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Block Toppling on an incline

haruspex

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Even without friction, the normal force and the weight can produce a non-zero couple when the angle is sufficiently great.
Sure, but if it's on the verge of toppling then the normal force acts through the lowest point of the block (A). Its moment about the c.o.m. opposes toppling, so as BvU says there is no moment about the c.o.m. to cause toppling. If you are taking moments about A then bear in mind that a moment is required to account for the acceleration down the plane. Toppling will only occur if the moment that way exceeds that required by the acceleration.
Once sliding, I believe the critical value of ##\mu_s## for toppling is block length / block height.
 

haruspex

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What will happen if ##\theta >60##. Will it topple and slide together?
If ##\tan(\theta) > \mu_k > \mu_s > \frac{block length}{block height}## then it will topple while sliding.
Correction: I have k and s crossed over there.
 
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591
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Sure, but if it's on the verge of toppling then the normal force acts through the lowest point of the block (A). Its moment about the c.o.m. opposes toppling, so as BvU says there is no moment about the c.o.m. to cause toppling.
Then that means that the block will not topple without friction because friction was providing a torque for rotation.



If you are taking moments about A then bear in mind that a moment is required to account for the acceleration down the plane. Toppling will only occur if the moment that way exceeds that required by the acceleration.
Once sliding, I believe the critical value of ##\mu_s## for toppling is block length / block height.
Are you talking about the torque due to pseudo force?
 
591
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If ##\tan(\theta) > \mu_k > \mu_s > \frac{block length}{block height}## then it will topple while sliding.
From the figure in #post24.
Finding torque about the com

## N\frac{l}{2}=\mu N \frac{h}{2}##

So ##\mu=\frac{l}{h}## (Let ##h## and ##l## be the height and length of the cube)

Did you find the value of ##\mu## in this way?
How did you come to know that at this value of ##\mu## the block will topple and slide simultaneously.
 

haruspex

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Then that means that the block will not topple without friction because friction was providing a torque for rotation.
That is the conclusion BvU and I have come to.
Are you talking about the torque due to pseudo force?
Maybe. Never been one for pseudo forces, but as I understand it the pseudo force associated with the linear acceleration down the slope would be equal and opposite to the resultant of the actual forces. Similarly, a pseudo torque. The pseudo force here has a pseudo torque about A. I believe adding that in to the torque equation about A will show there's no toppling without friction.
 
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Maybe. Never been one for pseudo forces, but as I understand it the pseudo force associated with the linear acceleration down the slope would be equal and opposite to the resultant of the actual forces. Similarly, a pseudo torque. The pseudo force here has a pseudo torque about A. I believe adding that in to the torque equation about A will show there's no toppling without friction.
I have a confusion:

If we choose our axis of rotation at the com of the body then we do not include pseudo force even if the c.o.m is accelerating. Right?

But if we choose our axis of rotation (other than c.o.m) on the body and if that point the body(through which the axis is passing) is accelerating then we have to consider pseudo force.

Where does the the pseudo force act on the body? Does it act on the c.o.m of the body?

Please reply to my #post30.:D
 

haruspex

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From the figure in #post24.
Finding torque about the com

## N\frac{l}{2} = \mu N \frac{h}{2}##

So ##\mu=\frac{l}{h}## (Let ##h## and ##l## be the height and length of the cube)

Did you find the value of ##\mu## in this way?
Yes
How did you come to know that at this value of ##\mu## the block will topple and slide simultaneously.
As you've just shown, if it is sliding then the toppling criterion is ##\mu_k > \frac{l}{h}##. (I had k and s crossed over before. I often do that. 'k' feels like 'stuck', while 's' feels like 'slipping'.) In order to slide we must have ##\tan(\theta) > \mu_s##. And necessarily ##\mu_s >= \mu_k##.
 

haruspex

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Where does the the pseudo force act on the body? Does it act on the c.o.m of the body?
The pseudo force is the equal and opposite of ma, not just as a vector but in all respects, so necessarily it acts through the c.o.m.
 
591
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The pseudo force is the equal and opposite of ma, not just as a vector but in all respects, so necessarily it acts through the c.o.m.
Is 'm' the mass of body and 'a' the acceleration of the point through which the rotational axis is passing?
 

haruspex

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Why should ##\mu_s >= \mu_k##?
Sorry if I am missing some thing obvious?
If kinetic friction were greater than static friction then what would happen if the force were just sufficient to overcome the static? As soon as the body moved a fraction it would be subject to the greater kinetic friction and stop again, so it can't actually slip. So by definition it has not overcome static friction.
 

haruspex

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Is 'm' the mass of body and 'a' the acceleration of the point through which the rotational axis is passing?
'a' is the linear acceleration of the body, so the pseudoforce for it passes through the c.o.m.
 
