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    Hey! Anyone know how I might go about part a of this question?
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    (Original post by marinacalder)
    Hey! Anyone know how I might go about part a of this question?
    What question?
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    the lamina is folded too tightly to see :teehee:
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    OMG!! hahaha!

    wow.. that pun... :lol::lol::lol:

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    (Original post by marinacalder)
    ...
    1) Find the c-o-m C relative to S by treating it as a rectangle of uniform density \rho and triangle of uniform density 2\rho.

    2) When it hangs in equilibirum, then C is vertically underneath S. It's just geometry/trig thereafter.
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    (Original post by marinacalder)
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    6 a) Folding reflects the triangle in the dotted line, so R lies on PQ axis, vert below T. RT is vert, RQ horiz, both length 3. Vertices at R(1, 0), Q(4, 0), T(1, 3).

    A rectangle of h = 3, w = 1, density p, joins the triangle of density 2p along RT (identical triangle lies beneath).

    Area (r) = 3 x 1 = 3 => Mass (r) = 3p. Area (t) = 0.5 x 3 x 4 = 6 => Mass (t) = 12p. Mass (lamina) = 15p.

    com (r) is (0.5, 1.5). com (t) = mean of vertices' (x, y) coords: x_t = (1 + 1 + 4)/3 = 2, y_t = (3 + 0 + 0)/3 = 1.

    For (X, Y) coords of com (lamina):

    (15p)X = (3p)(0.5) + (12p)(2) = 3p/2 + 24p = 51p/2
    => X = 51/30 = 17/10

    (15p)Y = (3p)(1.5) + (12p)(1) = 9p/2 + 12p = 33p/2
    => Y = 33/30 = 11/10

    com (lamina) is 17/10 units from PS, 11/10 from PQ.

    b) If you hang the lamina from S, the line joining S to com (lamina) is the vert. Use R-A triangle to calc @:
    opp = X = 17/10, adj = PS - Y = 3 - 11/10 = 19/10.

    So [email protected] = (17/10)/(19/10) = 17/19 => @ = 41.8°
 
 
 
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