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.

- anonymous

- schrodinger

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- anonymous

- amistre64

do you know how to define weight using mass?

- anonymous

hmm... good one.

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## More answers

- anonymous

w = mg

- 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

velocity . time graph

- amistre64

|dw:1349878299626:dw|

- anonymous

how come constant velocty?

- 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

\[a = \frac Fm\]
\[a = \frac {\frac59mg}m\]
\[a = \frac59g\]

- amistre64

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

- anonymous

yes lets see if the answer comes

- 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

i dont think, the answer is 7/3 mg

- amistre64

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

- amistre64

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

- amistre64

prolly a bad graph :/

- anonymous

hmm

- 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

i got the answer thanks :)

- anonymous

u wanna see my working

- anonymous

|dw:1349879825375:dw|

- anonymous

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

- 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

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

- anonymous

wht do umean? which part do u need to agree

- anonymous

...

- anonymous

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

- anonymous

yes :)

- anonymous

i substitues and got friction as 7/3 mg

- anonymous

wut.

- 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

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

- anonymous

yes i did

- anonymous

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

- anonymous

pretty funny.

- anonymous

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

- anonymous

you make no sense.

- anonymous

what acceleration are you using for the first part?

- anonymous

4/9 g

- anonymous

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

- anonymous

by negative I guess you mean upwards.

- anonymous

deceleration

- anonymous

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

- anonymous

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

- 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

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

no its correct...

- anonymous

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

- anonymous

yup

- anonymous

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

- anonymous

yes...

- anonymous

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

- anonymous

so second part.

- anonymous

yes how do i do tht

- 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

oh okay il try

- anonymous

btw which equation did u use

- anonymous

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

- anonymous

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

- 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

ok

- anonymous

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

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