## 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

1. zaphod Group Title

@Algebraic! pls hlp @Yahoo! @amistre64

2. amistre64 Group Title

do you know how to define weight using mass?

3. Algebraic! Group Title

hmm... good one.

4. zaphod Group Title

w = mg

5. amistre64 Group Title

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 Group Title

velocity . time graph

7. amistre64 Group Title

|dw:1349878299626:dw|

8. zaphod Group Title

how come constant velocty?

9. amistre64 Group Title

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 Group Title

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

11. amistre64 Group Title

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

12. zaphod Group Title

yes lets see if the answer comes

13. amistre64 Group Title

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

14. zaphod Group Title

i dont think, the answer is 7/3 mg

15. amistre64 Group Title

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

16. amistre64 Group Title

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

17. amistre64 Group Title

18. zaphod Group Title

hmm

19. Algebraic! Group Title

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 Group Title

i got the answer thanks :)

21. zaphod Group Title

u wanna see my working

22. Algebraic! Group Title

|dw:1349879825375:dw|

23. Algebraic! Group Title

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

24. zaphod Group Title

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! Group Title

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

26. zaphod Group Title

wht do umean? which part do u need to agree

27. Algebraic! Group Title

...

28. Algebraic! Group Title

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

29. zaphod Group Title

yes :)

30. zaphod Group Title

i substitues and got friction as 7/3 mg

31. Algebraic! Group Title

wut.

32. zaphod Group Title

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! Group Title

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

34. zaphod Group Title

yes i did

35. Algebraic! Group Title

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

36. Algebraic! Group Title

pretty funny.

37. zaphod Group Title

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

38. Algebraic! Group Title

you make no sense.

39. Algebraic! Group Title

what acceleration are you using for the first part?

40. zaphod Group Title

4/9 g

41. zaphod Group Title

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

42. Algebraic! Group Title

by negative I guess you mean upwards.

43. zaphod Group Title

deceleration

44. Algebraic! Group Title

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

45. Algebraic! Group Title

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

46. zaphod Group Title

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! Group Title

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 Group Title

no its correct...

49. Algebraic! Group Title

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

50. zaphod Group Title

yup

51. Algebraic! Group Title

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

52. zaphod Group Title

yes...

53. Algebraic! Group Title

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

54. Algebraic! Group Title

so second part.

55. zaphod Group Title

yes how do i do tht

56. Algebraic! Group Title

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 Group Title

oh okay il try

58. zaphod Group Title

btw which equation did u use

59. Algebraic! Group Title

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

60. Algebraic! Group Title

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

61. Algebraic! Group Title

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 Group Title

ok

63. Algebraic! Group Title

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