# CW/CCW Torques

1. Feb 10, 2009

### steveo0

1. The problem statement, all variables and given/known data
Find all cw and ccw torques. I don't know how to post the picture, but I was absent from my physics class today for AMC so I missed out. First problem: A horizontal massless rod of length 6 meters is pivoted about its center. There is a 2kg point mass on each end of the rod.

2. Relevant equations
torque = Fd

3. The attempt at a solution
So I looked at the textbook, but they only provide examples with cylinders. So I just need one example for this and I think I can pretty much do the rest. Do I calculate the Force for each one? or do I use the moment of Inertia?

2. Feb 10, 2009

### LowlyPion

Welcome to PF.

You mean it's a see-saw.

If it is in equilibrium then the sum of the Torques are 0.

The F*2m on one side is equal to the F*2m on the other.

3. Feb 10, 2009

### steveo0

oh. thanks. lol that's that only example where they're both equal. so uh.

1. The problem statement, all variables and given/known data
Find all cw and ccw torques. Find the net torque. Find the angular acceleration. A differential horizontal rod has a length of 4 meters and a mass of 5kg. It is pivoted about its center.

2. Relevant equations
torque = Fd
torque = (I)(alpha)

3. The attempt at a solution
I found Fg = -49 N
torque = Fxd = -98 N/m
angular acceleration = torque / moment of inertia = -98 / (20/3) = -14.7 m/s^2?
is this right?

4. Feb 10, 2009

### LowlyPion

Does the rod have mass?

5. Feb 11, 2009

### steveo0

yeah 5kg

6. Feb 11, 2009

### LowlyPion

If it is pivoted about the center where is the net force?

Is there a weight on one end as well?