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Previously on "Physics question - static equilibrium"

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  • Churchill
    replied
    Originally posted by AtW View Post
    Hmmm, sasguru isn't posting the solution.

    How peculiar...
    He's too busy blowing squaddies. Should've been NotAllThere's shift but he's too busy doing his kids homework.
    Last edited by Churchill; 10 November 2013, 21:19.

    Leave a comment:


  • AtW
    replied
    Hmmm, sasguru isn't posting the solution.

    How peculiar...

    Leave a comment:


  • DimPrawn
    replied
    Christ you lot are THICK.

    THICK.


    If you CRETINS hadn't wasted your time at school and got a proper education you would have this solved in seconds!

    Leave a comment:


  • mudskipper
    replied
    Originally posted by alluvial View Post
    Ah, but try hanging two masses on the string....
    Then you'd need the sine of d and e.

    Leave a comment:


  • alluvial
    replied
    Originally posted by mudskipper View Post
    Yay, so we're all right!
    Ah, but try hanging two masses on the string....

    Leave a comment:


  • suityou01
    replied
    Originally posted by mudskipper View Post
    Yes dear, but the opposite/adjacent thing isn't in relation to the hypotenuse, it's in relation to the angle you know - so your post was nonsense.

    Here's a picture for you.

    Good, apology accepted.

    Leave a comment:


  • mudskipper
    replied
    Originally posted by alluvial View Post
    This could be where the problem is.
    Yay, so we're all right!

    Leave a comment:


  • alluvial
    replied
    Originally posted by mudskipper View Post
    Where have we defined a and b?
    This could be where the problem is.

    Leave a comment:


  • mudskipper
    replied
    Where have we defined a and b?

    Leave a comment:


  • mudskipper
    replied

    Leave a comment:


  • alluvial
    replied
    Originally posted by mudskipper View Post
    Yes dear, but the opposite/adjacent thing isn't in relation to the hypotenuse, it's in relation to the angle you know - so your post was nonsense.

    Here's a picture for you.

    Sorry, but suity's post wasn't complete nonsense.
    His first statement is correct and agrees with my previous post.
    As the system is in static equilibrium, then the sum of the forces must equate to zero. As this is a 2D system, then you only need to consider the xy plane and so the sum of the forces acting upwards at the point of susppension must equal the downwards force created by the mass and so:

    F=19 cos(a) + 12 cos(b) This is the force that stops the mass from dropping downwards.

    The adjacent side in this case is the perpendicular and so cosine is used.

    His second statement was (almost) wrong as this will only be true if a=b.

    You then resolve the forces perpendicular to the direction in which the mass exerts a force. In this case, sine will be used and now, 19sin(a) = 12sin(b). This needs to be true to stop the mass from moving sideways.

    Once you can determine what a or b are, then you can solve the first equation and then determine M by dividing by g.

    Edit: BTW, I'm using a and b to be the angles obtained by subdividing the angle formed by the two suspension cables at the point of suspension.

    Leave a comment:


  • ctdctd
    replied
    Originally posted by mudskipper View Post
    Hypotenose?

    Leave a comment:


  • mudskipper
    replied
    Originally posted by ctdctd View Post
    Ooooooh, a picture, has it got kittens in it?
    Hypotenose?

    Leave a comment:


  • ctdctd
    replied
    Originally posted by mudskipper View Post
    Yes dear, but the opposite/adjacent thing isn't in relation to the hypotenuse, it's in relation to the angle you know - so your post was nonsense.

    Here's a picture for you.

    Ooooooh, a picture, has it got kittens in it?

    No, oh, as you were then.

    P.S. Suity, she called you Dear, but somehow I don't think you've pulled.

    Leave a comment:


  • mudskipper
    replied
    Originally posted by suityou01 View Post
    Oooh spooky, there's an echo.
    Yes dear, but the opposite/adjacent thing isn't in relation to the hypotenuse, it's in relation to the angle you know - so your post was nonsense.

    Here's a picture for you.

    Leave a comment:

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