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    I'm struggling with this:

    A uniform steel girder AB, of mass 40kg and length 3m, is freely hinged at A to a vertical wall. The girder is supported in a horizontal position by a steel cable attached to the girder at B. The other end of the cable is attached to the point C vertically above A on the wall, with <ABC=angle, were tan(angle)=3/4.
    A load of mass 60kg is suspended by another cable from the girder at the point D where AD =2m
    The girder remains horizontal and in equilibrium. The girder is modelled as a rod, and the cables as light inextensible string
    a) show that the tension in the cable BC is 980N
    b)Find the magnitude of the reaction on the girder at A

    Now for A I took moments around A:
    40g(3/2) + 2g(60)=3Tsin(angle)

    No problemos there,

    But in Question B apparently you have to consider the horizontal and vertical reactions. But I thought that reaction (N) is just perpendicular to the surface?

    If anyone could enlighten me please do.

    k10k
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    (a) M(A) : 60g+120g=1.8T

    => 180g=1.8T

    => T = 100g =980N

    (b) Vertical forces must be equal

    => 40g+60g-(0.6x980)=Vertical reaction at A = 392N

    Horizontally

    => (0.8x980) = 784N

    therefore, magnitude of resultant at A = √784²+392² = 876.5N = 877N

    From the force diagram, you should see that the reaction force is not just perpendicular to the wall, as the vertical forces do not add up..it is slightly different from the common ladder question
    • Thread Starter
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    'Find the magnitude of the reaction on the girder at A'
    In other words thats the reaction from the wall onto the girder?
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    yeah... you have to solve it by finding the two reaction components and then finding the magnitude ot these two components as KAISER MOLE did it above
 
 
 
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Updated: June 22, 2005
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