anonymous
  • anonymous
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.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@Algebraic! pls hlp @Yahoo! @amistre64
amistre64
  • amistre64
do you know how to define weight using mass?
anonymous
  • anonymous
hmm... good one.

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More answers

anonymous
  • anonymous
w = mg
amistre64
  • 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)?
anonymous
  • anonymous
velocity . time graph
amistre64
  • amistre64
|dw:1349878299626:dw|
anonymous
  • anonymous
how come constant velocty?
amistre64
  • 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
amistre64
  • amistre64
\[a = \frac Fm\] \[a = \frac {\frac59mg}m\] \[a = \frac59g\]
amistre64
  • amistre64
\[v=\frac{5}{9}t+v_o\] can we assume an initial velocity of 0?
anonymous
  • anonymous
yes lets see if the answer comes
amistre64
  • 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?
anonymous
  • anonymous
i dont think, the answer is 7/3 mg
amistre64
  • amistre64
i havent gotten to the end yet ... just wondering if my thought process made sense so far
amistre64
  • amistre64
assuming im correct :) the velocity for the first 3 quarters would then be:\[v=-\frac5{27}gt\]|dw:1349879118009:dw|
amistre64
  • amistre64
prolly a bad graph :/
anonymous
  • anonymous
hmm
anonymous
  • 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 \[t = \frac{ 3 }{ 2 }\sqrt{\frac{ 3h }{ 2g }}\]
anonymous
  • anonymous
i got the answer thanks :)
anonymous
  • anonymous
u wanna see my working
anonymous
  • anonymous
|dw:1349879825375:dw|
anonymous
  • anonymous
dude why are you posting questions you don't need help on?
anonymous
  • 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 :/
anonymous
  • anonymous
let's see your second part of the soln. I hope your first part agrees with mine?
anonymous
  • anonymous
wht do umean? which part do u need to agree
anonymous
  • anonymous
...
anonymous
  • anonymous
did you get the same results as I did for the first part of the descent, where a= -4g/9?
anonymous
  • anonymous
yes :)
anonymous
  • anonymous
i substitues and got friction as 7/3 mg
anonymous
  • anonymous
wut.
anonymous
  • 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
anonymous
  • anonymous
you got the force from friction = 7mg/3 ??
anonymous
  • anonymous
yes i did
anonymous
  • anonymous
ok. that means a net force upwards. since the force down is mg
anonymous
  • anonymous
pretty funny.
anonymous
  • anonymous
yeah..could u help me with the second part i posted right now :)
anonymous
  • anonymous
you make no sense.
anonymous
  • anonymous
what acceleration are you using for the first part?
anonymous
  • anonymous
4/9 g
anonymous
  • anonymous
and when the friction increases it more thn the weight, acceleration becomes - 4/3g
anonymous
  • anonymous
by negative I guess you mean upwards.
anonymous
  • anonymous
deceleration
anonymous
  • anonymous
you want to show me your work? because if you did the first part right, the second question is trivial.
anonymous
  • anonymous
it's just a matter of plugging in 'h'
anonymous
  • 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:)
anonymous
  • 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.
anonymous
  • anonymous
no its correct...
anonymous
  • anonymous
I mean, you're basically just assuming the whole trip takes 1 second...
anonymous
  • anonymous
yup
anonymous
  • anonymous
so you plotted v =( (3/9) g)*t
anonymous
  • anonymous
yes...
anonymous
  • anonymous
alright, I guess that's the way they want you to do it. bit ambiguous wording in the question, but w/e.
anonymous
  • anonymous
so second part.
anonymous
  • anonymous
yes how do i do tht
anonymous
  • 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
anonymous
  • anonymous
oh okay il try
anonymous
  • anonymous
btw which equation did u use
anonymous
  • anonymous
sorry about the confusion, I totally misinterpreted that problem I guess.
anonymous
  • anonymous
I used d= Vi*t + 1/2*a*t^2
anonymous
  • anonymous
d1= 1/2* 4g/9*(3t/4)^2 d2 = (3g/9)(t/4) - 1/2 (4g/3)*(t/4)^2 d1 + d2 = 60
anonymous
  • anonymous
ok
anonymous
  • anonymous
you get it? I got t=8 (using g=10) and V = 24/9*g

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