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So the way I'll structure will be as follows:

  1. Scenario
  2. Question(s)

The following would be conducted on earth with a gravitational field strength of 9.81N

Picture a slope at 45 degrees from horizontal. Sliding straight down it at any given velocity is a perfect cube with a given mass.

Considering the fact that an object will topple once its centre of mass is directly above the pivot, and there is some force that is acting upon the side opposite to the pivot... one must conclude that if the friction is able to keep the cube from sliding down the slope, then the object will topple (remember that the slope is at 45 degrees to the horizontal).

So with all these factors taken into account, if the object is again sliding down the same slope and it slides over a sudden change in material with greater traction, increasing the friction (correct me if I'm wrong), which gets subtracted from the velocity or momentum (neither of which are forces so this confuses me - would you use kinetic energy instead?) and as long as these forces are greater than the force of X then the cube can rotate about the pivot and topple)

So my question is: What force would X be, because I don't think it could be GPE. And would the cube even topple, since when the friction increased and the face of the cube came off the slope surface, the friction then decreases again because there is less surface area of the cube on the slope surface. And what would the formula be for finding force X.

Thanks.

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A lot of material in your question. Let me try to summarize:

You have an cube sliding down a frictionless slope of 45°, when it suddenly encounters a surface with friction $\mu$. Will it topple?

The answer is "yes" if the sum of two forces - gravity and deceleration - points along a line that does not pass through the base of the cube. Since the cube was on a 45 degree slope in your example, the line of action of the force of gravity is already pointing at the corner of the cube:

enter image description here

If you add any friction, that will result in an additional torque on the block that will topple it.

The problem is more interesting when you have a more stable situation - lower angle of slope (or wider block): in that case, the question of whether the block will topple (assuming that the coefficient of friction is unchanged with contact area - which is a reasonable assumption) has a more complex answer (but it's not what you asked for).

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  • $\begingroup$ Thanks for your answer, it's interesting to see all the forces that come into play in this scenario. Initially I was even more confused when wondering why moments don't factor in, and that's because the distance between the line of action of μ, and the pivot or leading edge which is in contact with the slope surface... would be 0. $\endgroup$ – jan macbean Dec 2 '16 at 18:51
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    $\begingroup$ What do you mean by the force of "deceleration"? $\endgroup$ – BowlOfRed Dec 2 '16 at 22:05

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