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    I must have made an error somewhere, but i can't quite see where it is.
    This is Q27 of Rev ex 2 of the Heinemann M2 textbook.

    My working is:
    Let S = the magnitude of the force, as the cylinder is smooth the force acts perpendicular to the rod at the point of contact.
    Let x be the distance along the rod from A to C.
    Let theta=t for simplicity.
    Let F be the frictional force and R the contact at the ground.

    Resolving upwards:
    R + Scos2t = mg

    Resolving horizontally:
    Ssin2t = F

    Taking moments at A gives:
    C = AC
    (Scos2t)(xcos2t) + (Ssin2t)(xsin2t) = (mg)(3acos2t)
    Sx(cos^2 2t + sin^2 2t) = 3mgacos2t
    Sx = 3mgacos2t.

    Resolving dimensionally upwards:
    xsin2t = 2a (diameter of cylinder).
    x = 2a/sin2t

    Substituting gives:
    2aS/sin2t = 3mgacos2t
    S = (3mg/2)(cos2t)(sin2t) = (3mg/4)(sin4t).
    But the book gives 3mgcos2ttant.

    Any help is much appreciated.
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    Gah... wish I could help, but I haven't got that far yet. M2 you say? I may be screwed. Still haven't finished FP1 let alone started M2 and M3.
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    Heh. No worries. You'll get them done ^^
    I have P2/P3/P4 to revise and P5/P6/D2 to cover from scratch before June, as well as Bio exams.
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    [QUOTE=Gaz031]

    Resolving dimensionally upwards:
    xsin2t = 2a (diameter of cylinder).

    [QUOTE]

    I don't see how you got this bit!

    I can't see a diameter in this triangle.

    Aitch
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    I have FP1 and P3 to finish and M2 M3 and S2 to do. Did the P2 exam yesterday.
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    [QUOTE=Aitch]
    (Original post by Gaz031)

    Resolving dimensionally upwards:
    xsin2t = 2a (diameter of cylinder).

    QUOTE]

    I don't see how you got this bit!

    I can't see a diameter in this triangle.

    Aitch
    Put a/x = tan t instead and you've got your answer...

    put a = x tant into Sx = 3amg cos2t

    get Sx =3x tant mg cos2t =>

    3 tant mg cos2t

    Aitch
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    [QUOTE=Aitch]
    (Original post by Gaz031)

    Resolving dimensionally upwards:
    xsin2t = 2a (diameter of cylinder).

    QUOTE]

    I don't see how you got this bit!

    I can't see a diameter in this triangle.

    Aitch
    You mean cylinder?
    It's 2.(the radius) = 2a, as the rod rests on the top of the cylinder (2a above the ground).
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    (Original post by Gaz031)
    You mean cylinder?
    It's 2.(the radius) = 2a, as the rod rests on the top of the cylinder (2a above the ground).
    It's not on top of the cylinder - If it were, it couldn't also touch the ground!

    Aitch
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    (Original post by Aitch)
    Put a/x = tan t instead and you've got your answer...

    put a = x tant into Sx = 3amg cos2t

    get Sx =3x tant mg cos2t =>

    3 tant mg cos2t

    Aitch

    Aitch
    Ahh, i see. Thankyou. The key being that at half the angle it will have only half the elevation.

    It's not on top of the cylinder - If it were, it couldn't also touch the ground!
    Oh oh my god i'm so stupid :rolleyes:

    Thanks for the help.
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    (Original post by Gaz031)


    Thanks for the help.
    No problem! I still owe you for all the exam questions!


    Aitch
 
 
 
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