Rotational Mechanics:Small Problems

[Urmi Roy]

I've just got some small problems this time.
The first being,do the particles at the axis of a rotating body really stand still??Don't they spin about themselves??

 

[ham]

I would think of it this way...If you think that a particle at the axis of rotation has a radius (dimension), then there would be rotation. However, since the axis has no dimension or radius, it is just a line, then basically there is no particle there hence no rotation. If there were rotation, the radius would be infinately small and of no consequence.

 

[Urmi Roy]

I think that means that this notion of the axis being completely still stems from the fact that we consider the particle to be infinitesimally small,which allows us to consider the axis as sttionary. Right?

Another thing that I was wondering about is that-I found that all the quantities that we discuss in rotational mechanics,can be obtained if we do a Vector product with the analogous linear quantities,like angular momentum is 'p (cross product) r' and then torque is ' F (cross product) r' , etc.

Is that just a coincidence, or does it have something to say about the physics of rotational motion?

 

[Urmi Roy]

Recently I observed something really strange. When I was toying around with my yoyo,the yoyo stopped at the end of the string and it started spinning around itself.

Simultaneously,the turns in the string seemed to be moving upward towards my hand.

I was wondering,is this what we mean by vectorial direction?

It might sound crazy,but I think that it kinda makes sense. I mean angular velocity has a direction,which is perpendicular to the plane of rotation--so it was in the yoyo!

 

[Urmi Roy]

Could someone please come back to this thread?

 

[gmax137]

Could someone please come back to this thread?

UrmiRoy - in the forum view (when you can see all the threads by title/subject) you can also see how many people have viewed the thread (right now this one has 140+ views). So you see, most people who view the thread do not post to it. Maybe they are too busy, maybe they think they have nothing useful to add. This is typical of the PF forum, fortunately. It is fortunate because it means most of the posts that people DO make are worthwhile. In other words, your patience will be rewarded.

 

[Urmi Roy]

I'm really sorry for being too impatient!

Please pardon me.

 

[jimcharoen]

Ha haaa.. this is really a philosophy question. Indeed, this is a question of either discrete or continuity. If you believe in Democritus (atom = undevidable) then there must be a very very small particle at the center of the object which stands still. If you, on the other hand, believe in continuity, there is no such particle i.e. even if the particle is extremely small, it is still devidable.

Cheers,
Charoen Peetiya

 

[BobbyBear]

I've just got some small problems this time.
The first being,do the particles at the axis of a rotating body really stand still??Don't they spin about themselves??

When you are studying rotational mechanics you are studying it within the branch of continuum mechanics in general, in which you're modeling matter as a continuum, and you can talk of there being matter at each point in space, and you can talk of velocity at each point, and two points can be infinitely near each other and their respective velocities will differ by an infinitesimal amount, and, and . . . and you can use calculus as your mathematical language to describe the phenomena you are studying. Remember though that this is just a model of reality, for matter is not continuous at all, but the model serves us well enough to describe and predict the behaviour concerning the motion of objects, as long as we don't get too close to the object so that we see that it is really not continuous at all (ie. when the continuum hypothesis is breached). I'm just trying to say, don't eat your head over what happens at a point (the axis of rotation), because it is a mathematical abstraction of reality - and for our particular interests, we don't really care what reality looks like, as long as with our model we can extract conclusions and make predictions about the particular aspect of reality that we are interested in.

I have no idea what really happens at the axis of rotation as far as the matter contained in it is concerned. Firstly, are you sure the axis would be occupied by matter at all (since in reality we know that only a very tiny percentage of the space even within an atom is really occupied by matter). And if there was matter exactly at the axis, would it remain there? aren't molecules always vibrating (if they are above the absolute zero temperature)? so wouldn't the centrifugal force, which would be very large near the axis, force them outward? Or maybe they'd get squished? I really don't know . . . but this is not what rotational mechanics is concerned about.

 

[Urmi Roy]

I get the point--the concept of a stationary axis is,number one an assumption,and number two,not that improtant when it comes to real life situations,thanks for the clarification!

Please let me draw your attention to posts number 3 and 4 of this thread.

 

[BobbyBear]

I get the point--the concept of a stationary axis is,number one an assumption,and number two,not that improtant when it comes to real life situations,thanks for the clarification!

Please let me draw your attention to posts number 3 and 4 of this thread.

