 2 years ago
  2. milliex51
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    How do I calculate the acceleration of the cart shown, given the following assumption: A frictional resistance of magnitude 2.0 N is acting on the cart. It first asks what the acceleration will be with no friction is acting on a cart and my formula is: T = m1a m2a = m2g - T T = m2g - m2a m1a+ m2a = m2g a = m2g / (m1 + m2)

    • 2 years ago
  3. milliex51
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    and the answer for my current problem stated above is 2.9 m/s^2... and I don't know how to solve it D;

    • 2 years ago
  4. milliex51
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    Someone? Help? :(

    • 2 years ago
  5. elica85
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    a=F/m F=driving force-resistant force what's the driving force in this case?

    • 2 years ago
  6. milliex51
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    driving force? It.. It doesn't say.

    • 2 years ago
  7. elica85
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    well i'm just asking you to think about it. it will help you in solving problems in the future. what force is causing the system to move?

    • 2 years ago
  8. milliex51
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    ohh.. Tension force.

    • 2 years ago
  9. elica85
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    is this moving to the left?

    • 2 years ago
  10. milliex51
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    to the right.

    • 2 years ago
  11. elica85
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    ok, so no it's not the tension force. it's the force from the hanging mass

    • 2 years ago
  12. milliex51
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    really? ohh.. okay.

    • 2 years ago
  13. elica85
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    gravity causes the masses to move

    • 2 years ago
  14. milliex51
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    ohh, i see it. 'Cause when the hanging mass falls the cart moves.

    • 2 years ago
  15. milliex51
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    yup.

    • 2 years ago
  16. elica85
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    so \[driving force=m _{hanging}*g\]

    • 2 years ago
  17. milliex51
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    alright.

    • 2 years ago
  18. elica85
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    now the resistant force is already given in the problem

    • 2 years ago
  19. milliex51
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    how? And is your equation the same as: a = m2g / (m1 + m2)?

    • 2 years ago
  20. elica85
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    what are you using for m2?

    • 2 years ago
  21. milliex51
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    2.0 kg

    • 2 years ago
  22. milliex51
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    and you said how the resistant force is already given in the problem-- how?

    • 2 years ago
  23. elica85
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    ok then yes, it will end up as that equation you have but i will break it down a=F/m resistant force = 2N F=2g-2N =19.6N-2N back to acceleration... a=F/mass of the system = (19.6N-2N)/(2kg+4kg)

    • 2 years ago
  24. milliex51
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    how did you get 2N as the resistant force? o.O 'cause the answer in the back of the text is 2.9 m/s^2

    • 2 years ago
  25. elica85
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    oops, actually different than yours but i'm confident in what i put up there

    • 2 years ago
  26. milliex51
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    do you know a way in which I can use my formula?

    • 2 years ago
  27. elica85
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    my final is 2.9m/s^2

    • 2 years ago
  28. milliex51
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    ohhh but do you know how to answer the problem with my formula? ...

    • 2 years ago
  29. elica85
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    tension cancels out for this system which is why i didn't use it in any of my equations

    • 2 years ago
  30. elica85
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    your formula seems a little off but i will double check

    • 2 years ago
  31. milliex51
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    ohh.. okay.

    • 2 years ago
  32. elica85
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    break down the system in 2 parts. hanging mass will be labeled 1. \[m _{1}g-T _{1}=m _{1}a\] ^^^derived from F=ma solve for T \[T _{1}=m _{1}(g-a)\] for second part... \[T _{2}-m _{2}g=m _{2}a\] solve for T... \[T _{2}=m _{2}(g+a)\]

    • 2 years ago
  33. milliex51
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    ohh..

    • 2 years ago
  34. elica85
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    so what you have to calculate a would work if there was no friction but since there is, the best way to do it is the way i did it

    • 2 years ago
  35. milliex51
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    ohh, cos that formula I showed you was for no friction & I thought maybe to switch a few places would be okay.

    • 2 years ago
  36. milliex51
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    Well then, thank you so much elica for putting up with me!! Thank you!!

    • 2 years ago
  37. elica85
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    np

    • 2 years ago
  38. milliex51
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    btw, why is it: F=2g-2N? Why is it 2g?

    • 2 years ago
  39. elica85
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    after you calculate 2g (2kg*9.8m/s^2) it will equal to 19.6N. the unit N=kg*m/s^2

    • 2 years ago
  40. elica85
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    just remember you can't add or subtract unless they have the same units

    • 2 years ago
  41. milliex51
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    ohhh, so it's like m2g?

    • 2 years ago
  42. elica85
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    yes

    • 2 years ago
  43. milliex51
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    ohh, okay. Thank you!

    • 2 years ago
  44. elica85
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    np

    • 2 years ago
  45. milliex51
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    so F = m2g - frictional resistance? (what's the variable for it?)

    • 2 years ago
  46. elica85
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    in this case, frictional force is denoted as \[F _{k}=\mu mg\] k for kinetic friction and the u-looking sign is the friction coefficient which is usually given in a problem. in this problem, the Fk was already solved (2N)

    • 2 years ago
  47. milliex51
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    ohhhh, I'm familiar with that! Oh, thank you!

    • 2 years ago
  48. elica85
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    yup

    • 2 years ago
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