Question 1062546: A workshop table has an equilateral triangular top, each side is 900mm, the legs being at the three corners. A load of 500N is placed on the table at a point distant 325mm from one leg and 625mm from another. What is the load in each of the three legs.
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! height(h) of equilateral triangle: 900*sqrt(3)/2 = 450*sqrt(3)
area of the equilateral triangle is sqrt(3)*900^2/4 = 202500*sqrt(3)
Let the point at 325mm from one leg and 625mm from another be P.
We draw three line segments from point P to each of the vertices of the equilateral triangle.
Now, let V(1), V(2), V(3) be the vertices of the triangular table top, x be the length of a side, L(1), L(2), L(3) be the loads on the three legs.
And V(1)P+V(2)P+V(3)P = x*sqrt(3)
V(1)P = 900*sqrt(3)-325-625
we have all three distances from P to the vertices
we have three triangles to consider, (V(2),P,V(3)), (V(1),V(2),P) and (V(3),P,(V(1))
select triangle (V(2),P,V(3)) to work with, the other two are done in a similar manner
L(1) will be the load*ratio of the area of triangle (V(2),P,V(3)) to the area of (V(1),V(2),V(3)
L(1) = 500*area of (V(2),P,V(3))/202500*sqrt(3)
we can use Heron's formula to determine the area of triangle (V(2),P,V(3))
T = sqrt(s(s-a)(s-b)(s-x)), where s = (a+b+x)/2, a=325, b=625, x=900
Do L(2) and L(3) the same way.
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