Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

zaphod Group Title

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.

  • one year ago
  • one year ago

  • This Question is Closed
  1. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @Algebraic! pls hlp @Yahoo! @amistre64

    • one year ago
  2. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    do you know how to define weight using mass?

    • one year ago
  3. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    hmm... good one.

    • one year ago
  4. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    w = mg

    • one year ago
  5. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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)?

    • one year ago
  6. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    velocity . time graph

    • one year ago
  7. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1349878299626:dw|

    • one year ago
  8. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    how come constant velocty?

    • one year ago
  9. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  10. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \[a = \frac Fm\] \[a = \frac {\frac59mg}m\] \[a = \frac59g\]

    • one year ago
  11. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \[v=\frac{5}{9}t+v_o\] can we assume an initial velocity of 0?

    • one year ago
  12. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes lets see if the answer comes

    • one year ago
  13. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  14. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i dont think, the answer is 7/3 mg

    • one year ago
  15. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i havent gotten to the end yet ... just wondering if my thought process made sense so far

    • one year ago
  16. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  17. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    prolly a bad graph :/

    • one year ago
  18. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm

    • one year ago
  19. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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 \[t = \frac{ 3 }{ 2 }\sqrt{\frac{ 3h }{ 2g }}\]

    • one year ago
  20. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i got the answer thanks :)

    • one year ago
  21. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    u wanna see my working

    • one year ago
  22. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1349879825375:dw|

    • one year ago
  23. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    dude why are you posting questions you don't need help on?

    • one year ago
  24. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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 :/

    • one year ago
  25. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    let's see your second part of the soln. I hope your first part agrees with mine?

    • one year ago
  26. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    wht do umean? which part do u need to agree

    • one year ago
  27. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    ...

    • one year ago
  28. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    did you get the same results as I did for the first part of the descent, where a= -4g/9?

    • one year ago
  29. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes :)

    • one year ago
  30. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i substitues and got friction as 7/3 mg

    • one year ago
  31. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    wut.

    • one year ago
  32. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  33. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    you got the force from friction = 7mg/3 ??

    • one year ago
  34. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i did

    • one year ago
  35. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    ok. that means a net force upwards. since the force down is mg

    • one year ago
  36. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    pretty funny.

    • one year ago
  37. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah..could u help me with the second part i posted right now :)

    • one year ago
  38. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    you make no sense.

    • one year ago
  39. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    what acceleration are you using for the first part?

    • one year ago
  40. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    4/9 g

    • one year ago
  41. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    and when the friction increases it more thn the weight, acceleration becomes - 4/3g

    • one year ago
  42. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    by negative I guess you mean upwards.

    • one year ago
  43. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    deceleration

    • one year ago
  44. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    you want to show me your work? because if you did the first part right, the second question is trivial.

    • one year ago
  45. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    it's just a matter of plugging in 'h'

    • one year ago
  46. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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:)

    • one year ago
  47. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  48. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    no its correct...

    • one year ago
  49. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I mean, you're basically just assuming the whole trip takes 1 second...

    • one year ago
  50. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yup

    • one year ago
  51. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    so you plotted v =( (3/9) g)*t

    • one year ago
  52. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes...

    • one year ago
  53. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    alright, I guess that's the way they want you to do it. bit ambiguous wording in the question, but w/e.

    • one year ago
  54. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    so second part.

    • one year ago
  55. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes how do i do tht

    • one year ago
  56. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
  57. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    oh okay il try

    • one year ago
  58. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    btw which equation did u use

    • one year ago
  59. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    sorry about the confusion, I totally misinterpreted that problem I guess.

    • one year ago
  60. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I used d= Vi*t + 1/2*a*t^2

    • one year ago
  61. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    d1= 1/2* 4g/9*(3t/4)^2 d2 = (3g/9)(t/4) - 1/2 (4g/3)*(t/4)^2 d1 + d2 = 60

    • one year ago
  62. zaphod Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

    • one year ago
  63. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    you get it? I got t=8 (using g=10) and V = 24/9*g

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.