# Homework Help: Torque lab

1. Dec 3, 2006

### chrisl

Hello, I am new here but am completley stuck on a lab that I need done for tommorow.

I am doing a torque lab, with a meter stick being used as a scale and I know the mass of it. You can basically think of the meter stick as a seesaw with only one person on one end. I have attached weights to one end, and found the new centre of gravity. My biggest problem is, It says: Use your knowledge of torque to find mass of unknown. I am then supposed to find out the real weight by using a mechanical scale. I just can't figure out how to find the mass of unknown without using the scale. Can anybody help me out with that.

Thanks alot!

2. Dec 3, 2006

### PhanthomJay

If you know the cg, you know the distance from the weight to the cg, and you should know the distance from the cm of the meter stick to the cg. So sum torques about the cg and use Newton 1 to solve for the weight.

3. Dec 3, 2006

### chrisl

Ok. The mass of the meter stick is 140g. The cg of meter stick is 50cm. cg of unknown mass and meter stick for my first object is 23.9. So that makes the distance from the CG of meter stick to the distane from the cg of the unknown mass object to the cg of the meter stick 27.1. Can somebody please help me find the mass of the first unknown? Basically, I totally forgot about the project until yesterday, and it was too late to ask me teacher for the formula, and I can not find it in my text book. The help is much appreciated. Thank you very much.

4. Dec 3, 2006

### chrisl

So is there anybody out there that can give me the formula for this? I am quickly running out of time, and would really appreciate it if somebody could help me out further. Thanks

5. Dec 3, 2006

### OlderDan

If I understand what you are doing, you are hanging a mass on one end of the stick and moving the pivot point to "balance" the mass with long side of the stick. In this configuration, the torque from the unknown is its weight times its distance from the pivot. The torque from the stick is its weight times the distance if its CG from the pivot. When the system is balanced, these torques are equal in magnitude and opposite in direction. You know where the pivot point is, so you know the distance from the unknown mass to the pivot and you know the distance from the CG of the stick to the pivot. You know the mass of the stick. If you equate the torques the only unknown in the equation is the unknown mass.

6. Dec 3, 2006

### chrisl

Thanks for the reply. I understand what you are saying, but I am still lost on what equation I need to use to find the unknown mass.

7. Dec 3, 2006

### chrisl

What is the "pivot"? Thanks.

8. Dec 3, 2006

### OlderDan

The pivot is where I assume you placed the stick support to find the CG for the system. When the CG of the system is at the pivot, the stick and mass are balanced. You said your first mass gave a mass plus stick CG at 23.9 (I assume that is centimeters). I assume you found that point by balancing the stick on some support at that location. That means the mass is 23.9cm from the pivot while the CG of the stick alone is (50.0 - 23.9)cm from the pivot point. The torque from the stick is 140gm*g*(50.0 - 23.9)cm. The torque from the mass wil also have a g (acceleration due to gravity) so when you equate the torques g will divide out. See if you can write the torque from the mass, equate it to the torque from the stick, and solve for the unknown mass.

9. Dec 3, 2006

### chrisl

Thank you very much for the help you provided. My mom and I figured it out about 20 minutes ago, so the lab is allmost complete now. Without your original help, I would have been stuck until your last post. Thank you very much. Chris