- #1
Jacob87411
- 171
- 1
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, here's what I did.
Each leg supports 12 KG, 117.6 N.
FD = Force down, one that's beeing applied.
C = Circumference of the table
Sum of the torque = FD(1/6C) - 117.6(1/3C) - 117.6(1/3C)
Since I don't 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
Im studying for my exam and going through problems I got wrong, trying to fix them, here's what I did.
Each leg supports 12 KG, 117.6 N.
FD = Force down, one that's beeing applied.
C = Circumference of the table
Sum of the torque = FD(1/6C) - 117.6(1/3C) - 117.6(1/3C)
Since I don't 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