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zaphod

  • 3 years ago

An acrobat of mass m slides down a vertical rope of height h. For the first three quarters of her descent she grips the rope with her hands and legs so as to produce a frictionals force equal to five-ninths of her weight. She then tightens her grip so that she come to rest at the bottom of the rope. Sketch a (t,v) graph to illustrate her descent, and find the frictional force she must produce in the last quarter.

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  1. zaphod
    • 3 years ago
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    @Algebraic! pls hlp @Yahoo! @amistre64

  2. amistre64
    • 3 years ago
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    do you know how to define weight using mass?

  3. Algebraic!
    • 3 years ago
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    hmm... good one.

  4. zaphod
    • 3 years ago
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    w = mg

  5. amistre64
    • 3 years ago
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    and the frictional force is said to be 5/9 of that right? oh, and what do t and v stand for in (t,v)?

  6. zaphod
    • 3 years ago
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    velocity . time graph

  7. amistre64
    • 3 years ago
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    |dw:1349878299626:dw|

  8. zaphod
    • 3 years ago
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    how come constant velocty?

  9. amistre64
    • 3 years ago
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    if its a constant velocity, it would be easier to determine the velocity at the start of the last quarter. Im just wondering if we should assume that its a constant velocity

  10. amistre64
    • 3 years ago
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    \[a = \frac Fm\] \[a = \frac {\frac59mg}m\] \[a = \frac59g\]

  11. amistre64
    • 3 years ago
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    \[v=\frac{5}{9}t+v_o\] can we assume an initial velocity of 0?

  12. zaphod
    • 3 years ago
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    yes lets see if the answer comes

  13. amistre64
    • 3 years ago
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    \[\Delta L=-\frac12\frac59gt^2\] \[\Delta L=-\frac5{18}gt^2\] \[\Delta \frac L3=-\frac{5gt^2}{54}\] just running thru an idea, any of this make sense?

  14. zaphod
    • 3 years ago
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    i dont think, the answer is 7/3 mg

  15. amistre64
    • 3 years ago
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    i havent gotten to the end yet ... just wondering if my thought process made sense so far

  16. amistre64
    • 3 years ago
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    assuming im correct :) the velocity for the first 3 quarters would then be:\[v=-\frac5{27}gt\]|dw:1349879118009:dw|

  17. amistre64
    • 3 years ago
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    prolly a bad graph :/

  18. zaphod
    • 3 years ago
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    hmm

  19. Algebraic!
    • 3 years ago
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    I think it goes something like: \[\frac{ 3h }{ 4 } = \frac{ 2g t^2 }{9 }\] solve for t use in \[V(t)= \frac{ 4g t }{ 9 }\] (not using signs here but everything is negative) you should get : \[V = \frac{ 3 }{ 4 }\sqrt{\frac{ 3gh }{ 2 }}\] and \[t = \frac{ 3 }{ 2 }\sqrt{\frac{ 3h }{ 2g }}\]

  20. zaphod
    • 3 years ago
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    i got the answer thanks :)

  21. zaphod
    • 3 years ago
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    u wanna see my working

  22. Algebraic!
    • 3 years ago
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    |dw:1349879825375:dw|

  23. Algebraic!
    • 3 years ago
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    dude why are you posting questions you don't need help on?

  24. zaphod
    • 3 years ago
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    i ws tryng this question for quite sometime, and i dint get a solution, so i posted it, but now i got the solution. sorry for causing u any inconvenience :/

  25. Algebraic!
    • 3 years ago
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    let's see your second part of the soln. I hope your first part agrees with mine?

  26. zaphod
    • 3 years ago
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    wht do umean? which part do u need to agree

  27. Algebraic!
    • 3 years ago
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    ...

  28. Algebraic!
    • 3 years ago
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    did you get the same results as I did for the first part of the descent, where a= -4g/9?

