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    Bit stumped on this question.

    Fig 2 (below) shows a vase placed on a uniform shelf that is supported by a steel wire (orange line below). The ma ss of the vase is 0.65kg and the mass of the shelf is 2.0kg. The shelf is hinged at A. The steel wire is attached to the shelf 0.30m from A and is at an angle of 30 degrees to the shelf. The other end of the steel wire is attached to the wall.

    Show, by taking moments about A, that the tension in the steel wire is about 50 N

    So I thought I would take moments about A as the question asked and I did

    A = (9,81*0.65*0.45) + (2*9.81*0.25) [0.25 because uniform beam so all mass acts here]

    Then I thought I would get the force on A and then do A/sin(30). I got a value that was no where near 50N. Anyone have any ideas? Thanks.
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    Your working is correct. But let's not forget that the string is anchored to the shelf 0.3m away.
    Therefore it is not \frac{A}{\sin 30}, it is the counterclockwise moment of \frac{A}{0.3\sin 30}
    Which will yield 51N
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    (Original post by The-Spartan)
    Your working is correct. But let's not forget that the string is anchored to the shelf 0.3m away.
    Therefore it is not \frac{A}{\sin 30}, it is the counterclockwise moment of \frac{A}{0.3\sin 30}
    Which will yield 51N
    Thanks for the reply

    I sort of understand why you did *0.3 but I don't know why you did it to the sin(30)?
    I thought the force acting on A would just be trig and not involve the 0.3? I'm still a bit confused! Sorry if I sound stupid! I just suck at the mech part of phys
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    (Original post by CrazyFool229)
    Thanks for the reply

    I sort of understand why you did *0.3 but I don't know why you did it to the sin(30)?
    I thought the force acting on A would just be trig and not involve the 0.3? I'm still a bit confused! Sorry if I sound stupid! I just suck at the mech part of phys
    Here, ill draw a diagram for you just a sec, ill edit this in a min.
    And no you dont sound stupid at all! its not easy stuff!
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    (Original post by The-Spartan)
    Here, ill draw a diagram for you just a sec, ill edit this in a min.
    And no you dont sound stupid at all! its not easy stuff!
    Wow thanks for that. Much appreciated
    So just to clarify, is the clockwise moment A the dotted arrow going upwards then?
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    This looks like a question from one of the Physics specimen papers. Nice
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    (Original post by CrazyFool229)
    Wow thanks for that. Much appreciated
    So just to clarify, is the clockwise moment A the dotted arrow going upwards then?
    No worries
    The clockwise moment is the one you found yourself. You labelled this A so i did too, maybe you have anticlockwise and clockwise mixed up?

    The anti-clockwise moment is the dotted line, acting upwards. This is T(0.3\sin{30}).
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    (Original post by TheTechnoGuy)
    This looks like a question from one of the Physics specimen papers. Nice
    Yup, it is. Just got round to finishing all the questions but I had to go back to this one because I couldn't get anywhere near 50N! Turns out it was that one silly step with the *0.3
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    (Original post by The-Spartan)
    No worries
    The clockwise moment is the one you found yourself. You labelled this A so i did too, maybe you have anticlockwise and clockwise mixed up?

    The anti-clockwise moment is the dotted line, acting upwards. This is T(0.3\sin{30}).
    Yep, my bad again getting it mixed up haha. Thanks a lot for your help btw!
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    (Original post by CrazyFool229)
    Yep, my bad again getting it mixed up haha. Thanks a lot for your help btw!
    I do it all the time
    And it's no problem! glad to be of help.
 
 
 
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