Shoutbox

Ok it's way late so I might be dead wrong.
But since there is a possible balance, shouldn't you take the distance from the thumble point into account?
So you would get:
If the chain is as longer on the left as it is less steep, there will be a balance, and if it is either longer or steeper it will fall an the left, shorter or less steep on the right...

From my experiences of balancing chains on corners of tables (), I think it would balance out, but I suppose it does depend on the exact values

A (alpha) and G (gamma) are shown above. AB is some line that crosses both AC and BC, and DC crosses BA perpendiculary and is in the same direction as the gravitional camp, which by convention points downside.
The chain will fall on the side of bigger cos*mass product:
if  (cos A)*(mass of chain above BC) > (cos G)*(mass of chain above AC), it will fall to the left.
if (cos A)*(mass of chain above BC) < (cos G)*(mass of chain above AC), to the right.

Or I need to revise it again =P.


P.S.: Codification: making non-latin symbols fail to renderize since the beggining.
Other P.S.: Swithced the sides after correcting it

I believe I can conclude that it'll fall to the left .

My teacher basically gave me this problem and said "ask anyone".

On the picture provided in class, it specifically said it could be done without calculating "and all that blah blah"

i think

quote:

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Originally posted by blessedguy

A (alpha) and G (gamma) are shown above. AB is some line that crosses both AC and BC, and DC crosses BA perpendiculary and is in the same direction as the gravitional camp, which by convention points downside.
The chain will fall on the side of bigger cos*mass product:
if  (cos A)*(mass of chain above BC) > (cos G)*(mass of chain above AC), it will fall to the left.
if (cos A)*(mass of chain above BC) < (cos G)*(mass of chain above AC), to the right.

Or I need to revise it again =P.


P.S.: Codification: making non-latin symbols fail to renderize since the beggining.

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* Chrono agrees

quote:

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Originally posted by Chrono
i think

quote:

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Originally posted by blessedguy

A (alpha) and G (gamma) are shown above. AB is some line that crosses both AC and BC, and DC crosses BA perpendiculary and is in the same direction as the gravitional camp, which by convention points downside.
The chain will fall on the side of bigger cos*mass product:
if  (cos A)*(mass of chain above AC) > (cos G)*(mass of chain above BC), it will fall to the left.
if (cos A)*(mass of chain above AC) < (cos G)*(mass of chain above BC), to the right.

Or I need to revise it again =P.


P.S.: Codification: making non-latin symbols fail to renderize since the beggining.

---

* Chrono agrees

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yeah that looks right, if not i didnt learn anything from my engineering degree

quote:

---

Originally posted by blessedguy

A (alpha) and G (gamma) are shown above. AB is some line that crosses both AC and BC, and DC crosses BA perpendiculary and is in the same direction as the gravitional camp, which by convention points downside.
The chain will fall on the side of bigger cos*mass product:
if  (cos A)*(mass of chain above BC) > (cos G)*(mass of chain above AC), it will fall to the left.
if (cos A)*(mass of chain above BC) < (cos G)*(mass of chain above AC), to the right.

Or I need to revise it again =P.


P.S.: Codification: making non-latin symbols fail to renderize since the beggining.
Other P.S.: Swithced the sides after correcting it

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Erm, if mass of chain is uniformly distributed, shouldn't the equations balance?
Mass being proportional to the hypotenuse of each side DC/cos(alpha) and DC/cos(gamma) respectively...

You can think of 2 blocks instead of a chain. It's the same thing:

W, W' = blocks' weights
T = tension in the rope

quote:

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Originally posted by blessedguy

if  (cos A)*(mass of chain above BC) > (cos G)*(mass of chain above AC), it will fall to the left.
if (cos A)*(mass of chain above BC) < (cos G)*(mass of chain above AC), to the right.

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This (as well as Chancer's) is 100% correct. Given that, I think Joa's probably got the answer your teacher is looking for with:

quote:

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Originally posted by Joa
so theoretically without all the exact info - right now one could even say that the chain could remain in place because perhaps the pull of gravity is evenly distributed on both sides due to the angle and length of chain ...

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This is because cos(A) is a smaller value but BC is greater in mass whereas cos(B) is a greater value but AC is of smaller mass, essentially cancelling out one another for no net movement. The only reason I think your teacher would have asked this question in such vague terms would be if that was the answer she was looking for even though technically that can't be concluded without assigning values to the lengths of the chain and angles.

quote:

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Originally posted by Volv
The only reason I think your teacher would have asked this question in such vague terms would be if that was the answer she was looking for even though technically that can't be concluded without the mass of the chain or angles assigned values.

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Heh, we'll find out soon when (and if) he tells us