Using torque to find mass of meter stick

[gake]

Homework Statement

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
Answer: D

Homework Equations

Torque== length of lever arm * Force

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

 

[haruspex]

Counter Clockwise Torque:
0.25*F1

• F1==Force between 0 and 25 cm==g(1+.25MS)

• MS== mass of stick

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

 

[gake]

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

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?

 

[mishek]

Isn't it 25 cm away, since mass is at 0 and fulcrum at 25 cm?

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

 

[gake]

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

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

 

[mishek]

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

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:

0.25*g(1+.25MS)==0.75*g(0.75MS)

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?

 

[Chestermiller]

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

 

[gake]

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

at 25cm?

 

[gake]

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 positioned from which those forces act.

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

Do they both act from the center of gravity?

 

[gake]

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

Or is it half way?

 

[Chestermiller]

Or is it half way?

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.

 

[gake]

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.

I got it! Thank you!

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