## anonymous 4 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.

1. anonymous

@Algebraic! pls hlp @Yahoo! @amistre64

2. amistre64

do you know how to define weight using mass?

3. anonymous

hmm... good one.

4. anonymous

w = mg

5. amistre64

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. anonymous

velocity . time graph

7. amistre64

|dw:1349878299626:dw|

8. anonymous

how come constant velocty?

9. amistre64

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

$a = \frac Fm$ $a = \frac {\frac59mg}m$ $a = \frac59g$

11. amistre64

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

12. anonymous

yes lets see if the answer comes

13. amistre64

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

14. anonymous

i dont think, the answer is 7/3 mg

15. amistre64

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

16. amistre64

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

17. amistre64

18. anonymous

hmm

19. anonymous

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. anonymous

i got the answer thanks :)

21. anonymous

u wanna see my working

22. anonymous

|dw:1349879825375:dw|

23. anonymous

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

24. anonymous

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. anonymous

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

26. anonymous

wht do umean? which part do u need to agree

27. anonymous

...

28. anonymous

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

29. anonymous

yes :)

30. anonymous

i substitues and got friction as 7/3 mg

31. anonymous

wut.

32. anonymous

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. anonymous

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

34. anonymous

yes i did

35. anonymous

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

36. anonymous

pretty funny.

37. anonymous

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

38. anonymous

you make no sense.

39. anonymous

what acceleration are you using for the first part?

40. anonymous

4/9 g

41. anonymous

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

42. anonymous

by negative I guess you mean upwards.

43. anonymous

deceleration

44. anonymous

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

45. anonymous

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

46. anonymous

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. anonymous

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. anonymous

no its correct...

49. anonymous

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

50. anonymous

yup

51. anonymous

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

52. anonymous

yes...

53. anonymous

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

54. anonymous

so second part.

55. anonymous

yes how do i do tht

56. anonymous

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. anonymous

oh okay il try

58. anonymous

btw which equation did u use

59. anonymous

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

60. anonymous

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

61. anonymous

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

62. anonymous

ok

63. anonymous

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