Question

Question deleted.

Answer 1

we start by stating Hook's Law for an elastic or spring

Answer 2

So, F =-kx ?

Answer 3

yes, where x is the distance stretched.
F is the force caused by gravity on the person
we can use the info in the problem to solve for the constant k

Answer 4

Okay. but we do not know how much of it is stretched..

Answer 5

rope is stretched twice its normal length by 75 kg
i.e. if we call its length L, and we want to be just at the water level,
2L= 71 m (it stretched to twice its length)
L= 71/2 m

and it stretched an additional 35.5 m (this will be x in the F= -k x)

Answer 6

Okay.

Answer 7

so far we have the rope being stretched by 35.5 m by 75 kg
next, find the force caused by gravity acting on 75 kg mass

Answer 8

Force = ma
Therefore, F = 75 x 9.8 (where acceleration is 9.8 m/s^-2)

Answer 9

yes:
75*9.8 = k 35.5
k= 75*9.8/35.5

Answer 10

we should be careful about the sign, but we know F acts opposite gravity

Answer 11

Okay.

Answer 12

Do i need to include the freefall? and can i apply integral calculus?

Answer 13

I don't think that is necessary (but I've never done this type of problem)

Answer 14

mhmm... okay.

Answer 15

you could come up with something like
\[ m x'' = - k x \]
if you want to model the "bounce" (harmonic motion) that occurs

Answer 16

I don't think that is needed, because the main thing I need to find is the natural length of the rope needed for a 75kg person. . .

Answer 17

that's my interpretation. I'll look over the other questions to see if we are being "too simple"

Answer 18

okay. My teacher suggested to do 1) freefall with acceleration being 9.8
2)Retardation of acceleration until it is 0. (When it is hanging; equilibrium point?)
Retardation until v=0.

Answer 19

i think that applies to questions 1-3.

Answer 20

that sounds like using calculus and integrating
something like F= ma ---> a= F/m and integrate x'' until you get 0

Answer 21

in other words, it almost sounds like deriving Hooke's Law

Answer 22

what would it give if it is derived?

Answer 23

Let me think about this. Based on the title "Dynamics" they do not want a static analysis.

Answer 24

mhmm.. i can't find a way to do the maths of finding the natural length.

Answer 25

thinking more about it, the rope stretches past the equilibrium point (where F= kx balances out F= mg), so we do have to solve
m x'' = -k x
the solution will be
\[ x= C \cos(\sqrt{k/m} t + \theta) \]
and the additional stretch past the equilibrium point is C
we need to identify the initial conditions to fill in the constants

Answer 26

where does the cos come from? and is x the stretch?

Answer 27

a 2nd order differential equation has a solution of cos (or sin)
so either you use that, or you learn how to solve the 2nd order from first principles (and get cos).

and yes, the x is the stretch.

Answer 28

Okay.

Answer 29

I'm a bit confused now..

Answer 30

I am thinking about these points
1) freefall with acceleration being 9.8
this will be until you reach the end of your rope (so to speak)
2)Retardation of acceleration until it is 0. (When it is hanging; equilibrium point?)
this should end up being where F= -kx balances F= mg
3) Retardation until v=0.
this will be the additional "stretch"

Now how to set up the equations ?

Answer 31

Well for the free fall, we integrate acceleration (I'm not sure if it can be with respect to time or displacement) to find velocity.

Answer 32

this is what I know about springs. But it's not clear to me if they want you to solve this using a 2nd order differential.

Answer 33

I hope you have time to do this. I'll post later if I can make sense of it.

Answer 34

I'll have a look at it and see if I can make sense of it as well. Thank you.

Answer 35

Here are some thoughts.

Answer 36

thank you. :)

Answer 37

A is the distance from x=L to x= L+A it takes for the velocity to slow from \(\sqrt{2gL} \) to 0

in other words, L+A is the length of the fully stretched rope

Answer 38

Thank you :)