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    I hate this topic, can do the ladders questions but the other ones bug me

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    The figure shows a pole of mass m and length l displaying a light banner. The pole is modelled as a uniform rod AB, freely hinged to a vertical wall at point A. It is held in horizontal position by a light wire inclined at angle & to the horizontal. One end of this wire is attached to the end B of the rod and the other end is attached to the wall at point C which is vertically above A and such that tan & = 1/3

    a) Show that the tension in the wire is 0.5mg x sqrt10

    Ok, I got this by doing moments about A. However I first tried resolving vertically and I got mg x sprt10. Why didnt resolving vertically work? Am i missing a force with a vertical component? I have the tension and mass but thats it.

    b) Find the magnitude of the force exerted by the wall on the rod at A

    Is this the horizontal force acting along the pole at A? I tried resloving horizontally but that didnt work. Am I missing any forces???
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    (Original post by Womble548)
    I hate this topic, can do the ladders questions but the other ones bug me

    3: (no picture )
    The figure shows a pole of mass m and length l displaying a light banner. The pole is modelled as a uniform rod AB, freely hinged to a vertical wall at point A. It is held in horizontal position by a light wire inclined at angle & to the horizontal. One end of this wire is attached to the end B of the rod and the other end is attached to the wall at point C which is vertically above A and such that tan & = 1/3

    a) Show that the tension in the wire is 0.5mg x sqrt10

    Ok, I got this by doing moments about A. However I first tried resolving vertically and I got mg x sprt10. Why didnt resolving vertically work? Am i missing a force with a vertical component? I have the tension and mass but thats it.

    b) Find the magnitude of the force exerted by the wall on the rod at A

    Is this the horizontal force acting along the pole at A? I tried resloving horizontally but that didnt work. Am I missing any forces???

    is this in the orange m2 book? if so what page?
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    Its in the 'revise for' book. Exam q3 at end
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    dont have it damn..sorry i cant visualize the pic
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    Hope thats an acurate enough diagram...

    Anyway... way to do it would be:

    So Moments about A:

    l.mg/2 = l.T√10

    so cancel l and solve for T.. and you get:

    T = 0.5mg * √10

    -------------------------------
    b) Resolve Horizontally for this (I think) - the number of marks for the question often gives you the hint.

    so Normal Reaction at A (R)= T.cos(α)

    So,

    R = 0.5mg*√10*3/√10

    R = 1.5mg

    Note, I'm not 100% sure about part B.. whats the answer in the book?

    MdSalih
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    lol I am doing revising for M2 by looking at people's posts on TSR :P
    I suck at M2 by the way, so dun trust my words, thank you~


    T = 0.5mg*sqrt10

    At hinge A there are two forces, one vertical and one horizontal (I think)

    Resolve vertical
    R = mg + 0.5mg*sin(α)*sqrt10
    = 15.mg

    Resolving horizontal
    F = 0.5mgXsqrt10X cos(α)
    = 1.5mg


    I got R = F = 1.5mg at the hinge..
    that doesn't sound right...check answer...

    and very uncertain on the following part as to whether correct or not and whether it is actually needed:

    magnitude (P) of force at hinge:

    P = sqrt ( (1.5mg)^2 + (1.5mg)^2)
    = sqrt (432.18m^2)
    = 20.78893937 X m
    so hinge force has magnitude = 20.8m at direction 45deg to horizontal?
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    The reaction at the wall is 0.5mg * √10

    Its a 7mark question to find the reaction so I guess there is more work than resolving horizontally I did that and got the same as you.

    Ill just have to hope that a ladders question comes up instead
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    (a)
    Vertically: Tsint + Y = mg
    Tcost = X

    C=AC, taking moments about A:
    (Tsint)(l) = (mg)(.5l)
    Tsint = 0.5mg

    T = mg/2sint
    T = mg/2sin(arctan1/3)
    T = 0.5mgrt10

    (b) X = Tcost = 3mg/2
    Y = mg-Tsint = 0.5mg

    Force = mgrt[1/4 + 9/4]^.5
    = 0.5mgrt10
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    Ahhh, got ya! There is more than just a horizontal force at A.

    So that explains why resolving vertically for Q1 didnt work!
 
 
 
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