load forces

A workshop table has an equilateral triangular top each side 900mm the legs are on the three corners . A load of 500N is placed on the table at point distant 325mm from one leg and 625mm from another . What is the load on each of the three legs
Andries ChakarisaAsked:
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☠ MASQ ☠Commented:
Around 300N on the leg nearest the weight.
Would you just like the answer or do you want some guidance in working this out for yourself?
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Andries ChakarisaAuthor Commented:
I want some guidance to understand its
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aburrCommented:
"I want some guidance to understand its "
Here are some question which will get you to the answer.
1. what is the sum of the forces on the legs?
2. Does the table tilt. Of course not, but that does lead you to the observation that the sum of the torques about the location of the mass must be 0.
3. You know the lever arm for two of the torques (325, 625). Use geometry to find the third lever arm.
4. label the three unknown forces u, v, w.
5. Try to get three equations for the three unknowns. u + v + w = 500.
6. Use the torques to get two more. (You may have to take the x axis and y axis components separately.)
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aburrCommented:
The first thing to do is to draw yourself a fairly accurate diagram.
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aburrCommented:
I was asked to expand on the solution outlined. The expansion is below
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Draw diagram with locations of  u and v determining the base of the triangle and w the apex.
Let x be the distance from w to p, the location of the load.
Let theta be the angle from the base at u to p.
Let phi be the angle from the base at v to p.
Let alpha be the angle from a horizontal at w to p.
Then the x components of the torques are
      -325 u cos theta
        625 v cos phi
        x w cos alpha
And the y components are
            325 u sin theta
            625 v sin phi
            -w x sin alpha
So we have six equations and seven unknowns, Get more information from geometry. Here we know that the triangle is 900 units on a side so that
325 sin theta + 625 cos phi = 900
325 cos (60 – theata) + x cos (alpha – 60) = 900
635 cos(60 – phi + x cos (120 – alpha) = 900
Solve these simultaneous equations being careful of the signs and converting degree to and from radians.
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☠ MASQ ☠Commented:
Visualised :)
:)
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aburrCommented:
requested expansion provided
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Math / Science

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