# Wait a Second, That Table Has No Legs!

Really, this doesn’t tell us much. All it says is that the upward-pulling tension has to be equal to the two downward forces (gravity and the other tension).

But what about the sum of the torques? If the object is in equilibrium, you can pick any point on the object to calculate the torque. I’m going to pick point o, where the upward-pulling string is attached. And I’ll say clockwise torques are negative values and counterclockwise are positive.

To get the torque resulting from each force, remember that τ = Fr. But since the distance (r) for T1 is zero, this tension results in zero torque.

So now, with only two other forces, the only way for their torques to offset is for one to pull clockwise and the other to pull counterclockwise. T2 is pulling down on the right side, which creates a negative torque around point o of T2 r2. But the gravitational force mg also pulls down—we can’t change that. That means the center of gravity of the top platform has to be on the other side of the central support string. So here’s our equilibrium torque equation:

That’s the key to the whole thing: The center of gravity of the “floating” tabletop and the downward force T2 need to be on opposite sides of the central suspension string. It’s actually not that complicated, right?