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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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zaphodBest ResponseYou've already chosen the best response.0
@Algebraic! pls hlp @Yahoo! @amistre64
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
do you know how to define weight using mass?
 one year ago

amistre64Best ResponseYou've already chosen the best response.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

amistre64Best ResponseYou've already chosen the best response.0
dw:1349878299626:dw
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
how come constant velocty?
 one year ago

amistre64Best ResponseYou've already chosen the best response.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

amistre64Best ResponseYou've already chosen the best response.0
\[a = \frac Fm\] \[a = \frac {\frac59mg}m\] \[a = \frac59g\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
\[v=\frac{5}{9}t+v_o\] can we assume an initial velocity of 0?
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
yes lets see if the answer comes
 one year ago

amistre64Best 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?
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
i dont think, the answer is 7/3 mg
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
i havent gotten to the end yet ... just wondering if my thought process made sense so far
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
assuming im correct :) the velocity for the first 3 quarters would then be:\[v=\frac5{27}gt\]dw:1349879118009:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
prolly a bad graph :/
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.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

zaphodBest ResponseYou've already chosen the best response.0
i got the answer thanks :)
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
dw:1349879825375:dw
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
dude why are you posting questions you don't need help on?
 one year ago

zaphodBest ResponseYou've already chosen the best response.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

Algebraic!Best ResponseYou've already chosen the best response.1
let's see your second part of the soln. I hope your first part agrees with mine?
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
wht do umean? which part do u need to agree
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
did you get the same results as I did for the first part of the descent, where a= 4g/9?
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
i substitues and got friction as 7/3 mg
 one year ago

zaphodBest ResponseYou've already chosen the best response.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

Algebraic!Best ResponseYou've already chosen the best response.1
you got the force from friction = 7mg/3 ??
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
ok. that means a net force upwards. since the force down is mg
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
yeah..could u help me with the second part i posted right now :)
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
what acceleration are you using for the first part?
 one year ago

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

Algebraic!Best ResponseYou've already chosen the best response.1
by negative I guess you mean upwards.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
you want to show me your work? because if you did the first part right, the second question is trivial.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
it's just a matter of plugging in 'h'
 one year ago

zaphodBest ResponseYou've already chosen the best response.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

Algebraic!Best ResponseYou've already chosen the best response.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

Algebraic!Best ResponseYou've already chosen the best response.1
I mean, you're basically just assuming the whole trip takes 1 second...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
so you plotted v =( (3/9) g)*t
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.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

Algebraic!Best ResponseYou've already chosen the best response.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

zaphodBest ResponseYou've already chosen the best response.0
btw which equation did u use
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
sorry about the confusion, I totally misinterpreted that problem I guess.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
I used d= Vi*t + 1/2*a*t^2
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.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

Algebraic!Best ResponseYou've already chosen the best response.1
you get it? I got t=8 (using g=10) and V = 24/9*g
 one year ago
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