Three legs placed equal distances apart on the edge support a 36-kg round table. What minimum mass, placed in the table's edge, will cause the table to turn over? Hint: Place the mass equally inbetween two legs. 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
When the table is just about to tip, the torque exerted by the applied force will equal the torque exerted by the weight of the table. The axis of rotation is the line joining two legs: Find the torques that the forces make about that line. Hint: find the distance the forces are to that line in terms of the radius R.