# Finding unknown mass from a meterstick, balanced equilibrium

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?

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SteamKing
Staff Emeritus
Homework Helper
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?
If the system is in equilibrium, what must the net torque be about the pivot?
(100*9.8 / 1000) = 0.980
What are the units here?

T = w*r = (0.98 * 0.35) = 0.343N Left
T = (0.98 * 0.65) = 0.637N Right
Are newtons the correct units for torque?
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
"N/cm" is how you write "newtons per centimeter". If you want to express a torque in newton-centimeters, "N-cm" is acceptable.
T = W*r
14.7 / .25 = 58.8 g?
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?

Simon Bridge