The assumption is an assumption about reality, in our case, about matter being continuous. For the disciplines of continuum mechanics, what's really happening at a microcopic level is not important (cannot be predicted). Just like singularities that arise in our mathematical description don't really exist, and we cannot predict with our model what is really happening at a singular point in space (eg, shockwaves are mathematical discontinuities, but in reality all magnitudes vary continuously, however large their gradients may be). And that our model is not valid for those points, doesn't necessarily mean that for other purposes it's not important to understand some aspect of it - but we need to study it using different models of reality.


About post 3, whether the equations for angular motion can be derived from those of linear motion by simply cross multiplying with the position vector . . . well, for indeformable bodies, I don't know what to say to you, perhaps so. But not in general: take for example, the equations of conservation of linear momentum and conservation of angular momentum. They are independent principles: one cannot be deduced from the other, however tempting it may be to try and do so (unless additional hypotheses are made - but actually these hypotheses are consequences of both principles having to hold). Remember that a principle is a statement that assumes a certain truth - and as such, no principle can be deduced from another.


I'm not sure I understand post 4. A vector quantity is simply one that is described by a direction as well as a magnitude. The direction of the angular velocity vector doesn't have a physical significance as far as I know, except that it is useful (for mathematical purposes) to define the angular velocity as a vectorial quantity whose cross product with the position vector equals the linear velocity vector.
I'm not sure what you mean by the angular velocity direction being in the yoyo.

 

[Urmi Roy]

Remember that a principle is a statement that assumes a certain truth - and as such, no principle can be deduced from another.

Well, actually the problem with me is that though I find every concept and theorem in linear mechanics very logical and easy to understand with each of them related to one another,even if they can't be deduced from each other,I can't find that lucidity in my concepts of rotational mechanics.

I have a strong intuition that every part of mechanics-be it linear or rotational,must be correlated,must be explanable on the basis of each other-but when it comes to some parts of rotational mechanics,I just can't find any explanation--these expressions of torque etc. etc. seem to have sprung out of nowhere!

Newton devised his second law of motion,F=ma, after observing the world around him,and having found a deep logic behind it--where does that apply to those expressions of rotational mechanics,with all that vector product and mathematical assumptions?

I'm not sure I understand post 4. A vector quantity is simply one that is described by a direction as well as a magnitude. The direction of the angular velocity vector doesn't have a physical significance as far as I know, except that it is useful (for mathematical purposes) to define the angular velocity as a vectorial quantity whose cross product with the position vector equals the linear velocity vector.
I'm not sure what you mean by the angular velocity direction being in the yoyo.

I'm sorry for being a little ambiguous about this topic.

Actually I wasn't saying that there was an "angular velocity direction in the yoyo," I meant that the direction that we would assign the angular velocity in the yoyo ( by the right hand thumb rule) would be pointing out of the plane of the yoyo,which is found to be same as the direction that the yoyo's string was seen to be coiling up (like when we twist a piece of string,we see sort of a movement up the string).
However,I understand that it was only my imagination,since,as BobbyBear said,the direction assigned to angular motion is only a mathematical assumption.

 

[Urmi Roy]

Recently I got into a tangle while doing rolling without slipping and rolling with slipping.

Well,in one of my textbooks it says that rolling without slipping involves rolling friction which provides the requisite amount of torque to stop a wheel from slipping and in another,it sates absolutely nothing about rolling friction and just states about static and dynamic friction (and that too very ambiguously).

How does a tyre slip with its linear velocity faster than its angular velocity would ideally allow it anyway? Can't really imagine that happenning.

I would be grateful if someone please gives me a description of how the forces that act and their individual roles on a wheel during rolling without slipping and rolling with slipping.

 

[D H]

If you have seen people laying rubber with a car, that is rolling with slipping. Step on the gas hard enough (assuming an adequately powered vehicle) and the torque applied by the engine will exceed the maximum torque possible from rolling friction. Result: The tire rotates faster than a rolling tire and leaves a strip of rubber on the road due to the higher sliding friction.

 

[sganesh88]

How does a tyre slip with its linear velocity faster than its angular velocity would ideally allow it anyway? Can't really imagine that happening.

