# Will a sliding box fall over when it stops?

Tibriel

## Homework Statement Not an actual homework problem but a discussion that came up in class while we were learning about torque.
A tall box is sliding across a surface with friction f, mass m, and velocity v. What equations would you use to figure out if the box would tip over while sliding to a stop.

T = Fd
f = (mu) N
?

## The Attempt at a Solution

I originally was thinking that you could solve this with just torque and force equations but now I'm thinking you would have to use rotational inertia or maybe angular velocity equations. I'm just not sure where to start with it.[/B]

#### Attachments

Homework Helper
Gold Member
2022 Award
you would have to use rotational inertia or maybe angular velocity equations.
Not really. It is fairly obvious that once it starts to tip conditions will favour tipping even more.
So post an attempt just based on torque balance.

Tibriel
So I've done this by using a dry erase marker as a stand in for the box. At low speeds the marker can slide to a stop. But go too fast and the marker tips once the friction goes from kinetic to static.

Homework Helper
Gold Member
2022 Award
So I've done this by using a dry erase marker as a stand in for the box. At low speeds the marker can slide to a stop. But go too fast and the marker tips once the friction goes from kinetic to static.
That's an interesting observation, but not quite what you asked. You wrote "while sliding to a stop".

Are you now asking under what circumstances it will tip as it stops (and not before)?
A simple view would say that s.f. cannot kick in until it has stopped, and once it has stopped there would be no reason for it to fall over.
But your experiment suggests there is more to it.

Here's a possibility... the k.f. varies along the surface, and it is almost enough to tip the block straightaway. At some point, the block reaches a place where the friction is a little higher. This starts to tip the block; as a result, the kinetic friction is now sufficient to stop the block too, so you observe the falling over coinciding with the stopping.

See if you can verify or dismiss that possibility with some equations.

Staff Emeritus
Homework Helper
Gold Member
Here's a possibility... the k.f. varies along the surface, and it is almost enough to tip the block straightaway.
Counter possibility: The friction does not vary as a step function with velocity. It should be relatively easy to set up a lab to test this.

Not really. It is fairly obvious that once it starts to tip conditions will favour tipping even more.
I do not think this is true. In order to tip for sure the CoM needs to be on the wrong side of the tipping point. Consider the limit of a very low flat box.

Homework Helper
Gold Member
2022 Award
I do not think this is true. In order to tip for sure the CoM needs to be on the wrong side of the tipping point. Consider the limit of a very low flat box.
I should have specified, so long as the block is sliding (which is a condition given in post #1).
While it is sliding, if it starts to tip then the restoring torque arm shrinks while the tipping torque arm increases. Once it stops sliding, tipping will only continue as long as there is some rotational momentum to sustain it.

Homework Helper
Gold Member
I have not worked out the details, but I suspect this is an example of stick-slip behavior. The block is sliding with friction force fk when all of a sudden the friction changes to a larger value fs. It would be a bit like hitting an unnoticed protruding part of the sidewalk with your toe while walking normally and tipping over.

Homework Helper
Gold Member
2022 Award
I have not worked out the details, but I suspect this is an example of stick-slip behavior. The block is sliding with friction force fk when all of a sudden the friction changes to a larger value fs. It would be a bit like hitting an unnoticed protruding part of the sidewalk with your toe while walking normally and tipping over.

Yes, this is what I was suggesting in post #4, but the interesting part is to show that sticking then leads to tipping, hence the coincidence observed.

Homework Helper
Gold Member
Yes, this is what I was suggesting in post #4, but the interesting part is to show that sticking then leads to tipping, hence the coincidence observed.
I think one can model this as moving and then suddenly stopping when sticking occurs (if not there will still be sliding.) What happens to the kinetic energy? I believe it becomes rotational kinetic energy about the leading corner of the block that acts as a pivot. So the block will rotate about this corner which means that the CM describes a circle of radius equal to half a diagonal. In order to tip over, the CM must have enough energy to reach a vertical height of ##\frac{\sqrt{b^2+h^2}}{2}## where ##b## and ##h## are the base and height of the block. The condition for tipping is $$\frac{1}{2}mv_{crit.}^2>mg \left( \frac{\sqrt{b^2+h^2}}{2}-\frac{h}{2} \right)$$where ##v_{crit.}## is the critical speed needed to go over the potential barrier. Clearly, the higher the ratio ##h/b##, the lower ##v_{crit.}##. I do not know how to estimate the speed below which slip-stick behavior occurs from information such as ##\mu_k##. Comparing that with ##v_{crit.}## should allow the prediction whether a sliding block with a given ##h/b## ratio will come to rest by tipping forward all the way, by teetering back and forth a few times or by continuing to slide.

Staff Emeritus
Homework Helper
Gold Member
I think one can model this as moving and then suddenly stopping when sticking occurs (if not there will still be sliding.) What happens to the kinetic energy?
In a ”sticking” kind of process, you are typically dealing with something inelastic.

