A community for students.
Here's the question you clicked on:
 0 viewing
zaphod
 2 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.
zaphod
 2 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.

This Question is Closed

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

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

amistre64
 2 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)?

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

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

amistre64
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0\[a = \frac Fm\] \[a = \frac {\frac59mg}m\] \[a = \frac59g\]

amistre64
 2 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?

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

amistre64
 2 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?

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

amistre64
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0prolly a bad graph :/

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1I 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 \[t = \frac{ 3 }{ 2 }\sqrt{\frac{ 3h }{ 2g }}\]

zaphod
 2 years ago
Best ResponseYou've already chosen the best response.0i got the answer thanks :)

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

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

zaphod
 2 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 :/

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

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

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

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

zaphod
 2 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

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

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

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

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

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

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

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

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

zaphod
 2 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:)

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1you 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.

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

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

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1alright, I guess that's the way they want you to do it. bit ambiguous wording in the question, but w/e.

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1I 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

zaphod
 2 years ago
Best ResponseYou've already chosen the best response.0btw which equation did u use

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

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

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

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1you get it? I got t=8 (using g=10) and V = 24/9*g
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.