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I'm wondering what makes a tall box rotate about its edge (the red point on the figure below) on a inclined plane (with or without slipping), when subjected to forces of gravity and friction only:

enter image description here

Both the box and the plane are idealized (no imperfections).

According to the answers to this question about a ball rolling an inclined plane, gravity force plays no role in rotating the ball, so I think that is the same case for the box: the gravity force is resolved into two perpendicular components: one opposite to the normal, and the other parallel to the plane, sliding the box down the plane (without rotating it).

In the case of the rolling ball, the friction force is what actually rolls the ball, by applying a torque about the center of the ball at a distance R (ball radius).

But in the case of the tall box, I see that the axis of rotation is not the center of gravity of the box, but the point of contact in the base of the box (the red point in the image above). The box rotates about this point. The problem is that the friction force has no tangential component with relation to the point of contact: its line of action passes through the point of contact. It would not be able to produce a torque about this point.

I can't see this box rotating about its center of gravity either (as in the case of the ball), because the box has edges which would "penetrate" the plane, as the image below:

enter image description here

The only situation where I can see the box rotating about the edge is if the inclined plane contains imperfections, so the box edge would be resting on this imperfection (the imperfection is depicted in green on the image below). But in this case rotation about the edge would be caused by the component of gravity parallel to the surface and not by the friction force, right?

enter image description here

So, is it correct to conclude that the perfect tall box, starting from rest and with its center of gravity pointing outside of this base, would not rotate at all on a perfect inclined plane (it would just slide)? Is there a scenario where it would rotate? And what would be the conditions for slip/no slip?

Additional thoughts:

As far as I know, in absence of friction, the box would just slide without rotating. I think there should be some minimum value for the friction force to make the box rotate without sliding (keep the corner static). And, if the plane is not able to supply that amount of friction, the box would slide while rotating about the corner. But I'm not able to work out these friction values.

In the case of the ball rolling down the plane, the center of gravity of the ball describes a straight line (parallel to the plane) as the ball accelerates down the plane, and the ball has no edges which would penetrate the plane as the ball rotates around its center. Thus I found it easy to see the ball motion as a combination of a translation of the center of gravity (through a straight line) plus a rotation about its center of mass. So I solve the equations for the angular acceleration of a point at the rim of the circle and for the linear acceleration of the center of gravity, and find the minimum friction force required to keep the ball rolling without slipping. If the surface was able to apply this amount of friction, the ball would make a pure roll movement. Otherwise, it would roll and slide.

But I'm not able to decompose the movement of the box as I did for the ball (translation about COM plus rotation about COM). First, because the center of gravity of the box is not translating in a straight line during the tipping (it is rotating about the corner), and second, because a point on the edge of box is not rotating about the center of gravity (it is rotating about the corner).

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ri_ri
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  • If friction is sufficient to stop the box from slipping and the center of gravity is to the right (and above) the corner, then yes, I'm pretty sure the box will rotate. I'm not sure if this changes at all when the high friction requirement is loosened. – raptortech97 Sep 17 '14 at 00:39
  • You may try [this] (http://physics.stackexchange.com/questions/95234/toppling-of-a-cylinder-on-a-block) – soumyadeep Sep 17 '14 at 04:49
  • How is a "perfect" box/plane different from a frictionless one? If there is no friction, then there is no torque and the box will not tip. – BowlOfRed Sep 17 '14 at 23:05
  • @BowlOfRed, sorry, I didn't say "perfect" meaning "frictionless". I just meant the plane contains no irregularities which would serve as a pivot for the box to rotate about (like on the last image on the question). Is there a better word to mean that? I'm considering there must be friction in order for torque to exist, but I don't understand why friction would cause torque about the bottom-right corner, because the corner is on the line of action of the friction force. – ri_ri Sep 17 '14 at 23:26
  • @raptortech97 Yes, if friction is strong enough to keep the bottom-right corner static, it is a situation similar to tipping a box on a horizontal plane (where the corner is static as well), right? But in the case of a inclined plane, I don't know how to work out the minimum value of friction force required in order to keep the corner static during tipping. – ri_ri Sep 18 '14 at 00:08
  • @ri_ri "But in this case rotation about the edge would be caused by the component of gravity parallel to the surface and not by the friction force, right?" This is not quite right since the rotating force must be perpendicular to the lever arm, which in this case is the line from the green corner to the center of gravity. – Daniel Genin Oct 20 '21 at 13:36

2 Answers2

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The problem with attempting to do the analysis with the forward point of contact on the box when it is sliding is that the box is accelerating. This makes a non-inertial frame and there's more moving parts. Besides the force of gravity on the center of mass, there will be fictitious forces.

