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Consider the above experimental design. At the end of the strip, at point $p$, a mass $m$ is attached, the stability of the wooden block above is determined by the angle $θ$ at which the block topples over. It seems that as the angle $θ$, approaches $0$ the block becomes easier to topple over. However, I am struggling to come up with a suitable reason for why.

At all values of the angle $θ$, the distance of the mass $m$ from the centre of the mass remains constant. Hence, the only thing that changes is the area of the strip over the top of the block. So then, is a suitable explanation, that the force downwards by the mass $m$ is distributed over a larger area, making the turning effect of the force less?

Another explanation I have come up with, is that the edge of the block that the strip is over, acts as the pivot for the mass $m$. As the strip angle increases, the distance between $p$ and the pivot actually decreases, therefore making the turning effect of the force less. Is this suitable?

I am looking for a sound physics explanation for the phenomena above. My attempts at explanations do not seem correct.

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  • $\begingroup$ Stability and toppling $\endgroup$
    – Farcher
    Commented May 3 at 9:33
  • $\begingroup$ "edge of the block that the strip is over" Yes, that is the reason. $\endgroup$ Commented May 3 at 11:48
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    $\begingroup$ This seems to be a conceptual question with a conceptual answer. Voting to re-open. $\endgroup$ Commented May 3 at 13:02

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Another explanation I have come up with, is that the edge of the block that the strip is over, acts as the pivot for the mass $m$

This is the correct explanation. If the block (or the bench that it sits on) is tilted, it will topple once a vertical line through the centre of gravity of the block/mass system moves outside of the base of the block. Decreasing the angle $\theta$ moves the mass at $P$ further away from the block, thus moving the centre of gravity of the block/mass system towards the right hand edge of the block, making the block easier to topple. If $m$ is sufficiently large compared to the mass of the block then there may even be an angle $\theta > 0$ at which the centre of gravity moves outside of the block and the block topples without being tilted.

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