Question

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.

Answer 1

@Algebraic! pls hlp @Yahoo! @amistre64

Answer 2

do you know how to define weight using mass?

Answer 3

hmm... good one.

Answer 4

w = mg

Answer 5

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)?

Answer 6

velocity . time graph

Answer 7

how come constant velocty?

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

yes lets see if the answer comes

Answer 12

\[\Delta L=-\frac12\frac59gt^2\]
\[\Delta L=-\frac5{18}gt^2\]
\[\Delta \frac L3=-\frac{5gt^2}{54}\]
just running thru an idea, any of this make sense?

Answer 13

i dont think, the answer is 7/3 mg

Answer 14

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

Answer 15

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

Answer 16

prolly a bad graph :/

Answer 17

hmm

Answer 18

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

\[t = \frac{ 3 }{ 2 }\sqrt{\frac{ 3h }{ 2g }}\]

Answer 19

i got the answer thanks :)

Answer 20

u wanna see my working