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# M1 Question - resolving forces?

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1. Hi there - I was wondering if someone could explain how to work out the two state questions at the end of this question (part iii). They are only worth a mark each, so I imagine they aren't too bad, but I just can't get my head around it.

2. (Original post by Funky_Giraffe)
Hi there - I was wondering if someone could explain how to work out the two state questions at the end of this question (part iii). They are only worth a mark each, so I imagine they aren't too bad, but I just can't get my head around it.
If P=0, then there's no force pulling the string to the side. So where's the ring going to hang?
3. (Original post by ghostwalker)
If P=0, then there's no force pulling the string to the side. So where's the ring going to hang?
It will be somewhere on the line AB, but I don't know how to use the weight of P to find out where it will be on that line.
4. (Original post by Funky_Giraffe)
It will be somewhere on the line AB, but I don't know how to use the weight of P to find out where it will be on that line.
But the ring is free to slide, and there's nothing to hold it up, hence....
5. (Original post by ghostwalker)
But the ring is free to slide, and there's nothing to hold it up, hence....
Ok, so below B? What about the value of T though?
6. (Original post by Funky_Giraffe)
Ok, so below B? What about the value of T though?
Yes, it will hang vertically below B.

Well, resolve vertically for the ring.

Spoiler:
Show

Tension will be half the weight of the ring.
7. If there is no horizontal force acting on the ring anymore (as P=0), then we must understand that the weight components effect is the greatest that acts on the ring, so the ring will be influenced by this force, therefore it will hang vertically below A.

The total length of the string is 0.7
AB is 0.24

0.7-0.24 = 0.46m diference.
As the weight hangs below PB at 0.23m, 2T = W as 0.46/0.23 gives a ratio of 2:1 seeing as the weight is supported by two strings with tension T.

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