ΜStatic = μDynamic: Easier to push a moving object?

[chudd88]

I was running into some weird behavior in a game (Unity/PhysX), and it got me to wondering whether the behavior was reasonably accurate, or whether it was a bug in the physics simulation. The issue related to friction, in a potentially unrealistic case where Static and Dynamic friction are equal. I realize PhysX isn't an accurate simulation, but in this case, I was curious about whether it's doing the right thing, or if this is incorrect.

First, the "weird" thing I'm seeing. My setup is a world where in addition to gravity applying -Y force, there is also a constant -Z force ("Wind") pushing a box into a wall. The wall is frictionless, but the ground is not. The box ("player") can move in four directions by applying force. Here's a top-down view of this world:

If the "Wind" force is equal to the "Player's" movement force, then walking into the wind results in no movement at all, and the player remains held against the wall. If the "Wind" force is significantly less than the player's movement force, the player can overcome the Wind, and walk forward. But if the Player's movement force is just slightly higher than the Wind force, odd things happen. Let's say the Movement force is 105, and the Wind for is 100. In that case, the player won't actually move if they try to move into the wind. My assumption is that the small amount of additional force the player can exert compared to the wind is not enough to overcome the frictional force.

But here's the weird part. If I start by making the player slide left/right, and only then do I try to move into the wind, I can make some forward progress. The image above shows this in action. I'm always applying the same forward force when trying to move forward. But while moving left/right, I move forward more easily than trying to move forward from a dead stop. This would make sense to me if Static and Dynamic friction were different, but in this case I have them set to be identical.

The specific behavior is that while the player is moving even a little but left/right, he can move forward. As soon as he stops sliding left/right, he is no longer able to make any more forward progress.

I was trying to reason through why this might be reasonable. My thought was that friction can only exert so much total force, and so by moving left/right, it somehow lowers the amount of frictional force that can be applied against my forward movement. Does that make sense? Or is this simulation behavior just wrong?

 

[jedishrfu]

This video describes the static vs dynamic frictional forces:

https://www.khanacademy.org/science...tion-ap/v/static-and-kinetic-friction-example

You might watch it first to see if what you have compares well to it.

Typically static friction is higher than dynamic friction. This means that as you push something first with a small force (pushing force < static frictional force) nothing will the box won't budge until you exceed the static frictional force (pushing force > static frictional force) and thereafter you must overcome the kinetic frictional force (pushing force > kinetic frictional force).

It seems reasonable if you were on a bike and the wind is pushing on you so strongly that you can't head into it. However you can ride across the wind, build up some momentum and then turn and attack the wind again making some progress.

 

[chudd88]

The video makes complete sense. The major difference (as I see it) between the video and the case I'm describing is the additional of the extra dimension. It makes sense to me that if I'm heading into a strong headwind, I could make some initial progress if I started with some momentum in the direction of the wind. But it doesn't make sense to me that moving perpendicular to the direction of the wind would help me at all in trying to make forward progress into the direction of the wind. That, however, appears to be happening in my simulation.

So reframing the question, assuming Static and Kinetic friction are equal (or very close), would inertia perpendicular to the direction of the wind help me at all in making progress into the wind? I assume the answer is no, and that this simulation result is inaccurate. But that's what I'd like to understand better.

Putting it another way, let's image a concrete block sitting on the ground. It's pretty difficult to move, and I can't move it at all if I push it North; the friction is just too great. Now let's say someone comes by and gives the block a strong push to the East, and the block starts sliding. I run along side it, matching its speed, and I try pushing the block North again while it's in motion. Assuming Static Friction equals Kinetic in this case, could I now possibly be able to get the block moving North? Or would the frictional force resisting northward motion be the same as it was before?

The object can only generate so much frictional force. It kind of seems like it has a friction budget. If the object is moving quickly East, a lot of its friction budget is "used up" resisting its Eastward motion. Does that effectively leave less force to resist its Northward motion? If I really launch the object East, so it's going thousands of meters per second (and it somehow stays nicely on the ground the whole time), does it generate the same amount of Northward friction when I push it as it does when the object is moving only very slowly Eastward?

