Physics Meterstick Question

A uniform meterstick pivoted at its center has a 90.0g mass suspended at the 24.0cm position.

1. At what position should a 95.0g mass be suspended to put the system in equilibrium?

2. What mass would have to be suspended at the 90.0cm position for the system to be in equilibrium?

Tried solving, but said program said it was wrong. Please show work so that I can learn. Thanks

Torque = Force * distance from the pivot point.
The force is the weight of the object.
Since the meter stick pivoted at its center, its weight does not produce any torque.
The 90 g mass is 26 cm to the left of the pivot point.

Counter clockwise torque = 0.090 * 9.8 * 0.26 = 0.22932
Clockwise torque = 0.095 * 9.8 * d = 0.931 * d
0.931 * d = 0.22932
d = 0.22932 ÷ 0.931
The distance is approximately 0.2463 meters = 24.63 cm to the right of pivot point.
Position = 50 + 24.63 = 74.63 cm

2. What mass would have to be suspended at the 90.0cm position for the system to be in equilibrium?
The counter clockwise torque is the same.
The 90.0cm position is 40 cm to the right of the pivot point.
Clockwise torque = m * 9.8 * 0.4 = m * 3.92

m * 3.92 = 0.22932
m = 0.22932 ÷ 3.92 = 0.0585 kg = 58.5 g

moments must be balanced,
F1 x1 = f2 x2
90 * 24 = 95 * x2
=> x2 = 90 * 24 / 25

The same method for the next part
90 * 24 = f2 * 90
f2 = 90 * 24 / 90 ( = 24 g)