A uniform .122kg rod of .90 m length is used to suspend two masses as shown below. At what distance x should the .20kg mass be place to achieve static equilibrium? Okay so I got the Torque of the 0.2kg mass T= (0.2)(9.8)(x) with x being the length I hafta find. Then I got the Torque of the 0.50kg mass T= (0.5)(9.8)(0.2) Then I plugged it into this.... sum of Torques CW= Sum of Torques CCW and the answer was 0.5m and I got that but originally for the 0.50 mass, I put in (0.25) for the length because that's what it is but I didn't get teh answer so I did trial and error with the numbers I had and I figured 0.2m was the right length to get the right answer. So I'm wondering why it's 0.2 m and if the mass of the rod matters.. cause I dind't use it.