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I've seen Consistent method for finding direction of static friction which was the closest question I could find, but I'd like something better. I need a method which :

1) Intuitively makes sense

2) Works for circular motion

3) Works for rolling without slipping

4) Works for most other simple cases

I'd like the method to work out the direction of friction in these cases :

1) A box is placed on a flat surface and pushed to the right.

2) A mass placed on a turntable rotates without slipping with respect to the turntable.

3) A mass is placed on a flat surface and a string is attached to it. It is then rotated so that it moves in a horizontal circle.

4) A wheel, driven by some force F, is moving to the right on a flat surface without slipping.

5) A wheel is rotating down an inclined plane without slipping.

6) A wheel is rotating up an inclined plane without slipping.

7) A car enters a banked curve with a speed greater than the required speed for circular motion with no friction (what I mean by 'no friction' is that even if there were no friction, the car would perform circular motion about the center as the speed is ideal).

8) A car enters a banked curve with a speed less than the required speed for circular motion with no friction.

I'm very weak when it comes to determining this direction if there is rolling without slipping, rolling with slipping, or circular motion. For one, I don't think I'm even clear on what slipping means.. I've always thought that I can think of static friction as something which opposes the relative motion between two bodies, but in these cases, it generally doesn't work.

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  • $\begingroup$ I just don't see how the answers to this question would be any different than the answer to the previous one. In other words I don't see an argument that this is a different question. In particular the answer to the previous question and the a(first) answer here both make exactly the same recommendation: that you figure out what motion would occur in the absence of the friction. $\endgroup$ – dmckee Oct 16 '17 at 23:25
  • $\begingroup$ Friction must have negative power, so $\vec{F} \cdot \Delta \vec{v} <0$ $\endgroup$ – ja72 Oct 16 '17 at 23:59
  • $\begingroup$ @dmckee That approach doesn't work in some of these cases, for e.g, in 3, if you somehow suddenly made friction 0 (pour oil on the turntable or whatever), then the mass would fly off on a tangent. Friction should oppose this motion, so it should act opposite the motion of this mass, but also tangential to the circle. What's even more confusing is, say in 7, the car is clearly moving relative to the banked curve, yet friction is down the incline rather than opposite the velocity of the car. How does it stop relative motion then? $\endgroup$ – John Oct 17 '17 at 1:15
  • $\begingroup$ I guess a better way to put it would be that no matter what you're doing, static friction wants you to keep doing it, i.e rather than saying it wants to stop relative motion, it should try to maintain the same relative motion. $\endgroup$ – John Oct 17 '17 at 1:16
  • $\begingroup$ John, it does work in those cases. I presume that you're just starting out here which probably means that you don't have the infrastructure to set up those problem clearly yet. But it is absolutely not true that static friction tries to maintain current behavior in general. That's a reasonable call for situation that remain as they are, but fails badly as soon as things get dynamic. $\endgroup$ – dmckee Oct 17 '17 at 1:18
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I've always thought that I can think of static friction as something which opposes the relative motion between two bodies

This is actually mostly correct. I would just change it to locally opposing the relative motion between two surfaces. Remember that (dynamic) friction is dissipative, meaning that it always acts to reduce the amount of (non-thermal) energy a system has by converting it into heat. That means if a system has kinetic energy (which may be linear or rotational), friction wants to reduce that kinetic energy, and will apply forces in the direction necessary to do this.

Static friction can be a lot trickier, especially for rolling objects. I think one of the most reliable / intuitive approaches here is to imagine "what would happen in this situation if there is slipping?" and then know that static friction (without slipping) acts in the same direction as dynamic friction would if there were slipping. For example:

  1. A car drives up a hill without slipping. In what direction does the static friction act on the tires? Well, if the hill were very icy and there were slipping, the car's wheels would spin around trying to move the car forward but would slip. The friction, trying to locally oppose the relative motion of the two surfaces (the car tire against the ice) will push upward on the tire and downward on the ice (equal & opposite forces). The direction is the same for the original case for static friction without slipping.
  2. A bowling ball rolls down a lane without slipping. In what direction does the static friction act on the ball? If the lane were just recently oiled and instead the bowling ball were sliding down the lane and not rolling (i.e. it was slipping), then dynamic friction would try to reduce its kinetic energy by pushing it backwards. So, friction pushes backwards on the bottom surface of the bowling ball. In the case where it rolls without slipping, static friction still points in the same direction that dynamic friction would: backwards.
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  • $\begingroup$ Thank you for the answer! A couple of questions : 1)How would this method work in example 3? If the turntable stopped rotating, then the mass would slip relative to the turntable, and so would move tangentially. Kinetic friction would try to stop this movement, so it would be in the opposite direction tangentially. 2) If the car drove downhill without slipping, won't this give us the same direction as in the case the car was driving uphill? 3) In the case of the bowling ball, won't there be no friction because if there were, the bowling ball would decelerate, causing slipping? Thanks! $\endgroup$ – John Oct 17 '17 at 1:05
  • $\begingroup$ 1) I'm not sure I fully understand the setup of example 3 from your description. If a mass is initially stuck to a rotating turntable without slipping (not sure where the string comes in?) then static friction is balancing centrifugal acceleration (without friction it would "slip" off the turntable). Your statement for the direction of (kinetic or static) friction if the turntable stopped is correct. 2) It depends: imagine the car is in neutral and being pulled down by gravity; now imagine the engine is pushing the car and it is accelerating downhill. 3) Can you explain your thinking here? $\endgroup$ – Sean49 Oct 17 '17 at 1:12
  • $\begingroup$ Ahh, sorry. I meant 2. For 3, by a lane, you mean a horizontal surface, right? If so, then the presence of static friction means the bowling ball will decelerate, so doesn't that imply it slips? But it's also true that unless there IS static friction, the bowling ball won't rotate since there's nothing to provide the torque.. Sorry, it's just that I'm not clear about slipping in general. $\endgroup$ – John Oct 17 '17 at 1:31
  • $\begingroup$ Static friction means there is no relative motion between the two contacting surfaces. That's it. If a bowling ball is rolling down a (horizontal) surface without slipping, it means the part of the ball touching the ground is instantaneously at rest with respect to the ground, but the ball overall can have forward momentum (just the part of the ball touching the ground does not move relative to the ground). If the ball slides down a slippery lane and is not rolling, it means the part of the ball touching the ground slides across it (relative motion of contacting surfaces), hence slipping. $\endgroup$ – Sean49 Oct 17 '17 at 2:47
  • $\begingroup$ Thanks. But in the case of a car going around a banked curve, or 3 ( mass attached to a string and rotated around a circle on a horizontal surface ), there is relative motion between the object and the surface. In both cases, static friction does not act tangentially i.e the direction which opposes this relative motion. $\endgroup$ – John Oct 17 '17 at 3:01

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