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rustynator
I am pretty sure you had to double the tension. I think it is just common sense...if there wasnt an equal force on the other cable the whole setup would just topple to one side.


But the equal force on the other side acts in the opposite direction.
Reply 81
SaccerZD
also, uranium mass, it sohuld be very small! i.e. 10^-4 kg or less



why...?
Reply 82
/\Shaz\/
But the equal force on the other side acts in the opposite direction.


That is exactly why you have to double the tension though. If they were acting in the same direction, the setup would topple over. There must be an equal an opposite force on each side of the wire...hence the overall force is double the tension on one side.
Reply 83
Troyes
The tension in the copper wire one. Did anyone else get 56.1 for the tension?

Yeah - me and a few otheres a know got that! :smile: What did you put for the next part, about the wire breaking?
Reply 84
I put the wire is not likely to break because the breaking stress was a lot less than the current stress on it. Than I worked out what force would need to be applied to get to the wire to the breking stress, can't remember the value though.
Reply 85
i just think the uranium mass should be low because of how energetic it is... a small mass of uranium could create a heating effect equal to 500mw, or 4k per hour, or whatever it was, couldn't it? not 10^13 kg, that is over half the weight of the earth?!
Reply 86
Its no where near half the weight of the Earth. Half the weight of the Earth would be 3 x 10^24 kg.
rustynator
That is exactly why you have to double the tension though. If they were acting in the same direction, the setup would topple over. There must be an equal an opposite force on each side of the wire...hence the overall force is double the tension on one side.


The overall, or resultant, force on the wire would be zero if two equal tension forces were acting in opposite directions. That's how tension works. The wire doesn't move; there's two equal tension forces of magnitude T acting in opposite directions, but the tension's still T, and not 2T.

I'm convinced you're not meant to double it, as there was no mention of neither the tension force on the other side being equal, nor that the dimensions on both sides were equal.
Reply 88
See thats whats really bugging me about this paper, theres not a lot of consensus on the answers! I mean normally its really obvious what ure supposed to do and its easily either right or wrong, but for this paper loads of people seem to have got answers in a grey area.
It was a weird paper, but I'm pretty much 100% certain on the not doubling tension thing, if my memory of basic Mechanics serves me correctly.
You do double the tension. It was 112 at the end I think. If you didnt double it, You're wrong as there were two cables.
/\Shaz\/
It was a weird paper, but I'm pretty much 100% certain on the not doubling tension thing, if my memory of basic Mechanics serves me correctly.


You're wrong because you've calculate a COMPONENT of the force not the force itself so two equal forces wont cancel out as they arent equal :smile:
Why
You're wrong because you've calculate a COMPONENT of the force not the force itself so two equal forces wont cancel out as they arent equal :smile:


Component of a force or whatever; it makes no difference. The horizontal components must cancel each other out, otherwise the wire would have a resultant acceleration and would MOVE. The tension in the central wire is equal to the horizontal component of the tension in the left support, so I have no idea what you're getting at.

I've just looked over my M1 textbook, and I'm right. If a string has tension T, there are two tension forces of equal magnitude acting in opposite directions (an action and a reaction), but the tension in the string is T, not 2T.

You do not double it, common-sensical or not. Check a Mechanics revision site or something with a basic problem involving a string.
You just cant admit you're wrong. Omg, A Wire is being pulled from one side but not from the other11!!
Why
You just cant admit you're wrong. Omg, A Wire is being pulled from one side but not from the other11!!


What? LOL. I couldn't care less if I was wrong on that question; I hardly need full marks to get into my chosen Uni. I'm just being honest. :smile:

Have you done Mechanics, i.e. for Maths, not the Physics AS module?

edit: just asked my dad, who's a mechanical engineer. You most definitely do not double it!

I remember my Mechanics now. When considering one particular point or particle in equilibrium, you should only consider the forces acting on that particular point. The tension on the right hand side should not be added onto the forces acting on the connecting point between the left support and the left end of the wire.
Reply 95
Yes, yes, you are all right, I made a schoolboy error...

me
3 x 10^12 kg is exactly half of 6 x 10^24 kg!



what an abject retarded fool.
If I recall correctly i got 98% in M1. According to your logic, If I cut the string on the right the string will remain taut which is by all means absurd.
Reply 97
Why
If I recall correctly i got 98% in M1. According to your logic, If I cut the string on the right the string will remain taut which is by all means absurd.


This argument is pointless. Exam's over.:cool:
Reply 98
no it's really interesting, I'd like to see it resolved :p: I would think that if you were meant to double it, they would have mentioned something about the other support being the same. Having said that, I 'forgot' to double it - intuitively I thought that I should.

However. Consider the top of the left mast. It's in equilibrium, and what's holding it there? The horizontal component of the tension in P, which is balanced out by the tension in the wire. So you do not double it.

this sounds right to me, I think I disagree with Why.


john

edit: I disagree, yes.
john-boro
no it's really interesting, I'd like to see it resolved :p: I would think that if you were meant to double it, they would have mentioned something about the other support being the same. Having said that, I 'forgot' to double it - intuitively I thought that I should.

However. Consider the top of the left mast. It's in equilibrium, and what's holding it there? The horizontal component of the tension in P, which is balanced out by the tension in the wire. So you do not double it.

this sounds right to me, I think I disagree with Why.


john

edit: I disagree, yes.


Clearly. The other mast and wire are not exerting any forces at all and are floating in free space.

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