  29. zaphod
    • 3 years ago
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    yes :)

  30. zaphod
    • 3 years ago
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    i substitues and got friction as 7/3 mg

  31. Algebraic!
    • 3 years ago
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    wut.

  32. zaphod
    • 3 years ago
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    now there is a second part in the question..it says , if the rope is 60m, calculate her greatest s[eed ii) the time she takes to descend

  33. Algebraic!
    • 3 years ago
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    you got the force from friction = 7mg/3 ??

  34. zaphod
    • 3 years ago
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    yes i did

  35. Algebraic!
    • 3 years ago
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    ok. that means a net force upwards. since the force down is mg

  36. Algebraic!
    • 3 years ago
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    pretty funny.

  37. zaphod
    • 3 years ago
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    yeah..could u help me with the second part i posted right now :)

  38. Algebraic!
    • 3 years ago
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    you make no sense.

  39. Algebraic!
    • 3 years ago
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    what acceleration are you using for the first part?

  40. zaphod
    • 3 years ago
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    4/9 g

  41. zaphod
    • 3 years ago
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    and when the friction increases it more thn the weight, acceleration becomes - 4/3g

  42. Algebraic!
    • 3 years ago
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    by negative I guess you mean upwards.

  43. zaphod
    • 3 years ago
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    deceleration

  44. Algebraic!
    • 3 years ago
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    you want to show me your work? because if you did the first part right, the second question is trivial.

  45. Algebraic!
    • 3 years ago
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    it's just a matter of plugging in 'h'

  46. zaphod
    • 3 years ago
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    ok i applied f = ma downwards mg - 5/9 mg = ma i found a = (4/9)g now this is the acceleration till the person increases his grip so using the acceleration i found the velocity after 3/4 time where he increases his grip v = u + at v = 0 + 4/9 g * 3/4 v = (3/9) g then i found the deceleration needed for the acrobat to come to rest v = u + at 0 = 3/9g + a * 1/4 a =( -4/3)g i applied f =ma downwards mg - Fr = ma mg + (4/3)mg = Fr Fr = 7/3 mg could u help me in the second one please:)

  47. Algebraic!
    • 3 years ago
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    you did 3/4 of the time... I think when it says " the first three quarters of her descent " it means 3/4 *h .... I could be wrong on that, but I kind of doubt it.

  48. zaphod
    • 3 years ago
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    no its correct...

  49. Algebraic!
    • 3 years ago
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    I mean, you're basically just assuming the whole trip takes 1 second...

  50. zaphod
    • 3 years ago
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    yup

  51. Algebraic!
    • 3 years ago
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    so you plotted v =( (3/9) g)*t

  52. zaphod
    • 3 years ago
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    yes...

  53. Algebraic!
    • 3 years ago
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    alright, I guess that's the way they want you to do it. bit ambiguous wording in the question, but w/e.

  54. Algebraic!
    • 3 years ago
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    so second part.

  55. zaphod
    • 3 years ago
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    yes how do i do tht

  56. Algebraic!
    • 3 years ago
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    I guess you have to solve 60 = 1/2* 4g/9*(3t/4)^2 +(3g/9)(t/4) - 1/2 (4g/3)*(t/4)^2 for t... and use that time in: v = 0 + 4/9 g * 3t/4

  57. zaphod
    • 3 years ago
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    oh okay il try

  58. zaphod
    • 3 years ago
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    btw which equation did u use

  59. Algebraic!
    • 3 years ago
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    sorry about the confusion, I totally misinterpreted that problem I guess.

  60. Algebraic!
    • 3 years ago
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    I used d= Vi*t + 1/2*a*t^2

  61. Algebraic!
    • 3 years ago
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    d1= 1/2* 4g/9*(3t/4)^2 d2 = (3g/9)(t/4) - 1/2 (4g/3)*(t/4)^2 d1 + d2 = 60

  62. zaphod
    • 3 years ago
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    ok

  63. Algebraic!
    • 3 years ago
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    you get it? I got t=8 (using g=10) and V = 24/9*g

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