A cube tipping over on an inclined plane

[ChessEnthusiast]

What's the minimal angle of a ramp such that a cube with side 10 cm placed on it will tip over? The coefficient of friction between the cube and the ramp is 0.6.

I have given it some thought and I have come to certain conclusions:

I think that the cube will tip over once the force of gravity has gone outside its base - the critical angle is plotted on my diagram.

The angle between the vertical component of the force of gravity and the force itself is 45 degrees, because the line containing the force vector also contains the diagonal of the square. The vertical component is parallel to the side of the cube, therefore, the angle must be 45 degrees.

This implies that the angle of inclination is also 45 degrees (triangle similiarity)

Does it mean that the minimal angle for a cube to collapse is always 45 degrees, regardless of the size of the cube and the coefficient of friction?

 

[Dale]

I think that the cube will tip over once the force of gravity has gone outside its base - the critical angle is plotted on my diagram.

This is not correct. Consider a frictionless ramp. It will slide rather than tip at any angle. And a high friction ramp will tip earlier. So the coefficient of friction must be a factor.

That means that your tipping condition is incorrect

 

[CWatters]

At what angle will it slide?

 

[ChessEnthusiast]

@CWatters
When $\mu = \tan(\theta)$ so that's around 26 degrees.

I have thought about this problem a little bit more and I've come up with another approach.

The cube will tip once it gets a non-zero torque in the clockwise direction with respect to the axis I marked.

There are several forces that contribute to the torque:

1. Normal force and the vertical component of the force of gravity - they act at the same perpendicular distance, have the same magnitude and act in opposite directions - their torque cancels out.

2. The horizontal component of the force of gravity - its torque is
$$\tau = mg \sin(\theta) \cdot \frac{a}{2} $$

3. The force of friction, but its torque appears to be zero.

Thus, I must have confused something, because my reasoning implies that this object will always tip.
Could somebody give me a hand?

 

[A.T.]

The cube will tip once it gets a non-zero torque in the clockwise direction with respect to the axis I marked.

Use the center of mass instead.

 

[ChessEnthusiast]

Why doesn't the one I chose work?

 

[Dale]

Normal force and the vertical component of the force of gravity - they act at the same perpendicular distance, have the same magnitude and act in opposite directions - their torque cancels out.

Check your assumption about where the normal force acts. Is it a reasonable assumption for this scenario?

 

[ChessEnthusiast]

@Dale
Intuition tells me that the normal force should be "shifted" to the right, but I am unable to determine the exact distance.

Will the normal force, right before tipping, act directly on the point I chose as the axis?

 

[Dale]

@Dale
Intuition tells me that the normal force should be "shifted" to the right, but I am unable to determine the exact distance.

Yes, that is correct. The distance changes depending on the slope, and it is the distance required to have 0 torque.

 

[A.T.]

Why doesn't the one I chose work?

If you use a reference point other than the center of mass, then a torque imbalance around that point doesn't necessarily imply that the the block starts rotating. A purely linear acceleration can also represent a change in angular momentum around some reference point.

 

[ChessEnthusiast]

So, since - in the critical position - the torque applied by the normal force is 0, there are still three forces that apply a nonzero torque
1. Vertical component of the force of gravity
2. Horizontal component of the force of gravity
3. Friction

How do I determine the torque of the friction force, based on my drawing?

 

[jbriggs444]

How do I determine the torque of the friction force, based on my drawing?

The torque from the friction force would depend on the magnitude and direction of the friction force and on the displacement of its point of application from your chosen axis of rotation.

Are you having problems computing the magnitude of the frictional force?
Are you having problems computing the direction of the frictional force?
Are you having problems computing the point of application of the frictional force?

 

[A.T.]

So, since - in the critical position - the torque applied by the normal force is 0, there are still three forces that apply a nonzero torque
1. Vertical component of the force of gravity
2. Horizontal component of the force of gravity
3. Friction

Gravity doesn't apply any torque around the center of mass.

 

[ChessEnthusiast]

I'm not sure where the frictional force is applied. It seems to me that it should be the rightmost point of the cube in contact with the ramp, but I'm not sure about it.

 

[jbriggs444]

since - in the critical position - the torque applied by the normal force is 0

Missed this. What axis of rotation are you using that makes you think the torque applied by the normal force is zero?

 

[jbriggs444]

I'm not sure where the frictional force is applied. It seems to me that it should be the rightmost point of the cube in contact with the ramp, but I'm not sure about it.

Given the direction of the frictional force, does it matter?

 

[ChessEnthusiast]

I was thinking about the rightmost point in contact with the ground, but now I know it was not a good choice.

No, the torque generated by the force of friction will equal zero, because it's parallel to the ground.

 

[jbriggs444]

No, the torque generated by the force of friction will equal zero, because it's parallel to the ground.

Yes. More generally the possible points of application are lined up in a direction parallel to the frictional force. It does not matter where the frictional force is applied along that line. The resulting torque is the same regardless.

 

[ChessEnthusiast]

So now, as I try to compute the net torque with respect to the center of mass, I end up with only one contributing force, the normal force. The force of friction is parallel to the ground and the force of gravity acts at a radius of 0, and so generates no torque.

The torque generated by the normal force is
$$\tau = \frac{a} {2}mg \cos(\theta) $$ counterclockwise
But this torque is always positive (90 degree angle makes little sense), therefore, how am I able to determine the critical angle?
I must be missing something.

 

[jbriggs444]

The force of friction is parallel to the ground

But the center of mass is not on the ground.

 

[A.T.]

So now, as I try to compute the net torque with respect to the center of mass, I end up with only one contributing force, the normal force. The force of friction is parallel to the ground and the force of gravity acts at a radius of 0, and so generates no torque.

The torque generated by the normal force is
$$\tau = \frac{a} {2}mg \cos(\theta) $$ counterclockwise
But this torque is always positive (90 degree angle makes little sense), therefore, how am I able to determine the critical angle?
I must be missing something.

With the center of mass on the ground it will never tip over. That's the point of having a low center of mass.

 

[Dale]

I'm not sure where the frictional force is applied. It seems to me that it should be the rightmost point of the cube in contact with the ramp, but I'm not sure about it.

Actually, it doesn’t matter where it is applied. All points on the bottom surface have the same perpendicular distance from the center of mass. So the frictional torque is the same regardless.

 

[ChessEnthusiast]

So, the perpendicular distance from the center of mass to the bottom surface is $a/2$.
Both, the normal force and friction act on the perpendicular arm of $a/2$.

Letting clockwise be the positive direction, the torque of the normal force is
$$\tau_N = \frac{a}{2}mg \cos(\theta)$$

Ant the torque generated by the force of friction is
$$\tau_F = -\frac{a}{2} \mu mg \cos(\theta)$$

And we have reached the critical point once these torques equal?

The only case when they do equal is when $\mu = 1$ which is not what I've been expecting..