Consistent method for finding direction of static friction

Question

I am having trouble coming up with a consistent method of determining the direction of static friction.

So far the best I have come up with is: it should oppose the relative acceleration the contact point would have parallel to the surface of contact in its absence.

This seems to work up to a point:

• A block sits on a rug. The rug is pulled to the left. The relative acceleration of the block to the rug is to the right. So static friction must act (in the frame of the rug) to the left.

• A cyclist makes a turn to the right and leans to the right. The torque that his weight puts about the contact point will make it accelerate to the left relative to the ground in the absence of friction. So the ground sees static friction pointing to the right.

• A coin rotates with a turntable. In the absence of static friction, the table sees a centrifugal force on the coin that will make it accelerate away radially from the centre of rotation. So friction must act radially towards the centre of rotation, in the frame of the table.

But then I run into some difficulties.

• (already solved; see first answer) A cyclist makes a turn to the right. As a result he leans to the right. The resulting torque around the centre of mass causes the point of contact to slide to the left in the absence of static friction, so static friction must act to the right. However, now the cyclist as a whole must make circular motion with static friction acting as the centripetal force. As there is no torque, this means the contact point must move along with the centre of mass and therefore act as something with an acceleration to the right. So according to my understanding of static friction, another layer of it must now be added acting to the left. Why is it not?

• A car enters a banked curve with speed $v$. Suppose the curve is banked at the right angle for the car to make circular motion in the absence of friction. In the absence of any static friction, therefore, the car's acceleration points towards the centre of the banked curve. However, in this case the car's acceleration still has a downward component along the slope. My understanding of the direction of static friction means therefore, when it is added in, it should act up the slope. But in reality no static friction is produced in such a case.

Can someone explain what is wrong with my method of finding the direction of friction? Or if there is nothing wrong, how have I misapplied it in this case to create this paradox?

Answer 1

You've basically got it; you can often analyze constraint forces like this by asking "what would happen if my constraint force disappeared?" For example, if you have a door which is constrained to be on two hinges, the bottom hinge must actually push the door outward, away from the door frame. This is easily seen if you imagine what would happen if that hinge disappeared: the door would pivot about the top hinge, swinging inward towards the frame where the hinge was. The lower hinge must provide the force opposite this which keeps the door upright.

Static friction only happens when the contact surface experiences no relative motion, so it is a constraint force that supplies whatever magnitude and direction is equal and opposite to all other forces acting on that contact surface. (More precisely, you decompose these forces into in-plane and normal components; it balances out the in-plane ones but not the normal ones: and it can only do this up to a certain point $\mu_s F_n$ where $F$ is the normal force.)

Cyclist mystery: this is not going to be so easy to analyze. It'll be much easier if you use a disc or a hoop -- those are gyroscopes; but the formidable bicycle resists easy analysis after people built bicycles which had extra flywheels that didn't make contact with the ground but could instead be rotated opposite the real wheels. If you do this to a hoop then it tends to just fall naturally; if you do this to a bicycle then it tends to follow a curve and stay stable. (Conjecture: these no-gyroscopic-effect bicycles would be really dangerous to pop a wheelie on.)

The "what would happen if the friction disappeared?" question says undisputably that the friction force points to the right: if it disappeared the contact point of the hoop would slip off to the left. In doing so, it helps to provide an extra centripetal force term. The origin of this force term can also be understood by simply standing a book on its edge and tipping it over: the center of mass wants to fall with angular acceleration $\frac{g}{h/2}~\sin\theta$ but this ultimately means that the topmost edge must fall with twice that acceleration, as if twice the gravity were active on it. This extra force must be provided by an internal strain, and that internal strain should therefore also exist on the other side in the center-of-mass frame.

Car mystery: This one is easier. You have designed the track to softly curve so that a certain car will describe a circle even if the track is made of ice: then our "what would happen if friction disappeared?" criterion says nothing would happen, so you have designed the track so that the static friction is 0. The direction of a force with 0 magnitude is totally ambiguous.

As for the paradox, I think this disappears in a corotating reference frame. To make the reference frame corotating, you add a centrifugal and Coriolis force; we're going to keep the car uniformly rotating, so its speed in the corotating frame is 0 and this kills the Coriolis force. You now have a force balance between three forces: the centrifugal force $m \omega^2 r = -m ~v^2/R~\hat x$ pointing "left" in the $-\hat x$ direction, the gravitational force $-m g\hat y$ pointing "down" in the $-\hat y$ direction, and the normal force $F_0 (\hat x~\cos\theta + \hat y~\sin\theta)$ pointing up and right, where $\theta$ is the angle of inclination of the ramp. By choosing $\tan\theta = g / (v^2 / R)$ you find this happy medium where the centrifugal force, projected into the plane of the ramp, perfectly balances the gravitational force, also projected into that plane, no matter what your radius of curvature for the turn $R$ is.