# Using torque to find mass of meter stick

1. Jun 10, 2013

### gake

1. The problem statement, all variables and given/known data

22) A 1-kg rock is suspended from the tip of a meter stick at the 0- cm mark so that the meter stick balances like a see-saw when the fulcrum is at the 25-cm mark. From this information, what is the mass of the meter
stick?
A) 1/4 kg
B) 1/2 kg
C) 3/4 kg
D) 1 kg
E) more than 1 kg

2. Relevant equations

Torque== length of lever arm * Force

3. The attempt at a solution

I keep on receiving B.....

Counter Clockwise Torque:
0.25*F1
• F1==Force between 0 and 25 cm==g(1+.25MS)
• MS== mass of stick

Clockwise Torque:
0.75*F2
• F2==Force between 25 and 100 cm==g(0.75MS)

Counter Clockwise Torque==Clockwise Torque (because it balances)

0.25*g(1+.25MS)==0.75*g(0.75MS)
0.25g+0.0625gMS==0.5625gMS
0.25g==0.5625gMS-0.0625gMS
2.5==5.0MS
MS==0.5

2. Jun 11, 2013

### haruspex

You're not calculating the moment of that portion of stick correctly. How far is that mass from the fulcrum?

3. Jun 11, 2013

### gake

Isn't it 25 cm away, since mass is at 0 and fulcrum at 25 cm? Or is it zero since the portion of the ruler accounted starts right after the fulcrum?

4. Jun 11, 2013

### mishek

Your meter is "split" in two by the fulcrum. Where is the weight of the each portion acting at, at which distance?

5. Jun 11, 2013

### gake

The 25 cm long part weighs 1\4th of the total and 75 cm weighs 3\4ths, doesn't it?

6. Jun 11, 2013

### mishek

Hello, yes, that would be correct, but that is not what I was pointing at.
If you look at the first post on this topic, you wrote following:

You are correct about the magnitude of the weights (0.25MS & 0.75MS), but you wrote a wrong position from which those forces act.

Could you tell where are 0,25MS and 0,75MS located?

7. Jun 11, 2013

### Staff: Mentor

Do you know where the center of mass of the meter stick is located?

8. Jun 11, 2013

### gake

at 25cm?

9. Jun 11, 2013

### gake

Do they both act from the center of gravity?

10. Jun 11, 2013

### gake

Or is it half way?

11. Jun 11, 2013

### Staff: Mentor

Yes. It is located at 0.5 meters from either end. This is where the total mass of the meter stick is taken to be concentrated (at least in a static situation like this). So the center of mass of the meter stick is situated 0.25 meters from the fulcrum.

12. Jun 11, 2013

### gake

I got it! Thank you!

.25mg==25g
m==1!