Re: a mechanical problem

[Herb Schulz <herbs@interaccess.com>]

> 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.
>                                          Ludwik Kowalski

Howdy,

I'll assume that we are talking about motion along the horizontal (i.e.,
perpendicular to the force of gravity on the object).

There are three forces acting on the object: Gravity (mag mg, dir down),
the Normal force (mag N, dir up) and the static friction force (arbitrary
mag!!! (the MAXIMUM value is set by mu*N) directed in the SAME direction as
the accelerating "truck" (i.e., the static friction force is what is making
the block accelerate with the surface below it: there is no RELATIVE motion
between the surface and the block)).  N and mg cancel out since there is no
acceleration in the vertical direction (Newton's 2nd Law applied to that
component of acceleration) so that the MAXIMUM value of the static friction
force is mu*N=mu*mg. In the horizontal direction Newton's 2nd Law gives
f=ma where f is the (variable magnitude) static friction force, m is that
mass of the block and a is the acceleration of the block (which must also
be the acceleration of the surface if there is no relative motion).  As a
gets larger, the f must also get larger until f=fmax=mu*mg. If you attempt
to make a larger than that the block will slip.

I hope this is clear.

Thanks,

Herb Schulz
(herbs@interaccess.com)

Herb Schulz
(herbs@interaccess.com)