A cube tipping over on an inclined plane

In summary: 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.Can you help me with an example?Can you help me with an example?Yes, I can provide an example. Consider a cube that is sitting on a frictionless ramp. If the cube starts to slide down the ramp, the normal force (pointing towards the bottom of the ramp) and the vertical force of gravity (pointing towards the cube) will both apply a torque to the cube. However, the horizontal force of gravity (pointing towards the center of the ramp) will not apply a torque, because it is always pointing towards the
  • #1
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?
 

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  • #2
ChessEnthusiast said:
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
 
  • #3
At what angle will it slide?
 
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  • #4
@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?
 

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  • #5
ChessEnthusiast said:
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.
 
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  • #6
Why doesn't the one I chose work?
 
  • #7
ChessEnthusiast said:
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?
 
  • #8
@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?
 
  • #9
ChessEnthusiast said:
@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.
 
  • #10
ChessEnthusiast said:
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.
 
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  • #11
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?
 
  • #12
ChessEnthusiast said:
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?
 
  • #13
ChessEnthusiast said:
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.
 
  • #14
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.
 
  • #15
ChessEnthusiast said:
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?
 
  • #16
ChessEnthusiast said:
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?
 
  • #17
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.
 
  • #18
ChessEnthusiast said:
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.
 
  • #19
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.
 
  • #20
ChessEnthusiast said:
The force of friction is parallel to the ground
But the center of mass is not on the ground.
 
  • #21
ChessEnthusiast said:
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.
 
  • #22
ChessEnthusiast said:
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.
 
  • #23
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..
 

1. How does the angle of the inclined plane affect the likelihood of a cube tipping over?

The steeper the angle of the inclined plane, the more likely the cube is to tip over. This is because gravity has a greater force pulling the cube downward, making it more difficult for the cube to maintain its balance.

2. Can the shape or size of the cube impact its stability on an inclined plane?

Yes, the shape and size of the cube can affect its stability on an inclined plane. A cube with a wider base will have a lower center of gravity and be more stable, while a taller or irregularly shaped cube may be more prone to tipping over.

3. What role does friction play in a cube tipping over on an inclined plane?

Friction can either help or hinder a cube's stability on an inclined plane. If there is enough friction between the cube and the inclined plane, it can prevent the cube from sliding and tipping over. However, if there is not enough friction, the cube may slide and lose its balance more easily.

4. Is there a specific angle at which a cube will always tip over on an inclined plane?

No, there is not a specific angle at which a cube will always tip over on an inclined plane. The tipping point depends on various factors such as the shape, size, and surface of the cube, as well as the angle and smoothness of the inclined plane.

5. How can the concept of torque be applied to a cube tipping over on an inclined plane?

Torque, or the rotational force that causes an object to turn, plays a significant role in a cube tipping over on an inclined plane. When a cube is on an inclined plane, the force of gravity acts on it at an angle, creating a torque that can cause the cube to tip over if it is not balanced properly.

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