 

[jedishrfu]

Sailboats can make progress going into the wind so this may be an example of that effect.

https://en.wikipedia.org/wiki/Tacking_(sailing)

 

[jbriggs444]

Let's say the Movement force is 105, and the Wind for is 100.

Wind is 100 in what direction? Always opposed to the current wind-relative velocity vector? How does wind force scale with the object's wind-relative velocity?

[In the model, not in reality -- we are trying to explain the model, not explain reality]

 

[chudd88]

I've probably introduced unnecessary complexity by calling it "Wind", as I didn't intend to introduce issues of wind resistance or anything like that. Instead of wind, call it an additional constant gravitational/magnetic force, which doesn't vary depending on the speed/position of the player. In the model, it's just arbitrary constant force being applied.

So, the player is able to exert 105N of force along the positive-Z axis (forward), while this strong constant force is pushing at 100N in the opposite direction (negative Z-axis, or backwards). That 5N of net force isn't enough to overcome friction, so the player doesn't move at all. But if some other force gets the player moving perpendicular to the Z-axis, the same 105N of force allows the player to move forward a bit, even though Static and Kinetic coefficients of friction are equal. So I'm trying to understand if that's reasonable. It would be reasonable if the kinetic coefficient of friction was less than the static coefficient, but that's not the case in this simulation.

 

[Tom.G]

1) Are you saying that while moving perpendicular to the Z axis you can instantaneously move, and continue moving, parallel to the Z axis?
-- OR --
2) Are you saying that while moving perpendicular to the Z axis you can turn and continue moving at some angle other than 0° to the Z axis?

• If 1) above, it is one of:
  • You do not understand the program settings (perhaps a documentation error)
  • Someone forgot that to accelerate from a stop, there is more traction force needed than is needed to maintain a constant speed
  • There is a bug in the program.
• If 2) above, it is called 'tacking.'

 

[A.T.]

So, the player is able to exert 105N of force along the positive-Z axis (forward), while this strong constant force is pushing at 100N in the opposite direction (negative Z-axis, or backwards). That 5N of net force isn't enough to overcome friction, so the player doesn't move at all. But if some other force gets the player moving perpendicular to the Z-axis, the same 105N of force allows the player to move forward a bit,...

There are two different factors here:

1) The propulsive force can be used to gain sideways momentum while the forces along Z from the wind and wall cancel. That stored momentum can be used to go along positive Z against a bit, while slowing down. Keep in mind: The net force determines acceleration not velocity.

2) While moving sideways, the friction has no Z component, so the net Z force can become positive and you can accelerate along Z for a while, until the friction's Z component becomes too large. This is why it requires less pull force to pull out a plug, when also turning it.

And forget all the comments about tacking. You have just confused people by calling the constant force "wind".

 

[chudd88]

You've said two very interesting things that both seem non-intuitive to me, but if true they would explain the behavior I'm seeing.

On your first point, you've said that sideways momentum can be used to go make some forward progress along the Z-axis. That's not the way I picture momentum working. If an object is moving East, it would take additional force to redirect it to start moving North. In the real world I can imagine a fast moving object "turning" north, and then moving northwards for a while due to the inertia. But it seems it would take a lot of force to rotate the object to get it to turn. And if you were going fast enough, and suddenly made a 90' turn towards north, you'd probably keep sliding East (like a car trying to make a high-speed turn too sharply). So, the idea of using X-axis momentum to move along the z-axis doesn't make any sense to me. Could you point me to what principle that would relate to? In my simulation, there's no turning or rotation.

On your second point, the plug thing is interesting, but also non-intuitive. As for the plug example, does that example still work when Static and Kinetic friction are equal? You've said the friction has no z-component because the object isn't moving along the z-axis, just along the x-axis. Does this mean that the faster the object moves along the x-axis, the less frictional force can be generated along the z-axis? So if an object were moving extremely fast along the x-axis, it could easily be pushed along the z-axis, as there isn't enough frictional force left? It seems we'rd to think of friction like that, where movement along one axis reduces the amount of friction that can be applied along another axis. Is that really how friction works?

