I am needing to be educated about one aspect of all this. One way to
determine the max value of µ(s) is to increase the slope until the mass
starts to slide. The algebra tell us that µ(s) = tanø. All is independent of
mass. So I am confused by the statements that one of the blocks will slide
before the other, based on mass. I have not been able to find that in all
the discussions. Maybe I missed it.That's really only true in that one situation. The trick here is that the ribbon (the surface) is accelerating. Just like the tablecloth, if you pull hard enough, static friction won't be able to pull the objects on top hard enough to keep up, and they'll slip. This is dependent upon the object's mass in this case, because the acceleration is dependent upon the mass in a more complex way than simple proportionality, so the mass doesn't cancel out.
The way that I did this was to look first at the case with no slippage and determine a. This situation works as long as the friction force required is less than u m g cos (theta) for EACH block. This means that the static friction coefficient must be greater than or equal to (2M/(m+M))* tan (theta).
This condition fails first for the small block, because of its smaller normal force. Note that the static friction forces on the two blocks are always equal, because of the 'light' ribbon condition.
If we examine the case where the small block slips and the large one doesn't, we can also find the acceleration.
That model's valid only if the acceleration returned by the solution is positive. That requires that the kinetic friction coefficient must be less than (M/m)tan(theta).
This range overlaps with the static friction coefficient range from the first case, and since mu_s > mu_k, this must cover all possibilities.
Josh
Ken Fox
On Sun, Nov 7, 2010 at 5:49 PM, brian whatcott <betwys1@sbcglobal.net>wrote:
> On 11/7/2010 4:59 PM, Carl Mungan wrote:
> >> Ah you are dealing with the consequences of imputing a particular method
> >> of holding the small mass and ribbon. You visualize a stretchable ribbon
> >> (shame on you, this is meant to be a physicist';s ribbon! :-) and
> >> the upward force is I suppose an upslope foce to balance the down slope
> >> gravity vector.
> >> For some reason, I imagined a clamp appling a normal force to the small
> >> mass, so the large block can stretch the ribbon as desired while the
> >> ramps grow more
> >> high-pitched, so to speak.... would that resolve the issue?
> >>
> >> Brian W
> > Well, I think that's a valiant effort. But not very stable. There is no
> friction between the ribbon and the prism. If the clamp provides ONLY a
> normal force, it will slip sideways and the block is likely to tilt the
> clamp over (I'm imagining it as consisting of a knife edge clamping down on
> top of the block) and sneak out. Realistically, I think that the blocks have
> to be help by a normal force perpendicular to the lower edge face of the
> block. Like a gate that swings open (sideways or upward) as in a ski race.
> -Carl
>
> Hmmm...here you are imputing the means by which one applies a force to
> render the smaller mass in a comparable situation to the larger mass for
> an initial period: no friction between ribbon and ramp, appreciable
> static friction between masses and ribbon:
> I can visualize a clamp (with a locating peg if you wish?) that applies
> a normal force to place the smaller mass in the position of providing
> greater normal force hence higher peak friction force to the ribbon
> than the larger mass.
>
> It took me a while to visualize your reference to the ribbon rippling up
> upslope of both masses. This would be a function of the stiffness of the
> ribbon: reminds me that climbers like compliant ropes like nylon for
> safety lines, that sudden jerk at a fall becomes manageable: Horse
> tethers made of stiffer fibers are preferred
> against the panic case, where an end latch lets go. Too much energy
> stored in the stretch becomes a missile to bystanders.
>
> Brian W
>
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