Homework Helper
Gold Member
2022 Award
What happens to the kinetic energy?
Angular momentum about a point on the surface would be more persuasive.

Homework Helper
Gold Member
Angular momentum about a point on the surface would be more persuasive.
Sure. That would also address the comment by @Orodruin #10. Conserve angular momentum about the pivot to find ##\omega## immediately after the collision and convert that to ##v_{crit.}##. The expression for ##v_{crit.}## will be different but the idea is the same.

Homework Helper
Gold Member
2022 Award
this is what I was suggesting in post #4,
Correction... that is not quite what I was suggesting.
Rather, I was thinking of the kinetic friction being a little higher at some point; not enough to stop it outright but enough to initiate the tipping. My question is whether that would lead quite quickly to the base stopping. In this way, the observer might see the two as simultaneous.

Homework Helper
Gold Member
It could be as you say. The usual approximation is that the acceleration is constant until the sliding object stops. The acceleration has to drop to zero from a (nearly) constant value. How to model that and make it stick (the pun is intentional) is beyond me.

Homework Helper
Gold Member
Your solution is appropriate the following situation: You push a block of base ##x## and height ##y## with horizontal force ##F## at height ##y/2##. What is the condition for the block to tip before it slides?
Answer: For the the block to tip before it slides, the maximum force of friction must be greater than the pushing force ##F## when the block tips. If we calculate torques about the pivot corner, ##\frac{mgx}{2}-\frac{F_{min}y}{2}=0##. Here ##F_{min}## is the minimum force that will initiate tipping. Now static friction ##f_s## must match ##F## in the horizontal direction otherwise there will be slipping. This means that ##f_s=F_{min}##. Substitute to get the force of static friction at the onset of tipping ##f_s=\frac{mgx}{y}##. That's all you can say. You cannot replace ##f_s## with its maximum value ##\mu_s mg##. That would be the case if you knew that the block is simultaneously on the verge of both tipping and slipping. Furthermore the case has not been made that this problem, with an external pushing force on a stationary block, is equivalent to the original problem of a sliding block that suddenly seizes and stops. I think you should consider conservation of angular momentum between just before and just after the block stops as suggested by @haruspex in #11 and then energy conservation as I suggested in #9. See where that gets you. I must admit I have not tried it, yet.

Homework Helper
Gold Member
2022 Award
I considered a uniform block of height 2h, width 2w (in the direction of motion) which is initially sliding and erect. It then encounters a region where the friction is a bit higher, enough to initiate tipping.
I am interested in the rapidity with which it shifts to be static friction.
Call the point of contact P and the mass centre C.
Writing ##\phi## for the angle CP subtends to the horizontal, I ended up with ##\ddot \phi/3=(\mu \sin(\phi)-\cos(\phi))(g/A+\ddot\phi\cos(\phi)-\dot\phi^2\sin(\phi))##, where A2=w2+h2.
However, I have reason to suspect an algebraic error - still checking.

We can replace g/A by 1 with a suitable change to the time base.
Note that the horizontal position (x) does not feature in this, so conservation of momentum doesn't help with this analysis.

Last edited:
Homework Helper
Gold Member
2022 Award
Following from post #17...

y is height of mass centre, z is its horizontal displacement "behind" the contact point.
A is the distance from contact point P to mass centre C: ##y=A \sin(\phi)##, ##z=A\cos(\phi)##.
##\ddot y=\ddot\phi A\cos(\phi)-{\dot\phi}^2A \sin(\phi)##.
N is the normal force. Friction is kinetic. I=mk2=mA2/3.
1. Moments about mass centre: ##I\ddot\phi=\mu N y-N z##
2. Vertical acceleration: ##m\ddot y = N-mg##
##mA^2\ddot\phi/3=(\mu y- z)(mg+m\ddot y)##
##=A(\mu \sin(\phi)- \cos(\phi))m(g+(\ddot\phi A\cos(\phi)-{\dot\phi}^2A \sin(\phi)))##
##\ddot\phi/3=(\mu \sin(\phi)- \cos(\phi))(g/A+(\ddot\phi \cos(\phi)-{\dot\phi}^2\sin(\phi)))##
##\ddot\phi(\frac 13-(\mu \sin(\phi)- \cos(\phi))\cos(\phi))=(\mu \sin(\phi)- \cos(\phi))(g/A-{\dot\phi}^2\sin(\phi)))##
We can normalise the time and distance units so that g=A=1:
##\ddot\phi(\frac 13-(\mu \sin(\phi)- \cos(\phi))\cos(\phi))=(\mu \sin(\phi)- \cos(\phi))(1-{\dot\phi}^2\sin(\phi)))##

But here's the problem... under reasonable conditions (##\dot\phi=0##, ##\mu=2##, ##\phi=\pi/3##), ##(\mu \sin(\phi)- \cos(\phi))\cos(\phi)=\frac 12\sqrt 3-\frac 14>\frac 13##, so ##\ddot\phi<0##.

