# Torque and Static Equilibrium

1. Nov 3, 2013

### tbdm

1. The problem statement, all variables and given/known data
Suppose we take a 1 m long uniform bar and support it at the 22 cm mark. Hanging a 0.29 kg mass on the short end of the beam results in the system being in balance. Find the mass of the beam.

2. Relevant equations
$\tau$=F*r

3. The attempt at a solution
I set it up so that I have the mass of the bar on each end time the gravity for the force of the bar on each end, and add the force of the added mass to the short end, all multiplied by their distance from the point the system is balanced on.
(.22 m)(.29 kg)(9.8 m/s^2) +(.22x)(9.8 m/s^2)(.22 m) = (.78x)(9.8 m/s^2)(.78 m)
And solve for x. Gravity cancels out, and I get x is 0.113928571 kg or approximately 0.11 kg. However this is wrong. I’m not sure what I’m not doing correctly

2. Nov 3, 2013

### TSny

Is all the mass of the short end of the stick located .22 m from the support?

3. Nov 3, 2013

### tbdm

That was what I had wondered, I wasn't sure if I could treat it as a point mass, but I'm not sure how else to write it.

4. Nov 3, 2013

### TSny

Have you studied the concept of center of mass (or center of gravity)?

5. Nov 3, 2013

### tbdm

I have, so the center of mass would be at .5 meters, or .28 meters from the balance point? And the other mass would be at .22 meters on the other side?

6. Nov 3, 2013

### TSny

Yes, you may consider all the mass of the entire bar to be concentrated at the center of mass of the bar. The only other mass will be the 0.29 kg mass at the end of the bar.

7. Nov 3, 2013

### tbdm

That makes sense, thank you so much for your help