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anonymous
 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 fiveninths 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.
anonymous
 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 fiveninths 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Algebraic! pls hlp @Yahoo! @amistre64

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0do you know how to define weight using mass?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0and 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
 3 years ago
Best ResponseYou've already chosen the best response.0velocity . time graph

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1349878299626:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how come constant velocty?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0if 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[a = \frac Fm\] \[a = \frac {\frac59mg}m\] \[a = \frac59g\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0\[v=\frac{5}{9}t+v_o\] can we assume an initial velocity of 0?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes lets see if the answer comes

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i dont think, the answer is 7/3 mg

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0i havent gotten to the end yet ... just wondering if my thought process made sense so far

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0assuming im correct :) the velocity for the first 3 quarters would then be:\[v=\frac5{27}gt\]dw:1349879118009:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0prolly a bad graph :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0i got the answer thanks :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u wanna see my working

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1349879825375:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dude why are you posting questions you don't need help on?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.0let's see your second part of the soln. I hope your first part agrees with mine?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wht do umean? which part do u need to agree

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0did you get the same results as I did for the first part of the descent, where a= 4g/9?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i substitues and got friction as 7/3 mg

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now 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
 3 years ago
Best ResponseYou've already chosen the best response.0you got the force from friction = 7mg/3 ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok. that means a net force upwards. since the force down is mg

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah..could u help me with the second part i posted right now :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what acceleration are you using for the first part?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and when the friction increases it more thn the weight, acceleration becomes  4/3g

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by negative I guess you mean upwards.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you want to show me your work? because if you did the first part right, the second question is trivial.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it's just a matter of plugging in 'h'

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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
 3 years ago
Best ResponseYou've already chosen the best response.0you 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
 3 years ago
Best ResponseYou've already chosen the best response.0I mean, you're basically just assuming the whole trip takes 1 second...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you plotted v =( (3/9) g)*t

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alright, I guess that's the way they want you to do it. bit ambiguous wording in the question, but w/e.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0btw which equation did u use

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry about the confusion, I totally misinterpreted that problem I guess.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I used d= Vi*t + 1/2*a*t^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0d1= 1/2* 4g/9*(3t/4)^2 d2 = (3g/9)(t/4)  1/2 (4g/3)*(t/4)^2 d1 + d2 = 60

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you get it? I got t=8 (using g=10) and V = 24/9*g
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