One of the cardinal rules here in the Colgate weight room is that there should never be a 3 plate (20kg weight) difference when an Olympic bar is on the rack. This is due to the serious possibility of injury that could result of this difference because it is at this weight that the light side of the bar will come flying in an arc and potentially injuring somebody. Since professor Metzler has corrupted all of us with her ability to force us into seeing physics everywhere, I saw this as a torque problem to be solved.
An Olympic bar with weighs 20kg, and is 2.13m long, and on the right side we are assuming that there are three 20kg plates, totaling 60kg 0.25m from the axis of rotation. There is no weight on the left side of the bar. It is also important to note that the axis of rotation is not at the end of the bar, it is roughly 0.45m from the end.
The net torque is then: Σ = (right) - (bar) when the right side is heavier.
Using the torque equation =(r)(F)(sinθ), the equation is as follows:
Σ =(.25m)(60kg*9.8m/s^2) - (.6m)(20*9.8m/s^2) = 147N-118N= 29N in the clockwise direction.
As you can see, a difference of three plates on a standard Olympic bar will cause a net torque to be created in the system, sending the bar swinging with potential for serious injury.