Can anyone spot my error?

Counter possibility: The friction does not vary as a step function with velocity. It should be relatively easy to set up a lab to test this.

.
What I observe on a polished granite countertop that should have minimal surface variation, is that a narrow object will coast smoothly to a stop up to some velocity at release; for slightly faster speed, it will wobble while slowing, but not fall; for higher yet release velocity, after coasting smoothly for a time, it will tip over in the direction of motion. To avoid any instabilities from pushing, I am using a large regular block taller than the target object to do the pushing. I think this supports your hypothesis, with the idea that as torque increases as velocity slows, how much momentum remains determines whether toppling will occur.

Homework Helper
Gold Member
2022 Award
after coasting smoothly for a time, it will tip over in the direction of motion
Were you able to determine whether it was still sliding when tipping commenced?

Homework Helper
Gold Member
2022 Award
for higher yet release velocity,
If the higher release velocity leads to toppling, consider release speeds v>u.
There are two possibilities:
• While slowing from v to u, a change takes place in the contact area, leading to a higher friction coefficient
• The toppling commences before slowing to speed u
The second says friction is greater at the higher speed, consistent with some research I see online. But it sounds like this not consistent with your observation.

Were you able to determine whether it was still sliding when tipping commenced?
My kitchen is hardly a lab, despite my efforts to be careful. I think it was still slidingwhen it began to tip. I cannot determine the speed when wobble begins (but no tipping over) compares to the speed when tipping over begins (in the cases where this happens). Could be different speeds, or the same, but it looks like it still sliding just before this.

My kitchen is hardly a lab, despite my efforts to be careful. I think it was still slidingwhen it began to tip. I cannot determine the speed when wobble begins (but no tipping over) compares to the speed when tipping over begins (in the cases where this happens). Could be different speeds, or the same, but it looks like it still sliding just before this.
Looking at this more carefully, from the side, as release happens, I think this is a red herring. The object I am using that is very near tipping point, has partly rounded bottom, and always has some small wobble on release. This implies maximizing sliding distance without tipping is more related to minimizing the initial wobble than anything else. The faster the release, the less able I am to to minimize the initial wobble (and the longer there is for chaotic effects to magnify the wobble).

Homework Helper
Gold Member
Can anyone spot my error?
I have not attempted to replicate your derivation yet, but can you justify choosing units so that ##g/A=1##? By dropping ##A## are you not losing the information about the height-to-base ratio which is important in stabilizing sliding before tipping when ##w>>h##? I would write ##A=w \sqrt{1+\eta^2}~~~(\eta \equiv h/w) ## and then choose time and distance units so that ##g/w=1##. That will provide an additional dimensionless parameter ##\eta## to adjust so that the sign of ##\ddot \phi## might shift one way or the other.

Homework Helper
Gold Member
2022 Award
can you justify choosing units so that g/A=1?
Choose the unit of distance such that A=1, then choose the unit of time such that, with that unit of distance, g=1.

Tibriel
Took a video with my phone but it keeps telling me the file is too large when I try to upload it.

Homework Helper
Gold Member
Choose the unit of distance such that A=1, then choose the unit of time such that, with that unit of distance, g=1.
I understand that. My point in #24 is that there are two distances given, height ##h## and width ##w##. Their ratio is important to the tipping condition. That ratio is not preserved when you set ##g/A = 1##. I think it is more correct first to consider that ##1/A = 1/(w\sqrt{1+\eta^2})## where ##\eta=h/w##. Then choose units such that ##g/A = g/(w\sqrt{1+\eta^2}) \rightarrow~1/\sqrt{1+\eta^2}.## Of course, one could just as well have chosen the height as the unit of length instead of the width. By assuming that ##g/A ~\rightarrow~1##, you are asserting that the box has equal height and width.

I replicated your derivation and I did not pick up any errors, but I have a question. Why did you not consider the horizontal equation of motion?
One starts from ##z=A \cos(\phi)~;~~\dot z=-A \dot \phi \sin(\phi)~;~~~~\ddot z=-A \ddot \phi \sin(\phi)-A \dot \phi^2 \cos(\phi).##
Then ##m\ddot z=-\mu N=-\mu(mg+m\ddot y)##
Putting in the expression for ##\ddot y## from #18 and for ##\ddot z## from above, I get with some algebra,
##\ddot \phi \sin(\phi)(1-\mu)+\dot \phi^2 \cos(\phi)(1+\mu)=\mu g/A##
The problem is that this equation derived from the horizontal equation of motion does not give consistent results with the one derived from in #18 from the vertical equation of motion for a specific value of ##\phi##. Is something over determined?