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Thank you haruspex for your help.
In post #19 voko said that "Even without friction, the normal force and the weight can produce a non-zero couple when the angle is sufficiently great". But your #post 26 seems to contradict his post.
I don't understand if a block can topple on a incline even without friction or not.
 

haruspex

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Thank you haruspex for your help.
In post #19 voko said that "Even without friction, the normal force and the weight can produce a non-zero couple when the angle is sufficiently great". But your #post 26 seems to contradict his post.
I don't understand if a block can topple on a incline even without friction or not.
I've provided my reasoning. I'd like to wait for voko to respond before commenting further.
 
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I've provided my reasoning. I'd like to wait for voko to respond before commenting further.
Thank you haruspex for helping me till here.:)
 
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Sure, but if it's on the verge of toppling then the normal force acts through the lowest point of the block (A). Its moment about the c.o.m. opposes toppling, so as BvU says there is no moment about the c.o.m. to cause toppling. If you are taking moments about A then bear in mind that a moment is required to account for the acceleration down the plane. Toppling will only occur if the moment that way exceeds that required by the acceleration.
Once sliding, I believe the critical value of ##\mu_s## for toppling is block length / block height.
Let’s recast our current discussion as a problem.

Given: a block of a particular height and a particular width, on an incline of a certain angle with the horizontal, with a certain coefficient of static friction between the block and the incline; the afore mentioned parameters are such that the block will tumble but not slide.

Question: is there a combination of the parameters satisfying the condition above and such that if the coefficient of static friction is perturbed, the block will slide but not tumble, all other parameters held fixed?

I interpret your passage above as a way of saying "yes" to the question. But can you demonstrate that directly?
 

haruspex

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Let’s recast our current discussion as a problem.

Given: a block of a particular height and a particular width, on an incline of a certain angle with the horizontal, with a certain coefficient of static friction between the block and the incline; the afore mentioned parameters are such that the block will tumble but not slide.

Question: is there a combination of the parameters satisfying the condition above and such that if the coefficient of static friction is perturbed, the block will slide but not tumble, all other parameters held fixed?

I interpret your passage above as a way of saying "yes" to the question. But can you demonstrate that directly?
As BvU observed, for the block (height h, length l) to tumble there must be a net torque to that effect about the c.o.m. of the block. At the point of tumbling, both the normal and frictional forces, N and F, can be taken as acting through the lowest point of contact (A in the diagram). For the net torque to be in the right direction, |F|/|N| > l/h.
What is your argument for saying it can tumble even with no friction?
 

ehild

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As for toppling even at zero friction:
If the blocks turns the CM must travel faster than the bottom edge around it topples. So something keeps back the edge in contact with the slope. If there is no friction, nothing keeps it back.

I would analyse the problem (sliding and toppling over) in an accelerating frame of reference. Just before the possible toppling, the CM travels with acceleration g(sinθ-μcosθ). There is an inertial force Fi acting backwards, parallel with the slope. Its lever arm is b/2.
The force of gravity G turns the block clockwise. The lever arm is c. You get it from the blue triangle. Gravity would topple the block, but the torque of the inertial force keeps it back. What is the condition that the block topples, with given a,b; θ, and μ, supposing that it slides, that is, μ<tanθ?

toppling2.JPG
 
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At the point of tumbling, both the normal and frictional forces, N and F, can be taken as acting through the lowest point of contact (A in the diagram).
That is how I used to think earlier. I demonstrated that in equilibrium, including on the verge of tumbling, the normal force acts at a point at the bottom, vertically below the CoM. I then stated that this was "regardless of friction". I now realize this was ungrounded, because my statement rests on an assumption of equilibrium. Without friction (or with insufficient friction), there is no equilibrium, so the implication is not valid.

When I think about this more, it is far from evident that in the frictionless situation the normal force is applied at the lowest edge.
 
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If the blocks turns the CM must travel faster than the bottom edge around it topples.
The block can be tumbling in two directions.
 

haruspex

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When I think about this more, it is far from evident that in the frictionless situation the normal force is applied at the lowest edge.
If it even starts to tumble, any contact forces must act at the contact point. The normal force will then give a moment that opposes tumbling. So it won't start.
The block can be tumbling in two directions.
You mean, it can go clockwise or anticlockwise? Same argument as above applies.
 

haruspex

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As for toppling even at zero friction:
If the blocks turns the CM must travel faster than the bottom edge around it topples. So something keeps back the edge in contact with the slope. If there is no friction, nothing keeps it back.

I would analyse the problem (sliding and toppling over) in an accelerating frame of reference. Just before the possible toppling, the CM travels with acceleration g(sinθ-μcosθ). There is an inertial force Fi acting backwards, parallel with the slope. Its lever arm is b/2.
The force of gravity G turns the block clockwise. The lever arm is c. You get it from the blue triangle. Gravity would topple the block, but the torque of the inertial force keeps it back. What is the condition that the block topples, with given a,b; θ, and μ, supposing that it slides, that is, μ<tanθ?

View attachment 74937
I must say this does not seem as straightforward an approach as taking moments about c.o.m., but having gone through it I get the same answer: ##\mu_k > a/b##. Do you?
 

ehild

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I must say this does not seem as straightforward an approach as taking moments about c.o.m., but having gone through it I get the same answer: ##\mu_k > a/b##. Do you?
Yes :)
 

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