First, lets assume friction is zero. If so, we can calculate the acceleration of the box. $$F_{x'} = mg\sin \theta$$ $$a_{x'} = g\sin \theta$$ In an accelerating frame, we get fictitious forces. $$ F_{fict} = -ma_{frame}$$ $$ F_{fict} = -mg\sin \theta$$ The sign indicates the force is opposite the direction of the acceleration. The case where we expect the box to tip most likely is where it is tall and not wide. Let's assume the width is minimal (a rod sliding on its end). If so, the torque from gravity will be the COM at half the height, and leaning forward $\sin \theta$ $$\tau_g = \frac12 hmg \sin \theta$$ Meanwhile the fictitious force is located at half the height and points opposite the acceleration vector, so it has a lever arm of exactly half the height. $$\tau_{fict} = F_{fict} d$$ $$\tau_{fict} = -\frac12 hmg\sin \theta$$

$$ \tau_{net} = \tau_{g} + \tau_{fict}$$ $$ \tau_{net} = \frac12 hmg \sin \theta - \frac12hmg \sin \theta$$ $$ \tau_{net} = 0$$

If you add friction, you will reduce the acceleration of the box, reducing the fictitious force in this frame. When that happens, the net torque would be sufficient to tip a rod, but would depend on the dimensions if a box would tip over.

I didn't do the math, but $\tau_{fict}$ should be pretty simple to calculate, and then $\tau_{g}$ becomes $\frac12h\sin \theta - \frac12w\cos \theta$ (I think)...

Sorry, I wanted to reply to this comment, and I couldn't do it in the comment space.

If we instead take the static plane to be the frame of reference (and a minimal width box), wouldn't we get the same results anyways?

The problem is that in the frame of the plane, the axis you want to consider in the question (the front corner of the box) is accelerating. I don't think that's okay to do. Let's imagine a pendulum being accelerated via a force on the axis. And we let it reach steady state so the angle isn't changing.

pendulum

The solid pendulum has a force accelerating it to the right, and it has gravity pulling it down. The force is applied directly at the axis and contributes no torque. Gravity pulls the weight down, so is applying a torque of $(mg \times L\sin \theta)$. Yet the pendulum does not rotate. Perhaps someone else has more information about this situation. My intuition is that this is simply the wrong technique for an accelerating axis, but I'm not sure what the correct analysis would be.

BowlOfRed
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  • I like your explanation. Let me try to understand the need of analyzing the movement of the box by using the box as the frame of reference instead of the static plane (inertial frame). It was just a means of "moving" the friction force point of action from the point of contact to the center of gravity, so it is able to generate torque around the corner of the box? – ri_ri Sep 20 '14 at 00:11
  • So is the total torque value around the corner is both influenced by both gravity and friction? By looking at the diagram of forces from a inertial frame of reference (this image here), the friction force would not contribute to the torque about the pivot (forward corner), since it has lever arm zero. – ri_ri Sep 20 '14 at 00:38
  • But since friction force contributes to reduce the acceleration of the center of gravity imposed by the component of gravity parallel to the plane $F_{x'}$, in a diagram of forces from a non-inertial frame of reference (this image here) there would be a fictitious force equals $-(F_{x'} + F_{friction})$ acting on the center of gravity, which does to contribute to the total torque around the pivot. Is that right? – ri_ri Sep 20 '14 at 00:39
  • Sorry, I think I was changing the comment to ask the questions above while you tried to reply to it. But the pendulum example you presented seems to be a good one too. In that case, both the axis and the ball are horizontally accelerating due to force $F$. I drew a diagram of forces for the non-inertial frame situated at the ball: it is here. The fictitious force is $-F$ (in red). – ri_ri Sep 20 '14 at 04:07
  • So, is it reasonable to expect the pendulum to achieve steady state when the component of force $-F$ perpendicular to the string $-F \times \cos \theta$ is equal to $mg \times sin\theta$ (so that the net torque around axis is zero)? What do you think? – ri_ri Sep 20 '14 at 04:07
  • You can use other methods to find the forces. Since the weight does not rise, $mg = T\cos \theta$ and $T\sin \theta =-F$. Substitute for T, rearrange and you get your $-F\cos \theta = mg\sin \theta$ – BowlOfRed Sep 20 '14 at 05:32
  • I think I see what you did there: you summed forces $F$ and $mg$ into a total (diagonal) force $T$, which, in order to cause no torque, should be parallel to the string, i.e., its angle should be $\theta$ as well. Thanks! – ri_ri Sep 21 '14 at 15:49
  • I didn't even consider torque, just translation. Since the pendulum mass does not change height, the vertical component of $T$ must equal the weight $mg$. Likewise, since there is no horizontal force to counter, the horizontal component of $T$ must equal the force applied. – BowlOfRed Sep 21 '14 at 17:22
  • That makes total sense. I still have some questions about how rewrite forces onto a non-inertial frame of reference, but I'm gonna write a new question about it. Thanks! – ri_ri Sep 22 '14 at 19:37
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Take a look on the forces diagram in the first picture. Let's ignore the force $N$ for now, it can only contribute to the tipping (due to its orientation), but since it is going to be confined to acting only on the red point as soon as the tipping occurs (if it does), it won't have any influence at all. E.g. it produces zero torque about the red point. Friction is disregarded as stated in the question.

