- #1

StephenHaw

- 5

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- TL;DR Summary
- A question on tipping, a cabinet is subject to a force greater than the tipping force to an angle of 10 degrees when the applied force is removed, how do I determine if momentum will cause the cabinet to fall over or fall back?

I am trying to determine if there is a set of formula to calculate if a cabinet will when subjected to a force on it's upper edge for an angular distance (not a duration) will gain enough momentum to continue over the balance point and fall over or whether the opposing moment due to mass and gravity will counter the momentum and return the cabinet to it's base.

I already know that for a turning moment that if F1D1 = F2D2 at the point at which tipping will begin: So a cabinet that is 1m tall, 0.5m wide and weight 60kg on a non slip floor would need just over 60*9.81*0.25/1 = 147.15 N to begin to move when the load is applied at the top edge of the cabinet.

The angle at which the CofG is to the pivot edge from the vertical plane is from trig tan-1 (0.25/0.5) = 26.56 degrees

However if a test required a 200N applied force on the cabinet we would know the difference in the applied turning moment would be 52.85N

causing the cabinet to tip. The force of 200N is constant and applied until the cabinet has tipped 10 degrees at which point the applied force is removed.

Now what I am trying to work out is how much K.E. is obtained by the cabinet at the point where the force is removed to understand its velocity to then work out how much further the cabinet will travel before either falling back onto its base or carrying on beyond the pivot point

and fall over.

In this respect perhaps I might be over thinking the problem, as I consider the turning moment applied by gravity is reducing as it approaches the balance point so the value for acceleration increases due to the increasing difference in the applied force and its opposing turning moment due to gravity.

The method to solve this I could then apply to different cabinets of different heights, widths and mass.

Any advice will be greatly appreciated.

I already know that for a turning moment that if F1D1 = F2D2 at the point at which tipping will begin: So a cabinet that is 1m tall, 0.5m wide and weight 60kg on a non slip floor would need just over 60*9.81*0.25/1 = 147.15 N to begin to move when the load is applied at the top edge of the cabinet.

The angle at which the CofG is to the pivot edge from the vertical plane is from trig tan-1 (0.25/0.5) = 26.56 degrees

However if a test required a 200N applied force on the cabinet we would know the difference in the applied turning moment would be 52.85N

causing the cabinet to tip. The force of 200N is constant and applied until the cabinet has tipped 10 degrees at which point the applied force is removed.

Now what I am trying to work out is how much K.E. is obtained by the cabinet at the point where the force is removed to understand its velocity to then work out how much further the cabinet will travel before either falling back onto its base or carrying on beyond the pivot point

and fall over.

In this respect perhaps I might be over thinking the problem, as I consider the turning moment applied by gravity is reducing as it approaches the balance point so the value for acceleration increases due to the increasing difference in the applied force and its opposing turning moment due to gravity.

The method to solve this I could then apply to different cabinets of different heights, widths and mass.

Any advice will be greatly appreciated.