The question

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The Solution

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Part (c) is where I have a doubt ...

How do I approach such problems?

They way I approached such problems (and most of them had sufficient friction so that that the block did not move/translate and only toppled) is by finding the torque about the corner, because intuitively, that would be the point about which the block would topple/rotate — it would be the (instantaneous?) axis of rotation.

(My book defines the Instantaneous Axis of Rotation (IOAR) as the point in the body (or extended body) which can be at rest momentarily — Is this definition correct?)

However, that logic broke down (or so the author thinks) in this problem since the corner would no longer be at rest (the cube is translating) and the author decided to calculate torque about a different point.

So, the textbook considers the torque about the c.o.m. of the cube. But, they do not explain how they arrived at the conclusion that we should use the c.o.m. point. This point is not at rest (neither momentarily or permanently) so the IOAR does not pass through it, right?

In other words, how can we decide the point about which we should calculate torque?

Also, let's say we found some good reason to consider torque about the c.o.m. instead of about the corner of the cube. Why doesn't the author (in part (c)) take the torque (about the c.o.m.) of friction into account?


I think I should add (just for reference) pictures of the pages where my book explains IOAR.

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enter link description here

^I'm having trouble uploading the second page. Will try again later. Please use this Google drive link to the second picture for now.

Any help will be appreciated. 👍

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    $\begingroup$ Hi, welcome to Physics SE! Can you convert the pictures of the text into typed-out, formatted text? It makes the content index-able by search engines, and shows up better on different devices' displays. For formulae, try MathJax instead. $\endgroup$ – user191954 Jul 22 '18 at 9:25
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    $\begingroup$ I'm voting to close this question as off-topic because this post requires formatting. $\endgroup$ – Nat Jul 22 '18 at 16:50
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    $\begingroup$ @AB. The fact that you disagree with the site policies does not exempt you from following them; there is an extremely strong site consensus that screenshots of text are not acceptable and if you wish to post here then you need to abide by it (or convince the community to change). However, if your arguments are "I value my time above that of others" and "I do not think X class of users matters" then you won't get very far. $\endgroup$ – Emilio Pisanty Jul 22 '18 at 17:34
  • $\begingroup$ Hi Anurag B. Linking to private clouds, dropbox, etc, is for various reasons not acceptable on SE, cf. this meta post. $\endgroup$ – Qmechanic Oct 6 '19 at 10:23

First of all: the point around which it rotates should be the centre of mass, because if the block starts sliding, the only forces acting on it are the external horizontal force $F$, gravity and the normal force. If an object is not attached to something else directly or by friction, the centre of mass is always the point at which rotation occurs.

So we can view this block as if it is in free space and at one point at the surface a force $F$ is acting on it, at the centre of mass gravity is acting on it, and the normal force acts on it (if we move our coordinate system along with the centre of mass). The external force and the normal force create a torque around the centre of mass. Keep in mind that the rotation is now the opposite direction from that in question a). When there is no external force the torque could be considered evenly distributed over the surface; when $F$ increases, the normal force gathers more to the left corner. Until $F$ is so big that the normal force should focus everything on the left corner - which is when the torque of the normal and external force are equal -, then the block starts turning around the centre of mass.

However, quickly after that the centre of mass starts increasing its distance from the table (as one would expect), which means the point around which rotation occurs changes (it goes to the lower left corner, I think). However, the torque of the external force is already so big, that toppling is inevitable. What I’m trying to say with this last part: it’s not so weird you get confused.

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  • $\begingroup$ Ok, if we are going to consider torque about the c.o.m. instead of about the corner (I'm still confused why), what about the torque of friction? That should be taken into account as well, right? $\endgroup$ – Anurag Jul 22 '18 at 7:55
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    $\begingroup$ There is no friction. The exercise reads: “If μ = μ_min” which is zero, since that is the answer of b). $\endgroup$ – Antaios Jul 22 '18 at 11:09
  • $\begingroup$ Thank you!! I feel so dumb now !! But I still don't understand why the axis should pass through the c.o.m. ... :( $\endgroup$ – Anurag Jul 22 '18 at 13:21
  • $\begingroup$ In empty space, if a force acts on an object, the object starts rotating around its centre of mass. I feel like that is intuitive, right? I wouldn’t know how to mathematically explain why exactly this should be, and it would probably not become more understandable. Only when an object is attached to some point it starts rotating around that point. $\endgroup$ – Antaios Jul 22 '18 at 19:19
  • $\begingroup$ In empty space, yes, about the c.o.m. But in our case (i.e., in the given question), if the cube wants to rotate about its c.o.m., the table will come in its way. The only way seems to be about the corner. $\endgroup$ – Anurag Jul 22 '18 at 20:10

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