Finding unknown mass from a meterstick, balanced equilibrium

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The discussion revolves around calculating the unknown mass suspended from a meter stick in equilibrium, with a known mass of 100g and a pivot at the 35 cm mark. The participant correctly identifies the need to calculate torques in both clockwise (CW) and counterclockwise (CCW) directions to find the net torque. The calculations yield a net torque of 14.7 N/cm, leading to an estimated unknown mass of 58.8g. Key points include the importance of consistent unit usage and the correct application of torque formulas.

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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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dioxy186 said:
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?
 
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.
 

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