DH, Urmi is bothered about linear velocity being greater than "what the angular velocity would allow"- meaning R X W(omega). I had a similar pre-occupation before. I thought that if a rolling wheel loses contact with the ground and gets effused into space, it will abruptly come to a stop because the liaison, friction, that "should" continuously act to equate V and "R X W" isn't present anymore.
The thing is that the wheel's linear velocity and angular velocity are actually independent. You can have any pair of values for these two quantities if friction doesn't interfere(so i was partly right in my misconception).
A wheel kept on a frictionless incline will just slide without any angular velocity. An automobile with its engine revved to even 12000 rpm on a frictionless surface will stay put with an enormous W (measured at its wheels) but zero linear velocity. As friction does its duty of preventing relative motion, it acts in a "less villainous" way when it comes to wheels. By modifying V and W such that V=R X W, friction stops functioning because in such a configuration, there is no point on the wheel that tries to slide or slip past the surface(observe a cycloid curve). Slipping and sliding arises when the max friction force possible isn't quite enough to bring about the synchronization. So in DH's example, as the accelerator pedal is pressed hard, wheels start rotating at an insane speed and friction is at a loss to sync V and W according to the above mentioned relation that quickly resulting in slipping. Same applies to sliding during hard braking.

Newton devised his second law of motion,F=ma, after observing the world around him,and having found a deep logic behind it--where does that apply to those expressions of rotational mechanics,with all that vector product and mathematical assumptions?

Ya. Rotational mechanics doesn't operate on a different set of laws. The same three laws suffice. Conservation of angular momentum can't arise if conservation of momentum is false.

 

[Urmi Roy]

The thing is that the wheel's linear velocity and angular velocity are actually independent. You can have any pair of values for these two quantities if friction doesn't interfere...

I think this explanation and the examples provided were excellent,thanks for the clarification!

Ya. Rotational mechanics doesn't operate on a different set of laws. The same three laws suffice. Conservation of angular momentum can't arise if conservation of momentum is false.

I read recently that the law of conservation of angular momentum can be derived directly (with quite meddlesome maths) from the expression of the law of conservation of momentum and also that both can be derived from something called the Noether's Theorem.

I was wondering if you all could perhaps throw some light on this issue and ofcourse I would love to receive more information relating to 'rotational mechanics' from all of you experts,especially pertaining to the last few posts in this thread(e.g. posts three and four perhaps).

 

[D H]

If you read that conservation of angular momentum derives from conservation of momentum, it is wrong. The two are fundamentally different. Conservation of momentum implies the weak form of Newton's third law. The strong form is needed for conservation of angular momentum. Alternatively, one could just use Noether's Theorem, but that is a topic of rather advanced (upper level undergraduate / lower level graduate) physics.

 

[Urmi Roy]

I'm sorry for making a blunder,I just remembered that it wasn't the conservation of momentum that the conservation of angular momentum is derived from,it was Newton's third law (every action has equal and opposite reaction,).

Since the two laws (conservation of angular momentum and conservation of momentum ) are analogues, they also can be derived from the same law-Newton's third law---that's what I read.

 

[Urmi Roy]

By the way, I've been pondering about what sganesh88 said about the rolling concept--well there are 2 things I want to confirm-

1. rolling friction seems to have no role here,it was probably just an alternate explanation in one of my books (H.C Verma).

By modifying V and W such that V=R X W, friction stops functioning because in such a configuration, there is no point on the wheel that tries to slide or slip past the surface(observe a cycloid curve). Slipping and sliding arises when the max friction force possible isn't quite enough to bring about the synchronization.

2. When the rolling does not have any slipping, the frictional froces don't work,because they don't need to.

 

[BobbyBear]

Rolling Friction

Recently I got into a tangle while doing rolling without slipping and rolling with slipping.

Lol for a minute I imagined you rolling with and without slipping on your living room carpet:P
Urmi, you are very sweet when you get all excited about physics :)

I think your queries on this have been addressed to, although you did ask about rolling friction. I'd just like to give my opinion that rolling friction is not some special or separate form of friction at all - I'd be more happy simply calling it 'rolling resistance'. Basically, it's the consideration of the distribution of forces involved within the contact area of the bodies, which is in reality a finite area, as all bodies are deformable. From these considerations one can conclude for example that a minimum torque is needed to set the wheel in rolling motion (due to the distribution of forces in general, not necessarily due to a dissipative force) - however, there'd be no way to explain this minimum torque if you are considering the bodies to be non-deformable. Can anyone please confirm that if we're considering the wheel to be perfectly rigid, we'd not be able to talk of a 'rolling friction force' associated with a 'rolling friction coefficient'?

