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    I am trying question 9 (the underlining means it "goes beyond the requirements of the specification and are provided as a challenge")

    I am attaching my most current workings.

    Can you tell me, please, is my diagram correct.

    Also can you give me a clue as to the significance of the radius being the same as half the length of the rod.

    Thanks
    Attached Images
  1. File Type: pdf Scan0015.pdf (391.4 KB, 484 views)
  2. File Type: pdf Scan0016.pdf (368.2 KB, 63 views)
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    (Original post by maggiehodgson)
    I am trying question 9 (the underlining means it "goes beyond the requirements of the specification and are provided as a challenge")

    I am attaching my most current workings.

    Can you tell me, please, is my diagram correct.

    Also can you give me a clue as to the significance of the radius being the same as half the length of the rod.

    Thanks
    Your diagram is perfectly correct.
    the signigficance of the radius of the hemisphere is that when you take moments about the point of contact of the rod and the ground you can express \sin\theta in terms of W and the reaction between the hemisphere and the rod without any lengths being involved.
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    (Original post by brianeverit)
    Your diagram is perfectly correct.
    the signigficance of the radius of the hemisphere is that when you take moments about the point of contact of the rod and the ground you can express \sin\theta in terms of W and the reaction between the hemisphere and the rod without any lengths being involved.
    OK.

    I'll work on that.

    Many thanks
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    (Original post by brianeverit)
    Your diagram is perfectly correct.
    the signigficance of the radius of the hemisphere is that when you take moments about the point of contact of the rod and the ground you can express \sin\theta in terms of W and the reaction between the hemisphere and the rod without any lengths being involved.

    Done some stuff and I've ended up with R = Wsin(Theta).

    Is that correct?
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    how hard is aqa maths/f.maths as compared to edexcel or ocr?
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    (Original post by maggiehodgson)
    Done some stuff and I've ended up with R = Wsin(Theta).

    Is that correct?
    Yes that's fine so far.
    It's now a matter of using horizontal and vertical resolutions and a bit of fiddlng to get the required result.
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    (Original post by brianeverit)
    Yes that's fine so far.
    It's now a matter of using horizontal and vertical resolutions and a bit of fiddlng to get the required result.

    Have been trying and I need more fiddling. Expect me to be asking again if I get bogged down but thanks so far.

    at the moment I've got
    mu cos ^2 t + mu cos t - mu sin t + cos t sint - sin^2 t = w sintcost

    I suspect that is beyond fiddling so I'll try something else.
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    (Original post by maggiehodgson)
    I suspect that is beyond fiddling so I'll try something else.
    You might find it useful to post your initial equations when resolving horizontally and vertically, so it's clear as to whether you have the starting point correct, or not.
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    (Original post by ghostwalker)
    You might find it useful to post your initial equations when resolving horizontally and vertically, so it's clear as to whether you have the starting point correct, or not.

    From my drawing a few posts ago I did this

    ***** F1 + F2 + R2sin(theta) = Wsin(theta)

    F1 = mu W cos (theta)
    F2 = mu((w( cos(theta) - sin(theta)))/cos(theta))


    and then I got what I said before.

    Sadly, I haven't had time to look at maths since my last post so I've not done a restart. Now that I'm back I'll continue.

    I'll assume my ***** line is correct until told otherwise.

    Thanks
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    (Original post by maggiehodgson)
    From my drawing a few posts ago I did this

    ***** F1 + F2 + R2sin(theta) = Wsin(theta)

    F1 = mu W cos (theta)
    F2 = mu((w( cos(theta) - sin(theta)))/cos(theta))


    and then I got what I said before.

    Sadly, I haven't had time to look at maths since my last post so I've not done a restart. Now that I'm back I'll continue.

    I'll assume my ***** line is correct until told otherwise.

    Thanks
    The equations you should have for your horizontal and vertical resolutions are in the attached pdf. Eliminate R_2 then use the R_1=W\sin\theta that you found earlier
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  3. File Type: pdf question.pdf (16.9 KB, 70 views)
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    (Original post by brianeverit)
    The equations you should have for your horizontal and vertical resolutions are in the attached pdf. Eliminate R_2 then use the R_1=W\sin\theta that you found earlier

    Oh yes, I see it now. Sometimes I'm resolving up and down the slope and mixing it up with horizontal and vertical. I was sure that the first line of your equation was mistyped until the algebra wouldn't work out. Then I actually redrew and labelled the horizontal forces and vertical forces and all became clear.

    Perhaps I should've stopped after M1.

    Thank you, all of you.
 
 
 
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