Calculating CW & CCW Torques for Point Masses on a Rod

[steveo0]

Homework Statement

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.

Homework Equations

torque = Fd

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?

 

[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.

 

[steveo0]

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

1. Homework Statement
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. Homework 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?

 

[LowlyPion]

Does the rod have mass?

 

[steveo0]

yeah 5kg

 

[LowlyPion]

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

Is there a weight on one end as well?