Three point masses 30 g each are placed at the vertices of anequilateral triangl

Question

Three point masses 30 g each are placed at the vertices of anequilateral triangle ABC and rigidly connected by three rods oflength 10cm. The rods AB and C have equal mass 50g while the massof the rod BC is 20g.

Question: calculate the distance from the center of mass to thevertex A.

I want to use coordinates for this equation (not vectors) and havereplaced the mass of the rods with point masses. I'm trying tosubstitute values into the equation I'm using: xmc=m1r1+m2r2+m3m3/M.

I think there's a way to use the fact that the triangle isequilateral and divide the normal 60 degree angle by 2 to make one30 degree angle. However I don't know where to go from there. Icould use sin or cosine to find an unknown side, but I'd need toknow at least one side in order to do that which is what I'm havinga problem doing....

I have the answer for the problem and it is 5.36 cm"(13/(143) * L" (* =multiplied by))"
The person who wrote the answer also put yac=yab= L3/4 and ybc= L3/2.Hopefully this helps in trying to explain it to me?

Would appreciate any help! x

Explanation / Answer

Ans.    Due to 3 point masses located at thevertex of the equilateral triangle,                     the centre of gravity = centroid of the triangle            But as the mass of the rod BC is different from others,                      the centre of gravity is not equal to centroid of triangle.           Now for each rod the centre of gravity will be at their midpoint.           Now if we join the mid points of each line(rod) then we will findanother equilateral triangle whose side=10/2=5 cm.            This can be found out from properties of triangle.           Now forthis new equilateral triangle,two 50 g is located at two vertices& 20 g is located at remaining vertices.           Now from the triangle,             50*5+50*5=(50+50+20)*x   [where x=distance of centre ofgravity from mid point of BC]            x=4.167 cm           distance of centre of gravity from vertexA=3/2*(length of side of original triangle)-x                                                                             =8.660-4.167                                                                             =4.493 cm    [Ans.]

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