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Physics question - static equilibrium

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    #21
    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.

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      #22
      Originally posted by ctdctd View Post
      Ooooooh, a picture, has it got kittens in it?
      Hypotenose?

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        #23
        Originally posted by mudskipper View Post
        Hypotenose?

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          #24
          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.

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            #25

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              #26
              Where have we defined a and b?

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                #27
                Originally posted by mudskipper View Post
                Where have we defined a and b?
                This could be where the problem is.

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                  #28
                  Originally posted by alluvial View Post
                  This could be where the problem is.
                  Yay, so we're all right!

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                    #29
                    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.
                    Knock first as I might be balancing my chakras.

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                      #30
                      Originally posted by mudskipper View Post
                      Yay, so we're all right!
                      Ah, but try hanging two masses on the string....

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