Hello! I was super confused about Question 95 from the Physics AAMC question pack and wanted to get some help understanding the answer.

Question
A 0.5-kg uniform meter stick is suspended by a single string at the 30-cm mark. A 0.2-kg mass hangs at the 80 cm mark. What mass hung at the 10-cm mark will produce equilibrium?

I'm still having trouble understanding the process and explanation. I understand that the stick is in rotational equilibrium so the torque clockwise = torque counter clockwise and that the torque expression can be simplified to torque = FL as all the forces are perpendicular. I am not sure how they considered the torque and force on the string being held at the 30 cm mark. First I thought that since this was the point of rotation for the stick, that it would not be considered in the equilibrium expression. Then when the answer considered it in the equilibrium expression, they used the distance to the centre of mass of the stick, while the other 2 masses distances were to the 30 cm point of rotation. As well the answer considered the string torque as contributing to the same direction of rotation as the mass being held at 80 cm, I thought that it would be in the same rotational direction as the mass being held at 10 cm.

Any help is appreciated thank you! :)

Hi Katrina! Thanks so much, I can see that I was mainly tripped up on how to include the ruler in the torque equation and didn't realize that the fulcrum and the centre of mass were different but now I understand :)