Urmi, I think you should concentrate first on understanding how the sliding friction force (the only kind of friction force I'd consider, ie. proportional to the coefficient of friction times the normal force and pointing in the direction so as to oppose impending relative movement) plays a part in producing or balancing the torques acting upon the wheel. Consider different scenarios: a wheel rolling down a hill (there is a gravitational force-component acting upon the centre of mass causing it to accelerate), a wheel rolling up a hill, a wheel connected to a shaft which is exerting a torque that tends to increase the angular velocity of the wheel, or a torque that is opposing the angular velocity of the wheel . . . in each of these cases, which must be the direction of the friction force? Consider what the movement would be like if there was no friction (completely slippery surface) and what role the friction force is playing in the movement of the wheel. It's always helpful to draw a free body diagram of the wheel with all the forces acting upon it, and simply apply Newton's second law for linear and rotational motion.

I'd also like you to consider, if you haven't already, that frictional force does no work when a body rolls without slipping. Do you see this clearly? (again, this assuming that there is no deformation).

 

[BobbyBear]

Um, I think I'm getting into a little tangle myself:P

Consdier an indeformable wheel rolling down an inclined plane. The wheel is subject to the force of gravity (F_g), which we can separate into a component normal to the plane and a component tangent to the plane, acting at the centre of mass of the wheel, say F_N and F_T. If there was no friction at all, then the wheel would not spin but slide downward with a constant linear acceleration given by Newton's law (a=F_T/m). But if friction is not zero it will set the wheel in angular motion. Let's suppose that the maximum static (sliding) friction is greater than F_T so that there is no slipping. Then the friction force at the point of contact, f, will be in the direction opposite to F_T and equal to F_T. So the net torque upon the wheel will be T=F_T*R, and the angular acceleration of the wheel ( $$\alpha$$ ) will be given by Newton's law: $$T=\alpha *I.$$

http://img150.imageshack.us/img150/2525/planef.gif

Obviously the angular velocity of the wheel is increasing, but since linear and angular quantities are related by $$v= R *\omega$$, the wheel has a linear acceleration too, or rather each point of the wheel - in particular, the centre of mass of the wheel has a linear acceleration given by $$a=r*\alpha$$ . . . no? If my calculations are correct,
I=3/2 * m * R² (with respect to the instantaneous centre of rotation) and the linear acceleration of the centre of mass is therefore 2/3 * F/m.

But if we apply Newton's second law to the wheel however it tells us that since the net force upon the wheel is zero, the acceleration of its centre of mass is zero.

Um, where am I going wrong?

 

[BobbyBear]

If you read that conservation of angular momentum derives from conservation of momentum, it is wrong. The two are fundamentally different. Conservation of momentum implies the weak form of Newton's third law. The strong form is needed for conservation of angular momentum. Alternatively, one could just use Noether's Theorem, but that is a topic of rather advanced (upper level undergraduate / lower level graduate) physics.

This is really beyond me, but does Noether's Theorem have something to say about the stress tensor being symmetric?

 

[BobbyBear]

2. When the rolling does not have any slipping, the frictional froces don't work,because they don't need to.

I think it's because the relative velocity at the point of contact is zero, thus the friction force is not doing work.

 

[Urmi Roy]

Basically, it's the consideration of the distribution of forces involved within the contact area of the bodies, which is in reality a finite area, as all bodies are deformable.

This is quite a good way of putting it.

however, there'd be no way to explain this minimum torque if you are considering the bodies to be non-deformable. Can anyone please confirm that if we're considering the wheel to be perfectly rigid, we'd not be able to talk of a 'rolling friction force' associated with a 'rolling friction coefficient'?

This is what I was confused about--we're doing rotational mechanics of rigid bodies,so ideally there should not be any such deformation considered.

 

[Urmi Roy]

Obviously the angular velocity of the wheel is increasing, but since linear and angular quantities are related by $$v= R *\omega$$, the wheel has a linear acceleration too, or rather each point of the wheel - in particular, the centre of mass of the wheel has a linear acceleration given by $$a=r*\alpha$$ . . . no?

In rolling without slipping v=R*w should be valid, but as we know the tyre may also slip,in which case this relation will not be valid, so we can't just assume that v=R*w is valid (and then explain the why there is no net acceleration of the tyre on this basis),----That's the whole point ,since I'm asking about what it is that makes it valid in the first place, for the special case of rolling without slipping,(in terms of forces acting therein).

