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  1. Oct 3, 2015 #1
    1. The problem statement, all variables and given/known data
    a thin uniform wire is bent to form two equal sides AB and AC of triangle ABC , where AB=AC=5cm . the third side BC , of length 6cm is made from uniform wire of twice the density of the first . the distance of centre of mass from A is ?

    2. Relevant equations
    x centre of mass = m1R1+m2r2/ sum of masses(where r= centre of mass of particle in sys.)

    λ=m/l ( where m= mass of object , l=length ) uniform density formula
    3. The attempt at a solution
    i was able to calculate the X c.o.m which is = 12λ*lenght of rod BC/2
    but was unable to calculate its y coordinate ?
  2. jcsd
  3. Oct 3, 2015 #2


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    I don't think that's the solution this problem is looking for.

    Once the wire is formed into the triangle, you are supposed to calculate the (x,y) coordinates of the center of mass from point A.
  4. Oct 4, 2015 #3


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    Make a drawing. Mark on it the centre of mass of each rod. One of the co-ordinates (say x) is trivial by symmetry. The other (say distance y from point A) needs some maths.

    Show your working (forum rules).
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