What am I doing wrong?

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discowisco

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A 1 kg meter stick is attached to the ceiling by a string at the 40 cm mark. If a 3 kg mass is attached to the meter stick at the end labeled 0 cm,what mass placed at the 70 cm mark would balance the meter stick?

Ans: 3.67 Kg

Okay so 1 meter = 100 cm,

Torque Clock Wise = Torque Counter CW

Pivot point is the center of mass, so at the 50 cm line.
Torque = Distance to pivot point x F
Gs Cancel out and I get...

1kg x (10 cm) + 3kg x (50cm) = X kg * (20 cm)

The ans i get is 8 kg??

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Remember that the meter stick is hanging by its 40 cm mark. That means its center of mass is 10 cm away. The 3 kg mass attached at the 0 cm mark has a distance of 40 cm away. The 70 cm mark is a distance of 30 cm away.
 
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1. Why are we using the 40 cm mark as a point of reference? And because we are, why is the distance from the meterstick (which is hanging at the 40 cm mark) 10 cm away instead of being 0?

Dont we use the center of mass of the meter stick as a point of reference? Thus the 50 cm mark?
 
40cm mark is the point where the meter stick will rotate about, so we set that as the point of reference. You can set the 50cm as point of reference, but it will complicate the calculation because the string isn't tied to the center of mass of the meter stick and have to calculate how much torque each unbalanced sides of the meter stick affects the system and incorporate that into the calculation. It is conceptually and calculation-wise easier to set the point of reference to the point where the system is rotating about.

10cm times 1kg is the torque of the meter stick itself (meter stick has 1kg mass, and 10cm distance away from the point of reference; its center of mass is at 50cm mark and point of reference is at 40cm, so it is 10cm distance away, specifically to the right of the point of reference).
 
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