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# M1 Tension help watch

1. HI
Im doing an M1 paper. Im on 7 part vi. I have worked through the question correctly, but cant seem to fiure out how one can assume that one string takes all the tension, whilst the other is slack. http://www.mei.org.uk/files/papers/m1_paper_jan13.pdf

if i start from the beginning and recalculate cos(alpha) and cos(beta) using lengths 0.5 and 2.3, i get T1 =41 and T2=9 which is of course incorrect
2. (Original post by starwarsjedi123)
HI
Im doing an M1 paper. Im on 7 part vi. I have worked through the question correctly, but cant seem to fiure out how one can assume that one string takes all the tension, whilst the other is slack. http://www.mei.org.uk/files/papers/m1_paper_jan13.pdf

if i start from the beginning and recalculate cos(alpha) and cos(beta) using lengths 0.5 and 2.3, i get T1 =41 and T2=9 which is of course incorrect
If you recalculated the cosines, what values did you get for them?
3. hi
i got cos(alpha)=0.25
cos(beta)=1.15
4. (Original post by starwarsjedi123)
hi
i got cos(alpha)=0.25
cos(beta)=1.15
Does having cos(something) > 1 make sense?
5. (Original post by Zacken)
Does having cos(something) > 1 make sense?
No. I dont think i can use SOHCAHToa because its no longer a right angled triangle i think?
6. I feel like im going about this question in a wrong way..........but surely i need the angles to work out the tension??
7. (Original post by starwarsjedi123)
I feel like im going about this question in a wrong way..........but surely i need the angles to work out the tension??
Working out the angles is a reasonable approach. However your cosines can't be correct, as Zacken pointed out.

Try the cosine rule and work out the angle for the shorter string. You should get a negative value, which tells you the angle needs to be greater than 90.

If these were rods, rather than strings, that would be feasible, but since they are strings, the maximum angle is 90 degrees, and the shorter string will hang vertically, whilst the longer one is slack. Hence tensions are....

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