But if we apply Newton's second law to the wheel however it tells us that since the net force upon the wheel is zero, the acceleration of its centre of mass is zero.

Um, where am I going wrong?

Whereas the net linear force is zero,the net torque is not zero.

BobbyBear stated that the net torque about the centre of mass of the wheel is T=(alpha)*I,due to the friction--the question is, which force balances this torque due to friction in order to have zero angular acceleration of the tyre?

 

[Urmi Roy]

I think it's because the relative velocity at the point of contact is zero, thus the friction force is not doing work.

The frictional force does work in rotating the wheel about the centre of mass,doesn't it?

See! There are so many confusions!

 

[Doc Al]

But if we apply Newton's second law to the wheel however it tells us that since the net force upon the wheel is zero, the acceleration of its centre of mass is zero.

Um, where am I going wrong?

For some reason, you assume that the friction force is equal to the component of gravity parallel to the plane. That's not the case--the net force is not zero.

 

[Doc Al]

Whereas the net linear force is zero,the net torque is not zero.

Neither the net linear force nor the net torque is zero.

BobbyBear stated that the net torque about the centre of mass of the wheel is T=(alpha)*I,due to the friction--the question is, which force balances this torque due to friction in order to have zero angular acceleration of the tyre?

The wheel accelerates.

The frictional force does work in rotating the wheel about the centre of mass,doesn't it?

No, the static friction force does no work.

 

[BobbyBear]

For some reason, you assume that the friction force is equal to the component of gravity parallel to the plane. That's not the case--the net force is not zero.

Yes I am assuming that! Shouldn't I? I'm supposing, as I said, that there is no slipping, so the friction force is a static friction force. The static friction force always equals (in magnitude) the force that wants to provoke movement, until it reaches its maximum possible value (which is the static friction coefficient times the component of the weight normal to the plane of movement). So, if instead of a wheel we had a rectangular block (of equal mass as the wheel), the block would not be moving (assuming an initial state of rest, and assuming, of course, that there's no toppling), because the friction force would be equal to (in magnitude but opposing) the component of gravity parallel to the plane.

The block would be in static equilibrium, because the sum of all forces would be zero, and the sum of all moments zero as well: the moment created by the pair of forces parallel to the plane (gravity component and friction) would be counteracted by an opposing moment created by the pair of forces normal to the plane (F_N, N) (their respectice resultants would be displaced by the necessary amount from the central point of the contact area to create this moment). But if the block had a small base so that the maximum opposing moment is less than the one created by the first pair, the block would topple over itslef, rotating about its bottom right corner.

It's the same static friction force that we're talking about for the wheel, which is continuously 'toppling' over its bottom right corner (which is reduced to the single point of contact). This would be what we're referring to rolling without slipping.

In both cases (toppling block or rolling wheel), there is no slipping, doesn't this mean (and is necessary for there to be no slipping) that the friction force has equal magnitude to the gravity component parallel to the plane?

Thank you for your comments,
Bobby

 

[BobbyBear]

Neither the net linear force nor the net torque is zero.

This is what I am not seeing (as far as the net linear force goes, that is) ... :(

 

[Doc Al]

Yes I am assuming that! Shouldn't I?

No.

I'm supposing, as I said, that there is no slipping, so the friction force is a static friction force. The static friction force always equals (in magnitude) the force that wants to provoke movement, until it reaches its maximum possible value (which is the static friction coefficient times the component of the weight normal to the plane of movement).

OK.

So, if instead of a wheel we had a rectangular block (of equal mass as the wheel), the block would not be moving (assuming an initial state of rest, and assuming, of course, that there's no toppling), because the friction force would be equal to (in magnitude but opposing) the component of gravity parallel to the plane.

OK. But that's true for a block but not a wheel.

In both cases (toppling block or rolling wheel), there is no slipping, doesn't this mean (and is necessary for there to be no slipping) that the friction force has equal magnitude to the gravity component parallel to the plane?

No.

In the case of the wheel, you must figure out (using Newton's laws) the friction force necessary to have the wheel not slip. It's not simply equal to the gravity component parallel to the plane.

Set up the torque/force equations for rotation and translation and solve for the needed friction force.

 

[BobbyBear]

No.
In the case of the wheel, you must figure out (using Newton's laws) the friction force necessary to have the wheel not slip. It's not simply equal to the gravity component parallel to the plane.

