# Finding unknown mass from a meterstick, balanced equilibrium

1. Oct 20, 2015

### dioxy186

Question: A uniform meter stick off mass Ms = 100g is supported by a pivot at its 35 cm mark. When an unknown mass is suspended from the 10 cm mark, the system is in equilibrium. What is the value of the unknown mass?

Given: Pivot point at 35 cm, and mass of meterstick = 100g.

Attempt:
Do I find the Torque in CW & CCW motion from 35cm and 65 cm, and then subtract my CW torque from CCW to get net torque. And then use the Torque = weight * r (distance from pivot) -> weight=T/r?

(100*9.8 / 1000) = 0.980
T = w*r = (0.98 * 0.35) = 0.343N Left
T = (0.98 * 0.65) = 0.637N Right
Left Torque = (0.343/2)* 35 = 6.003 N/cm
Right Torque = (0.637/2) * 65 = 20.703 N/cm
Net Torque = 20.703 - 6.003 = 14.7 N/cm

T = W*r
14.7 / .25 = 58.8 g?

2. Oct 21, 2015

### SteamKing

Staff Emeritus
If the system is in equilibrium, what must the net torque be about the pivot?
What are the units here?

Are newtons the correct units for torque?
"N/cm" is how you write "newtons per centimeter". If you want to express a torque in newton-centimeters, "N-cm" is acceptable.
Did you check the units of your calculation here? You have a torque in newton-cm and you are dividing by meters. Does that give you grams? Or something else?

3. Oct 21, 2015

### Simon Bridge

Sketch the situation - torques should be computed about the pivot point.

You have used cw/ccw in one place and left/right in another - best practise to be consistent with your notation.
You've done too much work converting to SI units - the conversion factors cancel .... same with the g.