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?
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 within 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.
To find the load on each of the three legs, we need to consider the equilibrium of forces acting on the table. The total load on the table is 500N, and it is applied at a point distant 325mm from one leg and 625mm from another leg.
To solve this problem, we can divide it into two components: the load sharing between the two legs that directly support the load, and the load distribution between these two legs and the third leg.
Let's solve the problem step by step:
Step 1: Determine the load sharing between the two legs directly supporting the load.
Let's assume that leg A supports the load at a distance of 325mm, and leg B supports the load at a distance of 625mm.
The load on leg A can be determined using the principle of moments. Moments are calculated by multiplying a force by its perpendicular distance from a reference point or axis. In this case, the reference point is the leg B.
The moment of the load on leg A = Force on leg A × Distance between leg B and the load
Since the load is in equilibrium, the sum of moments about any point should be zero. Therefore, the moment of the load on leg A should be equal to the moment of the load on leg B.
The moment of the load on leg B = Force on leg B × Distance between leg A and the load
Since the total load on the table is 500N, we can write the equation:
Force on leg A × 325mm = Force on leg B × 625mm
Step 2: Determine the load distribution between these two legs and the third leg.
The load distribution between these two legs and the third leg is determined by the principle of vertical equilibrium. In other words, the sum of vertical forces acting on the table should be zero.
Let's assume that the load on the third leg is C.
According to the principle of vertical equilibrium:
Force on leg A + Force on leg B + Force on leg C = Total Load
And we know that
Force on leg A + Force on leg B = 500N
Now we have two equations:
Force on leg A × 325mm = Force on leg B × 625mm
Force on leg A + Force on leg B + Force on leg C = 500N
We can solve these equations simultaneously to find the load on each of the three legs.