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1018

  • one year ago

please help with #s 3 and 4. thanks! (problems are in the picture below)

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  1. 1018
    • one year ago
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  2. welshfella
    • one year ago
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    Not sure about number 3 but i can help with number 4

  3. welshfella
    • one year ago
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    |dw:1442848903125:dw|

  4. welshfella
    • one year ago
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    let a be acceleration , t tension in the string, coefficient of friction = 0.45 equation of motion of block A_ Newtons 2nd law; T - 0.45*2.25g = 2.25 * a equation of motion of block B:- 1.3g - T = 1.3 * a

  5. welshfella
    • one year ago
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    solve these 2 equations to find T and a

  6. welshfella
    • one year ago
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    you can find the speed of the bodies after 3 cm by using one of the equations for constant acceleration

  7. welshfella
    • one year ago
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    you can use g = 9.81N

  8. 1018
    • one year ago
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    i got 2.97 for acceleration

  9. 1018
    • one year ago
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    is that correct? i got 2 equations from each block then combined those two equations

  10. welshfella
    • one year ago
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    i got 0.794 for acceleration

  11. welshfella
    • one year ago
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    and Tension = 11.72N

  12. 1018
    • one year ago
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    can i see your work? thanks

  13. 1018
    • one year ago
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    i have a question. in block B, i made the equation Tension - Weight = ma

  14. 1018
    • one year ago
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    T=W+ma

  15. welshfella
    • one year ago
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    T - 9.933 = 2.25a 12.753 - T = 1.3a adding eliminates T 2.82 = 3.55a a = 0.794 ms-2

  16. 1018
    • one year ago
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    yes, but shouldnt be tension positive, and weight is negative? or am i wrong with my analysis?

  17. welshfella
    • one year ago
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    you have to take account of the friction the frictional force = 0.45 * wieght

  18. 1018
    • one year ago
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    no, for block B. there's no friction

  19. welshfella
    • one year ago
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    the tension is positive for 2.25 block and negative for block B

  20. welshfella
    • one year ago
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    oh sorry block B force = weight - T

  21. 1018
    • one year ago
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    im still confused. haha. wait let me draw

  22. 1018
    • one year ago
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    |dw:1442851287028:dw|

  23. 1018
    • one year ago
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    the direction of the tension is upward right, and weight is downward. so i thought its summation of forces would be T-W=ma

  24. 1018
    • one year ago
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    or is this analysis wrong?

  25. welshfella
    • one year ago
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    the resultant force acts downwards weight acts down tensions acts up

  26. 1018
    • one year ago
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    so, tension is positive right, if it acts upward?

  27. welshfella
    • one year ago
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    No it w - t = ma

  28. welshfella
    • one year ago
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    don't worry too much about positive or negative the point is that the motion is downwards and there for the net force acts downwards

  29. 1018
    • one year ago
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    haha, ok. i'll just try it again. but when i get the acceleration, i can just sub it to one of the equations right? to get tension

  30. welshfella
    • one year ago
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    yes

  31. 1018
    • one year ago
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    and then for speed, just use one of the constant acceleration formulas?

  32. welshfella
    • one year ago
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    yea m use v^2 = u^2 + 2as

  33. welshfella
    • one year ago
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    u = iniital velocity = 0

  34. welshfella
    • one year ago
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    sorry i gotta go right now i think you can finish it OK

  35. 1018
    • one year ago
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    thanks!

  36. welshfella
    • one year ago
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    yw

  37. welshfella
    • one year ago
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    For number 3 , as the speed is constant I think the force applied will equal the tension in the string. In other words the forces are in equilibrium. what do you think?

  38. 1018
    • one year ago
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    wait let me check 3 again

  39. 1018
    • one year ago
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    the first thing i did is to make resultant forces = 0 for it is constant speed. is that right?

  40. welshfella
    • one year ago
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    yes thats what i meant by equilibrium

  41. 1018
    • one year ago
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    oh i think i see what you're saying. my solution is not like that. it has frictions

  42. welshfella
    • one year ago
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    well frictions must play a part

  43. 1018
    • one year ago
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    yep but the F is not equal to the tension in my solution

  44. welshfella
    • one year ago
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    the force must be just enough to overcome the frictions

  45. 1018
    • one year ago
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    but the strings on A and B should be equal right? they have the same tension

  46. welshfella
    • one year ago
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    yes

  47. welshfella
    • one year ago
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    is the force equal to the sum of the frictional forces on A and B?

  48. welshfella
    • one year ago
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    that would be 1.8*0.3 + 4.6*0.3?

  49. 1018
    • one year ago
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    here's what i did : Block B : Fx = -F + Tension + friction 1 + friction 2 = 0

  50. 1018
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    pretty much yes, but i included tension

  51. welshfella
    • one year ago
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    yea to be honest i'm not sure of this one.

  52. 1018
    • one year ago
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    haha, that's ok i'll try to review with other questions first. haha. thanks though! big help.

  53. welshfella
    • one year ago
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    yw

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