Help with understanding where equations came from Watch

Lehane
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#1
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For the first one I have:

P \times (\frac{r_0 - r_i}{r_o}) \times d where d is the lateral distance travelled to the left.

The top half of the fraction is distance from the rope to the ground, but I don't understand where the denominator comes from.

The second one:

[\sqrt {2 \times (1 + \cos \theta )} - 1]^2

I don't undertand what that calculates. All that is multiplied by \frac{1}{2} \times k \times L^2 which is the spring constant (the top edge of the triangle) (all corners are pivoted).

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ghostwalker
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(Original post by Lehane)
The second one:

[\sqrt {2 \times (1 + \cos \theta )} - 1]^2

I don't undertand what that calculates. All that is multiplied by \frac{1}{2} \times k \times L^2 which is the spring constant (the top edge of the triangle) (all corners are pivoted).

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There is a serious lack of information on that second diagram, and nothing else in your post.

Here's one interpretation:

The two shorter sides of the triangle both have length 1.

Then the formula you have quoted is the square of the extension of the spring (assuming that has length one too); derived from using the cosine rule with the angle in the triangle which is 180-theta.
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Lehane
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(Original post by ghostwalker)
There is a serious lack of information on that second diagram, and nothing else in your post.

Here's one interpretation:

The two shorter sides of the triangle both have length 1.

Then the formula you have quoted is the square of the extension of the spring (assuming that has length one too); derived from using the cosine rule with the angle in the triangle which is 180-theta.
Sorry! Poor diagram on my part. The shorter sides actually have length 0.4m. Unstretched, the spring is also 0.4m but it is at it's maximum extension in the diagram at the start. L is 0.4m
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ghostwalker
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(Original post by Lehane)
Sorry! Poor diagram on my part. The shorter sides actually have length 0.4m. Unstretched, the spring is also 0.4m but it is at it's maximum extension in the diagram at the start. L is 0.4m
In that case multiplying by L^2, gives you the square of the extension.
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Lehane
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(Original post by ghostwalker)
In that case multiplying by L^2, gives you the square of the extension.
Thank you so much! I am just a bit confused. Is the sqare root part a rearranged version of the cosine rule? Because I thought that was

L^2 + L^2 - (2 \times L^2 \times \cos(180 - \theta))
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ghostwalker
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(Original post by Lehane)
Thank you so much! I am just a bit confused. Is the sqare root part a rearranged version of the cosine rule? Because I thought that was

L^2 + L^2 - (2 \times L^2 \times \cos(180 - \theta))
In effect; they've just factored out the L^2.
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Lehane
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(Original post by ghostwalker)
In effect; they've just factored out the L^2.
But when you do that they get +cos as well as a square root, a -1 and a ^2 which I don't understand.
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ghostwalker
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(Original post by Lehane)
But when you do that they get +cos as well as a square root, a -1 and a ^2 which I don't understand.
The square of the length of the longer side is given by:

L^2 + L^2 - (2 \times L^2 \times \cos(180 - \theta))

=L^2( 2 - (2\times \cos(180 - \theta))

=L^2( 2 + (2 \cos(\theta))

Work out cos(180-theta) using the multiple angle formula, if you're still unsure.

So length is:

=L\sqrt{ 2 + 2\times \cos(\theta)}

Extension is:

=L\sqrt{ 2 + 2\times \cos(\theta)}-L

Square of extension is:

=(L\sqrt{ 2 + 2\times \cos(\theta)}-L)^2

=L^2(\sqrt{ 2 + 2\times \cos(\theta)}-1)^2

=L^2(\sqrt{ 2\times (1+  \cos(\theta))}-1)^2
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Lehane
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(Original post by ghostwalker)
The square of the length of the longer side is given by:

L^2 + L^2 - (2 \times L^2 \times \cos(180 - \theta))

=L^2( 2 - (2\times \cos(180 - \theta))

=L^2( 2 + (2 \cos(\theta))

Work out cos(180-theta) using the multiple angle formula, if you're still unsure.

So length is:

=L\sqrt{ 2 + 2\times \cos(\theta)}

Extension is:

=L\sqrt{ 2 + 2\times \cos(\theta)}-L

Square of extension is:

=(L\sqrt{ 2 + 2\times \cos(\theta)}-L)^2

=L^2(\sqrt{ 2 + 2\times \cos(\theta)}-1)^2

=L^2(\sqrt{ 2\times (1+  \cos(\theta))}-1)^2
Wow! You are a legend! Thank you so much! I will rep you in a couple of days because it won't let me now :p:
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ghostwalker
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(Original post by Lehane)
For the first one I have:

P \times (\frac{r_0 - r_i}{r_o}) \times d where d is the lateral distance travelled to the left.

The top half of the fraction is distance from the rope to the ground, but I don't understand where the denominator comes from.
DELETED.

Edit: Though my thinking it was iffy to start with.
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Lehane
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(Original post by ghostwalker)
My thinking on this is somewhat iffy, so don't ask me to justify it, also I am supposing that this is the same question you posted on a previous thread, and that diagram applies.

My interpretation is:

P represents a force, and not a couple (despite the title of that previous thread).

The reel moves a distance "d".

In so doing it turns through an angle of d/r0

So the rope unwinds a distance d/ro times ri, and the force moves (???? iffy) this distance away.

So the distance the reel moves relative to the force is d-\frac{dr_i}{r_0}

and this equals d\times\frac{r_0-r_i}{r_0}

and hence the work done is P times that.
Ohhh! I knew it had to be some sort of moment but the fraction seemed to be so random! Thank you so much for all your help!
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Lehane
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(Original post by ghostwalker)
My thinking on this is somewhat iffy, so don't ask me to justify it, also I am supposing that this is the same question you posted on a previous thread, and that diagram applies.

My interpretation is:

P represents a force, and not a couple (despite the title of that previous thread).

The reel moves a distance "d".

In so doing it turns through an angle of d/r0

So the rope unwinds a distance d/ro times ri, and the force moves (???? iffy) this distance away.

So the distance the reel moves relative to the force is d-\frac{dr_i}{r_0}

and this equals d\times\frac{r_0-r_i}{r_0}

and hence the work done is P times that.
Just realised I wrote the equation wrong... The fraction is actually a + not a - . Sorry! That suggests it's the total lateral distance + the amount the rope unwinds from the reel. This problem is annoying!
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ghostwalker
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(Original post by Lehane)
Just realised I wrote the equation wrong... The fraction is actually a + not a - . Sorry! That suggests it's the total lateral distance + the amount the rope unwinds from the reel. This problem is annoying!
I told you it was iffy!

Which just shows I'm not really understanding what's going on.

Work done is force times distance moved in the direction of that force; but by what? Could easily say the rope, and get the answer you want, but I don't feel any justification for it now; merde!!!

EDIT:

Alternatively:

It is the sum of the work done in moving laterally, plus the work done in rotating.

= Pd + Pd (ri/r0)

and I feel a lot happier with that.


Definitely threw me there, for a bit.
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Lehane
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(Original post by ghostwalker)
I told you it was iffy!

Which just shows I'm not really understanding what's going on.

Work done is force times distance moved in the direction of that force; but by what? Could easily say the rope, and get the answer you want, but I don't feel any justification for it now; merde!!!

EDIT:

Alternatively:

It is the sum of the work done in moving laterally, plus the work done in rotating.

= Pd + Pd (ri/r0)

and I feel a lot happier with that.


Definitely threw me there, for a bit.
Aaah! Perfect! Thank you again! Although now I have to do revision which is bad times!
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