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    I can find the moments for the others but i don't know what to do with X. i worked out that its length is 2root2 but then how do i find its moment, multiply by 2 to get the distance of X from D? i.e. my question is how to get the moment of X?
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    (Original post by AlvlVictim)
    You have to draw a diagram and stuff its complex check mark scheme
    I've checked the mark scheme and I can't really make sense of it.

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    For resolving forces questions is it ok to annotate the diagram? Or do you have to draw a new one out? Basically will you lose marks if you do your working on the diagram given?


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    do you think i can teach myself this unit and still revise for psychology unit 3 in 6 hours?? lmao help me
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    I saw a couple pages back someone said that if there is a part (i) and (ii) you lose marks if you do it in the wrong order...I'm 95% sure that's wrong, why would you be penalised anyway? You've still shown that you can do it?


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    (Original post by NickLCFC)
    I've checked the mark scheme and I can't really make sense of it.

    (Massive diagram ting)
    Someone on the interwebs explained it in a way which makes more sense, IMO:

    The pipeline runs north-south, going through point (6i, 0j + 0k). You can therefore create a vector equation out of this by doing
    r = (6i + 0j + 0k) + x(0i + 1j + 0k) (velocity is 0i + 1j + 0k because the pipeline is running in the j unit direction basically).

    Then you can create a second equation using the position of the mast, which goes through the point (0i + 2j + 0k). The "velocity" would be (-1i + -1j + 0k) because the radio mast is south-west to the walker (if you think about going one unit to the left and one down, the direction would be south west).

    So you get r = (0i + 2j + 0k) + y(-1i -1j + 0k)

    you then set these two equations equal to each other to find where the walker is, work out a value for x or y and sub it into the respective equation (you should get the same answer in either case) to get (6i + 8j)!
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    (Original post by NickLCFC)


    Can someone explain how the answer to part a) is (6i + 8j)?
    6i is pretty self explanatory from the question.
    8j however requires trig. Draw a diagram, and we know that south west is 45 in a downwards left direction. Use trigonometry on the triangle (using 6i) namely the tan function. Then add the 2j. This explanation becomes clear with a diagram.
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    Name:  maths2.png
Views: 88
Size:  2.4 KB Sorry.. i suck at paint

    (Original post by NickLCFC)
    I've checked the mark scheme and I can't really make sense of it.
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    (Original post by Someboady)
    6i is pretty self explanatory from the question.
    8j however requires trig. Draw a diagram, and we know that south west is 45 in a downwards left direction. Use trigonometry on the triangle (using 6i) namely the tan function. Then add the 2j. This explanation becomes clear with a diagram.
    Okay that makes so much more sense hahaha
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    (Original post by Someboady)
    Name:  maths2.png
Views: 88
Size:  2.4 KB Sorry.. i suck at paint
    Thanks mate :yy:
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    (Original post by Don Joiner)
    I saw a couple pages back someone said that if there is a part (i) and (ii) you lose marks if you do it in the wrong order...I'm 95% sure that's wrong, why would you be penalised anyway? You've still shown that you can do it?


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    You wouldn't be penalised. I personally do not label my questions. The marker must mark everything legible on your paper.
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    (Original post by n2697)
    I can find the moments for the others but i don't know what to do with X. i worked out that its length is 2root2 but then how do i find its moment, multiply by 2 to get the distance of X from D? i.e. my question is how to get the moment of X?
    Didn't think moments with non-perpendicular angles was in M1 but anyway, I guess we need to take moments about D
    So (5 x 2) + (3 x 2) = Xcos45 x 2
    (this is using force x perp distance)
    So 10 + 6 = 2X x rt2/2
    16 = X x rt2
    16 rt 2 = 2X
    so X = 8 rt 2 I guess
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    GUYS ANY LAST TIPS BEFORE THE EXAM? :please:

    BTW GOOD LUCK! HOPE EVERYONE DOES GOOD
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    (Original post by wenogk)
    GUYS ANY LAST TIPS BEFORE THE EXAM? :please:

    BTW GOOD LUCK! HOPE EVERYONE DOES GOOD
    Check if the question says:

    speed or velocity
    mass or weight
    acceleration or decceleration

    dont skip steps in long complex calculations, do one step at a time so when you are checking answers you can spot any errors more clearly

    good luck to you too!
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    Does anyone know how to use the information when something is north west or south west etc in a vectors question? I know it's 45 degrees but idk how that helps.
    I can do it when it's north or south or east or west as I know the i/j components are the same. But when it's like north west or something it confuses me
    Example question:
    Name:  GOLD Vectors Question.png
Views: 149
Size:  50.4 KB
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    Does this clash with FP2 again? Was a fun clash for me last year
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    (Original post by jtlmao)
    For resolving forces questions is it ok to annotate the diagram? Or do you have to draw a new one out? Basically will you lose marks if you do your working on the diagram given?


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    No you won't loose marks
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    (Original post by Tim73)
    Does anyone know how to use the information when something is north west or south west etc in a vectors question? I know it's 45 degrees but idk how that helps.
    I can do it when it's north or south or east or west as I know the i/j components are the same. But when it's like north west or something it confuses me
    Example question:
    Name:  GOLD Vectors Question.png
Views: 149
Size:  50.4 KB
    When Q is north west of P, the J component will stay the same, but you have to use the negative of the I component from the equation for PQ as it would be on the left. Then equate them.
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    (Original post by Tim73)
    Does anyone know how to use the information when something is north west or south west etc in a vectors question? I know it's 45 degrees but idk how that helps.
    I can do it when it's north or south or east or west as I know the i/j components are the same. But when it's like north west or something it confuses me
    Example question:
    Name:  GOLD Vectors Question.png
Views: 149
Size:  50.4 KB
    In this question you can use vector PQ from part b to help you. This vector must equate to k(-I + j) where k is just a constant.
    Now equate the i's and the j's to equal -k and k respectively.
    Then combine equations and solve for t
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    (Original post by candol)
    Didn't think moments with non-perpendicular angles was in M1 but anyway, I guess we need to take moments about D
    So (5 x 2) + (3 x 2) = Xcos45 x 2
    (this is using force x perp distance)
    So 10 + 6 = 2X x rt2/2
    16 = X x rt2
    16 rt 2 = 2X
    so X = 8 rt 2 I guess
    i gt da same answer too fr k
 
 
 
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