Homework Helper
Gold Member
2022 Award
Their ratio is important to the tipping condition. That ratio is not preserved when you set g/A = 1.
I don't see that.
A is the distance from mass centre to contact point, φ is the angle that line makes to the horizontal at some arbitrary time. With φ at some value, you no longer care about h and w. You just have a body with its mass centre at a known position in relation to its contact point and a known moment of inertia in relation to A.
You can change the h:w ratio, keeping A fixed, and then rotate the body so as to restore φ to its prior value. The result is the same equation even though the h:w ratio has changed.
Anyway, you can simply observe that in my equation, if it is correct, g, A and (implicitly) t are the only dimensioned entities. So time and distance are the only two dimensions. This gives two degrees of freedom in scaling. It is therefore possible to scale so as to fix two constant entities to have the value 1.
Why did you not consider the horizontal equation of motion?
Because it is irrelevant to the tipping process. So long as it is moving (sliding) it does not affect any of the forces.
If the tipping equation is solved it will be possible to find the friction as a function of time and deduce the horizontal motion.

Homework Helper
Gold Member
2022 Award
One starts from ##z=A \cos(\phi)~;~~\dot z=-A \dot \phi \sin(\phi)~;~~~~\ddot z=-A \ddot \phi \sin(\phi)-A \dot \phi^2 \cos(\phi).##
Then ##m\ddot z=-\mu N=-\mu(mg+m\ddot y)##
How are you defining z here? ##z=A\cos(\phi)## implies you are using it the way I did, but then you should have ##m\ddot x=-\mu N##. x is not z.

Homework Helper
Gold Member
2022 Award
I replicated your derivation and I did not pick up any errors
Thanks for checking it.
I now think there is no contradiction. The "reasonable conditions" I tried in post #18 might not lead to tipping. Indeed, the conclusion is that tipping will only start/continue if ##\cos(\alpha+2\phi)>\cos(\alpha)/3##, where ##\mu=\tan(\alpha)##.

Homework Helper
Gold Member
How are you defining z here? ##z=A\cos(\phi)## implies you are using it the way I did, but then you should have ##m\ddot x=-\mu N##. x is not z.
Yes, I see that now. I misunderstood the meaning of ##z##. I also understand the re-scaling. I now have to mull over the condition you posted in #30. Thanks.

Homework Helper
Gold Member
Indeed, the conclusion is that tipping will only start/continue if ##\cos(\alpha+2\phi)>\cos(\alpha)/3##, where ##\mu=\tan(\alpha)##.
I do not understand this inequality and how tipping will start/continue if it is satisfied. For ##\phi >\pi/4## the left side is certainly negative while the right side is positive and this will still be so as ##\phi## increases past that point. Isn't ##A## vertical and hence ##\phi=\pi/2## when the CM is at its highest point? What am I missing/misunderstanding?

Homework Helper
Gold Member
2022 Award
I do not understand this inequality and how tipping will start/continue if it is satisfied. For ##\phi >\pi/4## the left side is certainly negative while the right side is positive and this will still be so as ##\phi## increases past that point. Isn't ##A## vertical and hence ##\phi=\pi/2## when the CM is at its highest point? What am I missing/misunderstanding?
Ahem.. correcting my algebra..

For the condition for starting/continuing to tip we can take ##\dot\phi=0## and ##\ddot\phi>0##.
Will also assume φ≤π/2, since the development in post 18 might be invalid else.
From the equation in post #18:
##(\frac 13-(\mu\sin(\phi)-\cos(\phi))\cos(\phi))(\mu\sin(\phi)-\cos(\phi))>0##
Writing ##\mu=\tan(\alpha)## we can break this into two cases.
1. ##\mu\sin(\phi)>\cos(\phi)## (i.e. φ+α>π/2)
##\frac 13>(\tan(\alpha)\sin(\phi)-\cos(\phi))\cos(\phi)##
##\frac 23>\tan(\alpha)\sin(2\phi)-1-\cos(2\phi)##
##\frac 53>\tan(\alpha)\sin(2\phi)-\cos(2\phi)##
##5\cos(\alpha)>3(\sin(\alpha)\sin(2\phi)-\cos(\alpha)\cos(2\phi))##
##5\cos(\alpha)+3\cos(\alpha+2\phi)>0##
Looks weird, but..
2. φ+α<π/2
As above but with all inequalities reversed. I'd say the conclusion for this case is it can't happen.

Still looks mighty strange...

arydberg
I would approach this from an energy viewpoint. First find the center of gravity and weight of the box. Then draw a segment of a circle that the center of gravity followers when the box tips over. Then find the height that the center of gravity has to rise so a tip over is possible. find the energy of the box necessary for a tip over to the energy due to the velocity. Compute the velocity using e = 1/2 m v ^2