Set up the torque/force equations for rotation and translation and solve for the needed friction force.

Well this comes to me as a bit of a surprise, as I'm thinking of the case of the wheel as a limiting case of the block of n sides (regular polygon) as the No. of sides goes to infinity.

So okay, let me accept that for the case of the wheel the friction force is not equal to (in magnitude) the gravity component parallel to the plane. So what you're saying is I'd apply Newton's laws of rotation and translation to the wheel, so I'd get on the one hand the angular acceleration of the wheel as a function of the unknown friction force f, and on the other the linear acceleration of the centre of mass (as a function of f again), and then impose the no-slipping condition through which I can relate the angular acceleration of the wheel to the linear acceleration of the centre of mass (as I did previously except that I had assumed f to be known), and from there solve for f. I think this would be the procedure you're indicating?

The fact that f is not equal to the component of gravity parallel to the plane would certainly eradicate the contradiction I was facing about the linear acceleration of the wheel being nonzero while the net linear force was (as I was erroneously assuming to be) zero.

But I'm still a little confused then about the case of the rectangular block (or n-sided polygon in general). You agreed that in this case, the friction force does equal the gravity component parallel to the plane if we are assuming there is no slipping. So wouldn't I end up with the same contradiction as I was earlier with the wheel? The net linear force on the block is zero, but IF it is toppling because there is a net moment, then the centre of mass would be accelerating . . . yes?

 

[Doc Al]

Well this comes to me as a bit of a surprise, as I'm thinking of the case of the wheel as a limiting case of the block of n sides (regular polygon) as the No. of sides goes to infinity.

That's an interesting way of looking at it, which I haven't analyzed yet.

So okay, let me accept that for the case of the wheel the friction force is not equal to (in magnitude) the gravity component parallel to the plane. So what you're saying is I'd apply Newton's laws of rotation and translation to the wheel, so I'd get on the one hand the angular acceleration of the wheel as a function of the unknown friction force f, and on the other the linear acceleration of the centre of mass (as a function of f again), and then impose the no-slipping condition through which I can relate the angular acceleration of the wheel to the linear acceleration of the centre of mass (as I did previously except that I had assumed f to be known), and from there solve for f. I think this would be the procedure you're indicating?

Exactly. This is how one can solve for the acceleration as well as the friction force.

The fact that f is not equal to the component of gravity parallel to the plane would certainly eradicate the contradiction I was facing about the linear acceleration of the wheel being nonzero while the net linear force was (as I was erroneously assuming to be) zero.

Right.

But I'm still a little confused then about the case of the rectangular block (or n-sided polygon in general). You agreed that in this case, the friction force does equal the gravity component parallel to the plane if we are assuming there is no slipping.

I agreed that the net force is zero in the case of a block that neither slipped nor toppled.

So wouldn't I end up with the same contradiction as I was earlier with the wheel? The net linear force on the block is zero, but IF it is toppling because there is a net moment, then the centre of mass would be accelerating . . . yes?

If the block is toppling, the net force is not zero.

 

[BobbyBear]

If the block is toppling, the net force is not zero.

So, even if the static friction force is capable of opposing the component of gravity parallel to the surface (in the sense that the maximum static friction force is greater than that force), it doesn't oppose it fully because of the geometry and weight being such that toppling occurs. Hmm this is very curious indeed, and though it must be so to satisfy Newton's laws, I'm wondering if there is an intuitive and logical physical explanation for this phenomenon.

Could it be that perhaps during toppling, what's happening is that the friction force no longer obeys the law $$f_{max}=\mu N$$, and that the friction force is the maximum it can possibly be? I do realize that the law $$f_{max}=\mu N$$ is more of an approximate quantification of a complex phenomenon that may require a study of the principles of friction to understand better.

 

[Urmi Roy]

Well, I hope this doesn't sound too naive,but I remember that I once did a problem in which a block placed on a tilted ramp toppled over,and it seems that in that problem,it was considered that the block was put on a ramp,which was at a particular angle,such that considering the lowermost point of the block which is in contact with the ramp to contain the axis (going through that lowermost edge of the block), the net torque about that point due to the friction and the overall gravitational force was not zero.

This was mainly due to the angle at which the ramp was tilted--above a certain angle, the torques due to the friction and the overall gravitational forces are in such an orientation,that their torques no longer sum up to zero.