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    Hey guys
    Could someone please explain part c??

    I understand how to get the angle 90-alpha
    But don't understand what to do after that
    Please help!!
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    (Original post by CrystalPark)
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    Hey guys
    Could someone please explain part c??

    I understand how to get the angle 90-alpha
    But don't understand what to do after that
    Please help!!
    the resultant of the tension (the forces acting on the pulley) bisects the top right angle. Find half the top right angle, resolve the tensions on either side and add them together.

    Eg. If the tension was 10 N and the angle at the top right was 60 degrees then you would resolve the tension twice which would give you 2 lots of Tcos30.

    2Tcos(half of top right angle)



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    (Original post by KloppOClock)
    the resultant of the tension (the forces acting on the pulley) bisects the top right angle. Find half the top right angle, resolve the tensions on either side and add them together.

    Eg. If the tension was 10 N and the angle at the top right was 60 degrees then you would resolve the tension twice which would give you 2 lots of Tcos30.

    2Tcos(half of top right angle)



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    Wow this is definitely much harder than I thought
    But thank you once again

    If you don't mind me asking, in vectors when do you equal the components and when do you set them to 0? Especially when it's north or west? Do you equate or set to 0??
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    (Original post by CrystalPark)
    Wow this is definitely much harder than I thought
    But thank you once again

    If you don't mind me asking, in vectors when do you equal the components and when do you set them to 0? Especially when it's north or west? Do you equate or set to 0??
    Ok lets say that A has velocity of (2i-3j) and starting position of (-2j)
    Lets say that B has velocity of (3i+4j) and starting position of (i+j)

    To find the difference between them both, find r=r0+vt for both of them then do
    OB-OA as you can see why in this diagram. (at the bottom)

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    Okay so now you know the location of Q in relation to P.

    If Q is northwest. south west, etc to P then the angle it makes between the i and j axis will be 45 degrees either side. Therefore when something is northwest, say, its i and j coordinates will be -i and j as they have to have the same length either way.

    So if something is north west of something, the i components will be a multiple of -i and j component a multiple of j. they have the same multiple so set the i components = to -m and j components = to m then solve simultaneously to find t, which will be the time it is northwest

    if something is north of something else, its I component must be the same. If something is travelling north it must only go up so it only has a j value so i = 0
 
 
 
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