Calculating CW & CCW Torques for Point Masses on a Rod

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Homework Help Overview

The discussion revolves around calculating clockwise (cw) and counterclockwise (ccw) torques for point masses on a rod, specifically focusing on scenarios involving a massless rod and a rod with mass. The original poster describes a situation with a horizontal rod pivoted at its center with point masses at each end, and later introduces a different rod with mass.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to understand how to calculate torques and whether to consider forces or moment of inertia. Some participants question the conditions of equilibrium and the presence of additional weights on the rod.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the rod's mass and the conditions for equilibrium. Guidance has been offered regarding the relationship between torques and forces, but no consensus has been reached on the calculations or assumptions involved.

Contextual Notes

There is a mention of the original poster missing class, which may impact their understanding of the concepts. The problem involves both a massless rod and a rod with mass, leading to different considerations for torque calculations.

steveo0
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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?
 
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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.
 
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?
 
Does the rod have mass?
 
yeah 5kg
 
If it is pivoted about the center where is the net force?

Is there a weight on one end as well?
 

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