 

[chudd88]

I think I can reduce my question down to two very simple related questions, which are hopefully simple equations that can be solved. Both questions have the same initial conditions: A 1Kg block is resting on a smooth, flat surface. The block and the surface both have a kinetic coefficient of friction of 0.5. The block is given an initial push along the positive X axis, causing it to slide in that direction. Given these initial conditions.

1. Assume that after the initial push, the block is sliding in the +X direction at 1 m/s. How much force is required to move the block along the +Z axis, just enough to get it barely moving?
2. Same question, but now assuming the initial push got the block moving 100 m/s instead of 1 m/s. Now how much force is required to move the block along the +Z axis?

It seems, from observation, that the two questions will have different answers, which doesn't yet make sense to me.

The frictional force being applied here should be 4.9N. But I get confused about how that 4.9N is being applied when A) the object is just sliding under its own inertia, and B) The object is both sliding in one direction, and being pushed in a perpendicular direction. In the first case, if it were sliding at 100 m/s along +X axis, then it should have 4.9N of force pushing it in the -X direction. But if, at the same time that the block is sliding, I push it with +Z force, then what does that do to the friction vector? If the frictional force cannot exceed 4.9N, does that mean that pushing along +Z will actually cause it to slide further along the x-axis further, because not all of the 4.9N is going towards slowing its x-axis momentum?

This raises yet another question, or possible example: Imagine a hockey puck on ice, with an initial velocity of 10 m/s in the +X direction. We could compute how long it would take for friction to slow down the puck to a stop. While it's sliding on +X, all of the frictional force is directed at -X. Let's say, for example, the puck slides 50 meters before it completely stops. Now, do the same thing again, but blow some wind at the puck in the +Z direction, the entire time it travels. Now some of the frictional force will be "used up" trying to keep the puck from moving along the Z axis, which means that not as much of the friction is pushing in -X, implying that the puck will travel further along the X axis in this example. That sounds pretty weird to me. Is that actually what would happen?

 

[jbriggs444]

• Assume that after the initial push, the block is sliding in the +X direction at 1 m/s. How much force is required to move the block along the +Z axis, just enough to get it barely moving?

• Same question, but now assuming the initial push got the block moving 100 m/s instead of 1 m/s. Now how much force is required to move the block along the +Z axis?

Infinitesimal in both cases. You need to ask something more like "how much sideways force to maintain 0.1 m/s sideways motion at that downhill speed"?

Or "how much sideways force to maintain a 1 degree cross-slope deflection at that downhill speed"?

Both questions can be answered if you draw a free body diagram, identify the frictional force and split it into down-slope and cross-slope components.

 

[A.T.]

In my simulation, there's no turning or rotation.

There is obviously turning because the object changes direction. Whether the object also rotates while turning is irrelevant.

It seems we'rd to think of friction like that, where movement along one axis reduces the amount of friction that can be applied along another axis. Is that really how friction works?

Friction is complex and can modeled in different ways. But in the simple Coulomb model this is the case, because the direction of friction is velocity dependent, but its magnitude isn't.

Since you have the simulation, why don't you just visualize all the force vectors or output and plot them, to see what is going on.

 

[chudd88]

Thanks for all the help on this. I believe I understand the behavior now mathematically, and it's starting to sink in intuitively as well. I think the behavior I initially showed is reasonably accurate, and represents how friction actually works. It still "feels" a bit odd, but I think I can get past it. Thanks again for the help.

 

[Hiero]

This is why it requires less pull force to pull out a plug, when also turning it.

Nice insight! Never thought about that before, thanks!

I’ll just reiterate what’s been said: @chudd88 The key idea here is that the direction of friction depends on the direction of velocity (specifically, they are opposites).