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M3 Statics of Rigid Bodies watch

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    Attachment contains the question.

    I needed some help in understanding how the triangle in the solution part has an angle of 20degrees. And also, how it was deduced that triangle OGS has an angle alpha on the dashed line produced by SG.

    If anyone could offer further diagrams of how these angles were worked out it would GREATLY be appreciated.

    Question and solution are in word attachment below.

    MANY THANKS!
    Attached Files
  1. File Type: docx statics of rigid bodies.docx (155.0 KB, 79 views)
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    can anyone offer any advice please? mainly on the point of how alpha is placed on the right hand side triangle in the solutions.

    Thanks!
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    (Original post by Dalek1099)
    ...
    Hi, I've heard you are a somewhat of a genius at maths

    Would you be able to shed some light on the solution provided in the first post please(its in the word attachment). Mainly on how they deduced triangle OGS has angle alpha on the dashed line SG produced.

    Thanks
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    (Original post by Coral Reafs)
    Hi, I've heard you are a somewhat of a genius at maths

    Would you be able to shed some light on the solution provided in the first post please(its in the word attachment). Mainly on how they deduced triangle OGS has angle alpha on the dashed line SG produced.

    Thanks
    I Don't do M3 till next year sorry M2 is hard enough-the mock was so hard that I think my class are going to have another one.
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    (Original post by Dalek1099)
    I Don't do M3 till next year sorry M2 is hard enough-the mock was so hard that I think my class are going to have another one.
    ah ok, would you maybe know anybody else who can explain it?

    thanks anyway
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    (Original post by Coral Reafs)
    ah ok, would you maybe know anybody else who can explain it?

    thanks anyway
    This guy is virtually the god of maths:
    http://www.thestudentroom.co.uk/member.php?u=937760
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    (Original post by DJMayes)
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    Hi, very sorry to bother you, but I'm stuck on understanding the method of a question.

    The question and method are in the first post, in the word document attachment.
    My confusion was, how in triangle OGS the dotted line produced by SG had an angle of alpha. I understand that alpha is the angle that the rectangular plane face makes with the horizontal, but I don't know how to put this to use.

    Thanks in advance!
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    (Original post by Coral Reafs)
    Hi, very sorry to bother you, but I'm stuck on understanding the method of a question.

    The question and method are in the first post, in the word document attachment.
    My confusion was, how in triangle OGS the dotted line produced by SG had an angle of alpha. I understand that alpha is the angle that the rectangular plane face makes with the horizontal, but I don't know how to put this to use.

    Thanks in advance!
    Firstly, note that the angle between the plane and the radius of the semicircle touching it is 90, as the plane is tangent to a circular arc. This allows you to deduce angle OGS as 20 degrees.

    Now, you define alpha as the angle between your horizontal and the flat face of the semi-circular cross-section. Call the point from which you measure the angle as B. Then, OBSG is a quadrilateral, with angles BSG as 90 degrees (Obviously, as this is horizontal-vertical) and OBG 90 degrees (As the centre of mass of the object is in the middle by symmetry). Using the sum of the angles of a quadrilateral as 360, you deduce angle OGS as 180 - alpha, giving the angle you mentioned as alpha (Angles in a line).
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    (Original post by DJMayes)
    Firstly, note that the angle between the plane and the radius of the semicircle touching it is 90, as the plane is tangent to a circular arc. This allows you to deduce angle OGS as 20 degrees.

    Now, you define alpha as the angle between your horizontal and the flat face of the semi-circular cross-section. Call the point from which you measure the angle as B. Then, OBSG is a quadrilateral, with angles BSG as 90 degrees (Obviously, as this is horizontal-vertical) and OBG 90 degrees (As the centre of mass of the object is in the middle by symmetry). Using the sum of the angles of a quadrilateral as 360, you deduce angle OGS as 180 - alpha, giving the angle you mentioned as alpha (Angles in a line).
    Thank you for taking the time to produce a very detailed post, this has helped me immensely and I can't thank you enough for responding this late at night too!!

    very much appreciated and I gave you a +1
 
 
 
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