Question 1062547: A workshop table has an equilateral triangular top, each side of which 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 rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! Euler did the original work on solving reaction problems using rigid legs and tables
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the height(h) of equilateral triangle is 900 * sqrt(3) / 2 = 450 * sqrt(3)
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the area(A) of the equilateral triangle is sqrt(3) * 900^2 / 4 = 202500 * sqrt(3)
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lets say the point P is where we placed the load on the table and we draw three line segments from point P to each of the vertices of the equilateral triangle table top
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we have to introduce some notation to understand how to solve this problem
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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
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we use the law of cosines twice to determine the length of V(1)P
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625^2 = 325^2 + 900^2 - (2*325*900*cos(PV(2)V(3)))
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cos(PV(2)V(3)) = (325^2 + 900^2 - 625^2) / (2*325*900) = 0.8974
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cos^(-1) (0.8974) = 26.18 degrees
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angle V(1)V(2)P = 60 - 26.18 = 33.82 degrees
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(V(1)P)^2 = 900^2 + 325^2 - (2*900*325*cos(33.82)) = 429612.71
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V(1)P = square root (429612.71) = 655.45 approximately 655 mm
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we have all three distances from P to the vertices
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we have three triangles to consider, (V(2),P,V(3)), (V(1),V(2),P) and (V(3),P,(V(1))
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lets select triangle (V(2),P,V(3)) to work with, the other two are done in a similar manner
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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)
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L(1) = 500 * area of (V(2),P,V(3)) / 202500 * sqrt(3)
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we can use Heron's formula to determine the area of triangle (V(2),P,V(3))
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T = sqrt(s(s-a)(s-b)(s-x)), where s = (a+b+x)/2, a=325, b=625, x=900
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We do L(2) and L(3) similarly
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