Question

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 on each of the three legs?

Answer 1

Take a piece of paper, and sketch the geometry as describe below to make a sketch that you can follow the steps.

1. Sit the triangle on the x-axis, with the left vertex A at origin, so that (0,900) is the other vertex B.
The third vertex C has coordinates (450, 450√3)

2. The loading point D forms another triangle with a common base of the first, assuming point D is with<i>in</i> the triangle with sides 900, 325 and 625.

3. Assume mAD=325, hence mBD=625.

4. Drop a perpendicular from D to AB, meeting AB at E.
Denote height mDE as h, and mAE as x, then mEB=900-x.

5. Using Pythagoras Theorem, we have two equations:
h^2+x²=325²................(1)
h^2+(900-x)²=625²..........(2)

6. Rewrite (1) as h²=325²-x² and substitute in (2). Solve for x, and hence h.
You should get x=875/3, h=50√(74)/3.

7. Denote leg reactions as Ra, Rb and Rc respectively, and the load P=500N.

8. Take moments about x-axis.
P*h-Rc*(450√3)=0
Solve for Rc=500√(74)/3^(7/2);

9. Take moments about the y-axis.
P*x-Rb*900-Rc*450=0
Solve for Rb=(4375*3^(3/2)-750*sqrt(74))/3^(9/2)

10. Finally, Ra=P-Rb-Rc=500-(500*sqrt(74))/3^(7/2)-(4375*3^(3/2)-750*sqrt(74))/3^(9/2)

You are expected to be able to solve the simple algebraic equations using a calculator, which is much simpler than the expressions which I gave for your verification.

Hint:
You will find that the proportions of Ra, Rb and Rc are approximately 3:1:1, which is quite reasonable, since D is closer to A than B.

I will be glad to check your answer if you would post it.