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Im studying for my exam and going through problems I got wrong, trying to fix them, heres what I did.

Each leg supports 12 KG, 117.6 N.

FD = Force down, one thats beeing applied.

C = Circumference of the table

Sum of the torque = FD(1/6C) - 117.6(1/3C) - 117.6(1/3C)

Since I dont have direct measurements of the table, the distance I put in terms of circumference, using one of the legs that the force is being put in between as a rotation point. I came out with:

1/6(FD) = 78.4

FD = 470.4 N

Mass = 48 Kg

The answer is 36, any help is appreciated