Re: a mechanical problem

[John Mallinckrodt <ajmallinckro@CSUPomona.Edu>]

On Sun, 5 Oct 1997, LUDWIK KOWALSKI wrote:

> A wooden block is placed on a horizontal plate which is at rest. The 
> coefficient of static friction, mu, is given. What force, applied to the 
> block will initiate sliding? We know the answer, F=mu*m*g. And now a 
> different question. What is the minimum acceleration, a, of the supporting 
> plate which will result in sliding (of the block with respect to the plate)? 
> Everybody knows that the block will slide backward when the acceleration of 
> the plate is too large. How do we explain this? We say that the net force 
> acting on the non-sliding block must be m*a and that for very large a this
> exceeds the static frictional force mu*m*g. Thus the answer is a=mu*g. 
> I am not sure this is correct, even in elementary mechanics.
>
> In a fixed frame of reference I recognize two objects, an accelerating 
> plate and a block above it. A free body diagram for the block shows three
> forces acting on it: (1) Weight, mg, pointing down, (2) the reaction force 
> from the plate and (3) horizontal force due to "static friction". I am not
> sure that the reaction force (plate acting on the block) is vertical when
> the plate is accelerating. The answer a=mu*g is correct only when the
> first two forces cancel each other. 

Perhaps your confusion is a result of confusing the "reaction force" from
the plate with the normal component of that force.  There are only two
forces acting on the block: 

 (1) the force that the earth exerts on it due to gravity, (m*g_vec)

 (2) the force that the plate exerts on it due to contact

By Newton's second law, force (2) *must* equal (m*a_vec) + (-m*g_vec).

If you'd now like to invoke the standard Eulerian model of friction (one
of the *least* adequate models we talk about in intro physics), then force
(2) is broken down into "normal" and "frictional" components which are
perpendicular and parallel to the contact surface respectively by
definition. The model says that the frictional component is related to the
normal component through an empirical coefficient. 

In your case, if the block is accelerating horizontally, then the normal
componentis -m*g_vec and the frictional component is m*a_vec.  If the
block is not sliding, the model says that the frictional component is not
greater in magnitude then mu_static*(magnitude of the normal component). 
Thus, necessarily, the maximum acceleration before slipping occurs is
mu_static*g.

John
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