zaphod
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.
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zaphod
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@Algebraic! pls hlp @Yahoo! @amistre64
amistre64
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do you know how to define weight using mass?
Algebraic!
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hmm... good one.
zaphod
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w = mg
amistre64
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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)?
zaphod
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velocity . time graph
amistre64
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|dw:1349878299626:dw|
zaphod
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how come constant velocty?
amistre64
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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
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\[a = \frac Fm\]
\[a = \frac {\frac59mg}m\]
\[a = \frac59g\]
amistre64
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\[v=\frac{5}{9}t+v_o\]
can we assume an initial velocity of 0?
zaphod
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yes lets see if the answer comes
amistre64
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\[\Delta L=-\frac12\frac59gt^2\]
\[\Delta L=-\frac5{18}gt^2\]
\[\Delta \frac L3=-\frac{5gt^2}{54}\]
just running thru an idea, any of this make sense?
zaphod
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i dont think, the answer is 7/3 mg
amistre64
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i havent gotten to the end yet ... just wondering if my thought process made sense so far
amistre64
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assuming im correct :) the velocity for the first 3 quarters would then be:\[v=-\frac5{27}gt\]|dw:1349879118009:dw|
amistre64
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prolly a bad graph :/
zaphod
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hmm
Algebraic!
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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
\[t = \frac{ 3 }{ 2 }\sqrt{\frac{ 3h }{ 2g }}\]
zaphod
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i got the answer thanks :)
zaphod
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u wanna see my working
Algebraic!
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|dw:1349879825375:dw|
Algebraic!
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dude why are you posting questions you don't need help on?
zaphod
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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 :/
Algebraic!
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let's see your second part of the soln. I hope your first part agrees with mine?
zaphod
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wht do umean? which part do u need to agree
Algebraic!
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...
Algebraic!
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did you get the same results as I did for the first part of the descent, where a= -4g/9?
zaphod
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yes :)
zaphod
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i substitues and got friction as 7/3 mg
Algebraic!
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wut.
zaphod
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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
Algebraic!
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you got the force from friction = 7mg/3 ??
zaphod
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yes i did
Algebraic!
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ok. that means a net force upwards. since the force down is mg
Algebraic!
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pretty funny.
zaphod
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yeah..could u help me with the second part i posted right now :)
Algebraic!
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you make no sense.
Algebraic!
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what acceleration are you using for the first part?
zaphod
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4/9 g
zaphod
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and when the friction increases it more thn the weight, acceleration becomes - 4/3g
Algebraic!
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by negative I guess you mean upwards.
zaphod
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deceleration
Algebraic!
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you want to show me your work? because if you did the first part right, the second question is trivial.
Algebraic!
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it's just a matter of plugging in 'h'
zaphod
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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:)
Algebraic!
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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.
zaphod
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no its correct...
Algebraic!
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I mean, you're basically just assuming the whole trip takes 1 second...
zaphod
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yup
Algebraic!
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so you plotted
v =( (3/9) g)*t
zaphod
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yes...
Algebraic!
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alright, I guess that's the way they want you to do it. bit ambiguous wording in the question, but w/e.
Algebraic!
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so second part.
zaphod
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yes how do i do tht
Algebraic!
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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
zaphod
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oh okay il try
zaphod
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btw which equation did u use
Algebraic!
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sorry about the confusion, I totally misinterpreted that problem I guess.
Algebraic!
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I used
d= Vi*t + 1/2*a*t^2
Algebraic!
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d1= 1/2* 4g/9*(3t/4)^2
d2 = (3g/9)(t/4) - 1/2 (4g/3)*(t/4)^2
d1 + d2 = 60
zaphod
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ok
Algebraic!
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you get it? I got t=8 (using g=10)
and V = 24/9*g