Predicting the distance where torque should be applied?

[JazzyJones]

Homework Statement

Finding the mass of the meterstick!
Distance was found through experimentation
I am given a mass hanger of 50g to place on a 100cm ruler held up on its center of mass at 49.8cm
I displace the clip holding the ruler by 15cm to the left. (assuming the ruler counts up from left to right) The experimental value in practice was 13.1cm for a balanced ruler. (How am I to calculate the assumed value?) With this information how am I to calculate the mass of the meterstick?

If not sufficient information is given or hard to imagine I could draw a picture and upload it.

Homework Equations

Mass of ruler: 95.1 g
τ = rfsinθ
Ʃ F = 0
Ʃ τ= 0

The Attempt at a Solution

I don't know what to plug in, I also think I am missing an equation. Trying to visualize the problem, I see how moving the clip to the left that I am increasing the torque in the problem, and to balance the meterstick, I must place the 50g mass at a certain distance to achieve equal amount of torque on both sides of the meterstick thus achieving equilibrium and a balanced meterstick. Am I missing something for finding the mass?

 

[tiny-tim]

Hi JazzyJones !

i] what is the torque of the 50 g mass?

ii] what is the distance of the centre of mass (from the pivot)?

iii] so what must the meterstick's mass be?

 

[JazzyJones]

Hi JazzyJones !

i] what is the torque of the 50 g mass?
τ= mass*a*(distance from pivot)
τ= .05kg*9.8m/s^2*.0217m
τ= 0.010633
ii] what is the distance of the centre of mass (from the pivot)?
15cm or .015m
iii] so what must the meterstick's mass be?

I'm taking a guess at this but what if I do this,
.05kg*9.8m/s^2*.0217m = m*9.8m/s^2*.015
right? because the net torque will equal 0

 

[tiny-tim]

Hi JazzyJones!

(try using the X² button just above the Reply box )

i] what is the torque of the 50 g mass?
τ= mass*a*(distance from pivot)
τ= .05kg*9.8m/s^2*.0217m
τ= 0.010633
ii] what is the distance of the centre of mass (from the pivot)?
15cm or .015m

I'm taking a guess at this but what if I do this,
.05kg*9.8m/s^2*.0217m = m*9.8m/s^2*.015
right? because the net torque will equal 0

Where does the 21.7 cm come from?

Apart from that, it looks right …

the LHS is the torque of the 50g mass, and the RHS is the torque of the unknown m,

so now you just divide out to get m.

btw, you should leave g as "g" (instead of 9.8), since it'll cancel out anyway,

and you should leave everything in gram and cm (instead of converting to kg and m), since that'll cancel out also! ​

 

[Nickie]

To calculate the mass of the meterstick, you can use the equation τ = rfsinθ, where τ is the torque, r is the distance from the pivot point (in this case, the center of mass of the meterstick) to where the force is applied, f is the force applied, and θ is the angle between the force and the lever arm (r). In this case, you know the values of τ (0), r (49.8 cm), and f (50 g). Therefore, you can rearrange the equation to solve for the unknown mass of the meterstick:

m = τ / (rfsinθ)

Substituting in the known values, you get:

m = 0 / (49.8 cm * 50 g * sin(90°))

Since sin(90°) = 1, the equation simplifies to:

m = 0 / (49.8 cm * 50 g * 1)

m = 0 g

This means that the mass of the meterstick is negligible and can be ignored in this experiment. However, if you want to calculate the mass of the meterstick including the mass of the 50g hanger, you can simply add 50g to the calculated mass. Therefore, the total mass of the meterstick would be 50g.

It is important to note that this calculation assumes that the meterstick is uniform and the mass is distributed evenly along its length. If the meterstick is not uniform, or if the mass is not evenly distributed, the calculation may be different.