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Let's say an object is sliding on a slope and is the object has a velocity of (0,0,5). The friction would be acting in the opposite direction of motiong, being (0,0,-1).

However, gravity is also affecting the object on the slope. The gravity is doing an impulse of (-1,-1,0) every frame in the direction of the slope tangent. Should the object also experience friction in the direction of (1,1,0)? Or would it only experience friction in the direction of (0,0,-1)?

I find the notion that friction always opposes the direction of motion strange, because this means that when an object is sliding downwards slightly on a slope, but not much, due to friction acting in opposite direction of gravity, when you add an impulse in a perpendicular direction, suddenly the object will be sliding downwards much more because friction is no longer resisting gravity. Is that correct? i.imgur.com/hapvc9b.png

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  • \$\begingroup\$ I think this problem would be much easier to understand if you used a 2d environment rather than a 3d environment. \$\endgroup\$ – Vaillancourt Apr 27 at 19:28
  • \$\begingroup\$ @Vaillancourt I would, but I don't think this case can be simplified to 2D \$\endgroup\$ – xcrypt Apr 27 at 19:31
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    \$\begingroup\$ In your figure, the impulse would cause the object not to be in contact with the surface anymore. No contact => no friction. \$\endgroup\$ – Sacha Apr 28 at 8:24
  • \$\begingroup\$ @Sacha no, it would still have contact. It is a 3D slope and it extends in the Z direction, where the impulse goes. \$\endgroup\$ – xcrypt Apr 28 at 8:25
  • \$\begingroup\$ The impulse goes sideways? (oriented towards or coming from the screen)? In this case you can use the following representation: commons.wikimedia.org/wiki/… \$\endgroup\$ – Sacha Apr 28 at 8:27
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Friction is directly related to velocity.
Gravity and other forces might cause an object to accelerate, changing its velocity, so they indirectly affect friction.
But they do not appear in the friction equation, for example in the simplest case you have:

\$ \vec{F} = k.\vec{v} \$ with \$k\$ a constant depending on the object's and contact surface's properties.

However, in more complex models, the coefficient \$k\$ can also depend on the normal force at the surface, which is itself proportional to the object's weight.

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  • \$\begingroup\$ I find this strange, because this means that when an object is standing still on a slope due to friction acting in opposite direction of gravity, when you add an impulse in a perpendicular direction, suddenly the object will be sliding downwards because friction is no longer resisting gravity. Is that correct? \$\endgroup\$ – xcrypt Apr 27 at 18:59
  • \$\begingroup\$ @xcrypt Keep in mind that the "initial force" required to make an object move from a static state is greater than the force required to keep the object moving. \$\endgroup\$ – Vaillancourt Apr 27 at 19:14
  • \$\begingroup\$ @Vaillancourt Yes, but for the sake of the argument you could say that we can ignore static friction if the object is moving slightly due to the projected gravity being slightly greater than the kinetic friction. It would suddenly start moving downwards much more if it got a perpendicular impulse to the side, because the direction of friction would change according to the model. I just find that really strange. \$\endgroup\$ – xcrypt Apr 27 at 19:20
  • \$\begingroup\$ @xcrypt The object undergoes static friction when at rest and kinetic friction when moving. Static frictions prevents the object from moving and is parallel to the surface of contact. Kinetic friction is colinear with the velocity vector, which means also parallel to the surface of contact. The direction of friction does not change. \$\endgroup\$ – Sacha Apr 27 at 19:34
  • \$\begingroup\$ @Sacha But the impulse I'm talking about would cause the velocity direction to change, which also means the direction of friction would change \$\endgroup\$ – xcrypt Apr 27 at 19:36
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Friction acts in the opposite direction of the force vector. When gravity is the only force acting on the box, friction is in the opposite direction of gravity. Adding an impulse perpendicular to gravity does not negate gravity or the friction acting against it on the plane of the slope, it simply adds another component to the force vector. Add all force vectors together and then subtract the friction.

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  • \$\begingroup\$ Let's take a hockey puck on a level ice rink. When I hit the puck with my stick, I impart a force and friction opposes that force. But after that strike, and the puck coasts across the ice, the only force acting on the puck is straight down, but friction still opposes its velocity, right? So I think there's another factor we need to consider in addition to force here. \$\endgroup\$ – DMGregory Apr 29 at 20:54

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