A very small cube of mass m is placed on the inside of a funnel rotating about a vertical axis (the central axis of the funnel) at a constant rate of v revolutions per second. The wall of the funnel makes an angle [tex]\theta[/tex] with the horizontal. The coefficient of static friction between cube and funnel is [tex]\mu_s[/tex] and the center of the cube is at a distance r from the axis of rotation. Find the (a) largest and (b) smallest values of v for which the cube will not move with respect to the funnel.

For reference, this is problem 53 chapter 6 of Halliday Resnick Krane vol. 1.

Am I correct in thinking that friction, gravity and normal force are the only forces acting on the box? And that maximum v corresponds to minimum friction force (0?)? And that minimum v corresponds to maximum friction force ([tex]\mu_sN[/tex])?

Spoiler

If it helps, the answers to the above are (a)[tex]\frac{1}{2\pi}\sqrt{\frac{g(\tan(\theta)+\mu_s)}{r(1-\mu_s\tan(\theta))}}[/tex] and (b) [tex]\frac{1}{2\pi}\sqrt{\frac{g(\tan(\theta)-\mu_s)}{r(1+\mu_s\tan(\theta))}}[/tex] (from the answers in the back of my book ). I'm hoping for an explanation of these answers. It might help to notice that the expressions inside the square roots are (a) [tex]\frac{g}{r}\tan(\theta+\tan^{-1}(\mu_s))[/tex] and (b) [tex]\frac{g}{r}\tan(\theta-\tan^{-1}(\mu_s))[/tex]

1. The problem statement, all variables and given/known data

u r somewhat correct, but not fully. consider the tendency of the box movement when the funnel is rotating slowly. the centrifugal force pressing the cube with the wall will be less and hence the friction will also be less and gravity will pull it down for slow enough force. so the friction will act upward for the lower limit of rot speed.
but for a large rot speed the centrifugal force will be large and the component of it along the the incline of funnel may 'beat' the gravity and pull it upward along the incline and friction will be downward in this case, i.e, for higher limit of rot speed the friction is downward.
so it is not the case of max or min friction. it is actually the case of direction of limiting friction along the incline.

Static friction acts against relative motion. That cube could slip down the wall when the rotation is slow: then friction points upwards, parallel to the wall; or it would go up at too fast rotation: then the friction points at the opposite direction.

thinking about it again, for the case of maximum v, im actually imagining the normal force to be 0...
I'm visualizing the funnel spinning too fast and the box flying off the plane entirely, which would correspond to N=0. For the case of minimum v, yes thats what I meant, maximal friction pointing upwards.

My justification: the centripetal force is provided by the normal and friction forces (gravity is perpendicular to the radius of rotation). Frictional force points outwardly and normal force inwardly, at all times, so Fc=Nx-Ffx (centripetal force is the normal force along the radial direction minus the friction along that direction). At a certain point, the centripetal force will have to become so great (meaning v is great, and N will have to be great to provide the force) that the y component of the normal force is greater than Fg, and thus the block accelerates off the plan into free-fall.