The only remaining force is gravity. Let's say the box has mass $m$, height $h$ and width $w$ and the plane is inclined by angle $\alpha$ (the rightmost angle in the top drawing). Gravity can now be separated into the parallel to the plane part $m g\sin(\alpha)$ and perpendicular to the plane part $m g\cos(\alpha)$.

Their torques about the red point are obtained by multiplying them by the perpendicular distance from the red point to the center of mass (that is where the forces act, since it is gravity we are talking about). The torques are $m g h \sin(\alpha)/2$ in the clockwise direction (due to the part that is parallel to the plane) and $m g w\cos(\alpha)/2$ in the counter-clockwise direction (due to the perpendicular part).

The box will tip if clockwise torque exceeds the counter-clockwise, that is when $\tan(\alpha)>w/h$. This also happens to be the condition that center of mass points outside of the box.

Red Act
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DoctorJAM
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    If the box is sliding, then it is accelerating and this is not an inertial frame. The fictional force present in the frame will oppose the rotation. Otherwise the same analysis would suggest that the same block in free-fall would rotate in the same manner. – BowlOfRed Sep 17 '14 at 06:42
  • I admit my mistake. Going to research this. – DoctorJAM Sep 17 '14 at 22:49
  • I appended some additional thoughts at the end of the question. It may help you to know where I'm struggling. – ri_ri Sep 18 '14 at 00:21
  • @BowlOfRed I am not allowed to comment on your answer due to my low reputation (joined recently), so I will comment here, hope you don't mind. What bothers me is that when solving a very similar problem with a ball (or cylinder) accelerating down the slope, the torque about the point of contact is calculated without taking the fictitious force into the account. This refers to Berkeley Mechanics by Kittel et. al, somewhere around page 145 (not sure since I don't use the english edition). – DoctorJAM Sep 18 '14 at 01:42
  • In that case, is the ball rolling or sliding? If the ball is rolling, then the point of contact is not accelerating and there is no problem calculating the torque about that point directly. – BowlOfRed Sep 18 '14 at 03:43
  • Rolling, and the center of mass is accelerating, so the point of contact has to be as well (the ball doesn't get deformed). – DoctorJAM Sep 18 '14 at 18:12
  • @DoctorJAM, I think that in the case of the ball rolling down the plane, the point of contact has zero velocity, and its acceleration is in the direction of the center of the ball (centripetal only), not parallel to the plane. – ri_ri Sep 19 '14 at 23:45
  • That's right. This question is different because the block is sliding. – BowlOfRed Sep 21 '14 at 18:51
  • I think @DoctorJAM analysis is correct if the green corner of the box is pinned, e.g., static friction exceeds weight component in the direction of the plane. In this case we can think of the corner contact point as a hinge. The box won't be sliding and so its frame of reference remains inertial. Still need to think more about how this squares with the BowOfRed counter argument about unbalanced torque on the free falling block. – Daniel Genin Oct 20 '21 at 13:33
  • @BowlOfRed The free-falling box will not rotate because in its frame of reference there are no forces acting on it. Weight is cancelled out by the pseudo force induced by the accelerating frame of reference. – Daniel Genin Oct 20 '21 at 13:49