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    Hi! A little stuck on finding the force at A... anyone have any ideas?

    Many thanks in advance!
    Attachment 618860
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    You doing moments around a point, like CW and ACW ?

    Attachment doesn't seem to work, but maybe it takes time to authorise or something
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    (Original post by Casisalive)
    You doing moments around a point, like CW and ACW ?

    Attachment doesn't seem to work, but maybe it takes time to authorise or something
    Yes! Hope this works this time Name:  IMG_2080.jpg
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    (Original post by marinacalder)
    Also,Name:  IMG_2083.jpg
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    Any worked solutions would so greatly appreciated! I've been stuck on these for literally days
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    (Original post by marinacalder)
    Any worked solutions would so greatly appreciated! I've been stuck on these for literally days
    Post what you've got so far and we'll help.

    We don't DO questions for you.
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    Ok, for the first one

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    For Q7 it looks as though you've started in the right way. Similarly, you can find the horizontal force acting at A by taking moments about B. Equally, you could resolve horizontally and conclude that the horizontal force at A is minus that at B (as the gate is in equilibrium and there are no other horizontally acting forces in the picture).

    The next step is to resolve vertically, noting the clue in the question that the force at B is horizontal (meaning that the pin at B bears no weight, meaning in turn that all the weight must be borne by the pin at A).

    Can you take it from there?
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    (Original post by old_engineer)
    For Q7 it looks as though you've started in the right way. Similarly, you can find the horizontal force acting at A by taking moments about B. Equally, you could resolve horizontally and conclude that the horizontal force at A is minus that at B (as the gate is in equilibrium and there are no other horizontally acting forces in the picture).

    The next step is to resolve vertically, noting the clue in the question that the force at B is horizontal (meaning that the pin at B bears no weight, meaning in turn that all the weight must be borne by the pin at A).

    Can you take it from there?
    EEP! I've solved it! I couldn't have done that without your help!! Thank you ever so much!! Feel so excited having finally understood this question! , resultant force at A = 15(square root 562) g N.
 
 
 
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