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On the above question, for the upper bound I took moments about the right bottom corner, resulting in the upper bound: $$ H_{UB} = mg\frac{b}{2F} $$ For the lower bound, I took moments about the COG: $$ H_{LB} = mg\frac{\mu a}{Fa-1}* $$ Do these bounds satisfy the toppling conditions?
My thoughts are:

1) I think friction should come into account in the upper bound i.e. I should be taking moments about COG here as well.
2) However, I'm not even sure if I am allowed to take moments about the COG because I don't think it is a pivot.
3) For the lower bound, I think I should be taking an inertial reference frame, and using the dynamic friction coefficient (reduced resistance to toppling backwards). However, I'm finding that difficult to visualise, what force causes the toppling backwards due to movement forwards (F=ma?)

PS. These questions are set deliberately vague. You are free to make assumptions and inferences, as long at they're outlined in the solution.

*EDIT: I spotted an arithmetic error in the lower bound. Here are my new calculations, assuming lower bound for line of action of F is below the COG:

$$ \Sigma M_{COG} = 0 $$ $$ F(\frac{a}{2}-h)-F_f(\frac{a}{2}) = 0 $$ $$ H_{LB} = \frac{mg\mu a- Fa}{2F} $$

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  • $\begingroup$ The phrase PS. These questions are set deliberately vague. You are free to make assumptions and inferences, as long at they're outlined in the solution. sounds a lot like a homework assignment. Is it? $\endgroup$ – Floris Dec 20 '16 at 19:37
  • $\begingroup$ @Floris No, it's a past exam paper question. It's an optional exam, one which pushes the boundaries of our syllabi and thus they set questions vaguely to force us to flesh out our thought processes on paper. $\endgroup$ – user120568 Dec 20 '16 at 20:03
  • $\begingroup$ Dimensionally, $Fa-1$ doesn't make any sense... $\endgroup$ – Floris Dec 20 '16 at 21:49
  • $\begingroup$ @Floris Thank you. Do you think that is down to arithmetic or the fact that I'm taking moments about the centre of gravity? $\endgroup$ – user120568 Dec 20 '16 at 21:57
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    $\begingroup$ @sammygerbil I am aware of this. If you reread my question, you'll see I have highlighted 3 parts of this problem that I don't intuitively understand. These are concepts that I really want to get a grasp of. $\endgroup$ – user120568 Dec 21 '16 at 10:30
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I believe you should use both conditions of static equilibrium: that the sum of forces vanishes and that the sum of moments of forces vanishes.

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