Finding the Mass of a Meter Stick (Torque and Balance)

[Katelyn]

Homework Statement

Attach a 216.82 g mass at the 90 cm position. Red-adjust the position of the knife edge such that the meter stick is once again balanced. Determine the mass of the meter stick without using the balance.

2. Homework Equations
Torque = force * distance from center
F=ma
T1+T2 = 0

The Attempt at a Solution

I started by plugging in all of the numbers into T1+T2=0:

(m1) (9.8) (80) - (216.82) (9.8) (10) = 0

(m1) (784) - 21,248.36 = 0

784 (m1) = 21,248.36

m1 = 27.1025

This is only the mass of the meter stick on the side without an added mass, so I then found the mass of the whole meter stick by doing this:
27.1025 = 0.8m

m = 33.88

The actual mass is supposed to be more than twice the mass that I found. That is the only hint I have gotten from my teacher. I'm not sure what I am doing wrong here, but any help would be greatly appreciated! Does it have something to do with my distances? That is what I suspect is wrong. Thank you in advance.

 

[Isaac0427]

You are definitely missing some key information in this problem. Can you share the entire problem with us please? For example, what position is the fulcrum at now?

 

[scottdave]

Draw a picture. What makes up the torques on each side of the pivot point? There is something missing from your equation (summing the torques to zero). It looks like the pivot point is 10 cm from the suspended mass?

 

[Katelyn]

Unfortunately, this is all of the information that I have for the problem. As a part of the lab, we were supposed to find where to position the fulcrum so that the meter stick would stay balanced with the mass placed at 90 cm. We found that it balanced when the fulcrum was placed at 80 cm. The torque on the left side (where there is no added mass) would be made up of the mass of the meter stick and 80 cm of distance from the fulcrum? I think that is probably what I am getting wrong. I know that the torque on the right side is made up of the mass of the meter stick plus the mass of the 216.82 g and the distance of 10 cm from the fulcrum. Is that right?

 

[Isaac0427]

We found that it balanced when the fulcrum was placed at 80 cm

That’s what I was looking for.

(m1) (9.8) (80) - (216.82) (9.8) (10) = 0

Your problem is with that 80 cm term on the left. We can pretend that m1g (gravity of the Earth on the meterstick) is acting at one point on the meterstick. What mark would that be? Also, how far is that mark from the 80 cm mark?

EDIT:
The part about the units was removed by me. It does not make a difference for your final answer. But, SI units would still be recommended by me.

ANOTHER EDIT:
That last part has awful grammar—there is a problem using the letter “eye” on my phone when editing the post. Sorry if it made me sound bad.

 

[Katelyn]

Would the distance on that side equal zero? I guess I am just confused about the distance on that side because it seems like the force of gravity would be acting on infinitely many points all across the meter stick. Does that make any sense or am I totally wrong?

 

[Isaac0427]

Would the distance on that side equal zero? I guess I am just confused about the distance on that side because it seems like the force of gravity would be acting on infinitely many points all across the meter stick. Does that make any sense or am I totally wrong?

You are right—but, it turns out that gravity can be thought of as acting on only one point: the center of mass. Assuming that your ruler is uniform, what mark will the CM be at?

 

[Katelyn]

Ahhh, that makes sense. I should have remembered that! It would be at the 90 cm mark where the added mass is correct? The center of mass is at the point where the mass is most concentrated?

 

[Isaac0427]

Ahhh, that makes sense. I should have remembered that! It would be at the 90 cm mark where the added mass is correct? The center of mass is at the point where the mass is most concentrated?

That is incorrect. Note here that we are analyzing two separate objects here. We are only talking about the center of mass of the meterstick. The weight is separate. Also, that is not the correct definition of the center of mass. It could be explained here, but there are some really really good videos on YouTube that explain the concept great. I’d try one of those for a refresher.

 

[Katelyn]

I found this definition: For simple rigid objects with uniform density, the center of mass is located at the https://www.khanacademy.org/science/physics/linear-momentum/center-of-mass/a/math/geometry/triangle-properties/medians-centroids. That would mean that the center of mass for the meterstick would be at 50 centimeters, correct?

 

[Isaac0427]

I found this definition: For simple rigid objects with uniform density, the center of mass is located at the https://www.khanacademy.org/science/physics/linear-momentum/center-of-mass/a/math/geometry/triangle-properties/medians-centroids. That would mean that the center of mass for the meterstick would be at 50 centimeters, correct?

Exactly! What does this make your distance that you are looking for?

 

[Katelyn]

If I take the location of the fulcrum at 80 cm and subtract the center of mass at 50 cm, that would mean the distance would be 30 cm?

 

[Isaac0427]

IMO right now you have the perfect ability to finish this problem, so I’ll let you go at the rest of it yourself.

 

[Katelyn]

Okay, great! Thank you so much for your help!