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    http://papers.xtremepapers.com/Edexc...elphis%206.pdf

    In question 5 a) it asks to draw the forces. I'm know concerned about the force acting on the hinge. I know the reaction force has 2 components the vertical (or parallel to the wall) which is the force of friction and the horizontal (or perpendicular to the wall) which is the normal reaction. Since the question here states the rod is smoothly hinged i THINK there is no vertical force am I right? if so, what balances out the weight?
    thanks
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    A smooth hinge just means that it can rotate easily - just like the 'smooth pulleys' you probably met in M1. There are therefore both horizontal and vertical components of the force exerted on the rod by the hinge (and on the hinge by the rod) (note: not friction, the rod is HINGED to the wall, not resting on a rough wall).
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    (Original post by dragonkeeper999)
    A smooth hinge just means that it can rotate easily - just like the 'smooth pulleys' you probably met in M1. There are therefore both horizontal and vertical components of the force exerted on the rod by the hinge (and on the hinge by the rod) (note: not friction, the rod is HINGED to the wall, not resting on a rough wall).
    so to get this straight the vertical force isn't friction and it is existing if even if it is smoothly hinged am I correct?
    Plus what about the horizontal component im going to assume it is not the normal reaction force or what? thanks again
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    (Original post by >>MMM<<)
    so to get this straight the vertical force isn't friction and it is existing if even if it is smoothly hinged am I correct?
    Plus what about the horizontal component im going to assume it is not the normal reaction force or what? thanks again
    Yes - it may help you to visualise holding a door (ok, a small door...) in your hand by the hinge, with the rope attached as in the picture. Try to imagine where the door would fall if you suddenly let go - this is the direction in which the force of the door on your hand is going. As every force has an equal and opposite reaction force, your hand must therefore be exerting a force in the exact opposite direction in the door - i.e. sort of diagonally.

    There is actually just one force (the normal force), but you tend to split it into horizontal and vertical components to make calculations easier.
    EDIT: maybe it's sort of two forces... I tend to just label them N (perpendicular to the surface, so in this case horizontally) and F (which could stand for friction if it was just resting on a rough surface, or just force (as in this case)). The examiner can decide what they mean... (I doubt you will be asked 'what is the normal force' anyway...)
 
 
 
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