[Phys-L] Re: Free body diagram misconception

[jsd at AV8N.COM (John Denker)]

Gene Gordon wrote:
> Block 1 is at rest atop Block 2.  Block 2 rests atop a frictionless
> surface.  A force is applied to Block 1 pulling it to the right.   The
> coefficient of friction between blocks 1&2 is some non-zero value less
> than 1. What is the Net force equation for Block 2?
>
> We drew the free body diagrams for both blocks and came up with the
> following equation for Block 2.
>
> (mu)m1g - F(pull) = m2a2               [1]
>
> Now all of us agreed on this, but we also saw a problem.

I don't think I agree.

The overwhelmingly most common conception about coefficient of
friction problems is the notion that you can caclulate "the"
force from "the" coefficient of friction.

For starters, we should note that there exist more than one
coefficient of friction.  I assume that the "mu" in the question
is the coefficient of _static_ friction.  This should have been
specified in the statement of the problem.  If my assumption is
wrong, please re-ask the question.

Secondly, there cannot be any equations (i.e. equalities) of
the form [1] above --- only inequalities.  The basic static
friction equation is

           F_parallel <= mu F_perp              [2]

Now, since we are assuming static friction, there is really
only one possibility, namely that block1 and block2 move as
a unit.  The free body diagram is trivial.

         a = F_applied / (m1 + m2)              [3]

         F2 = m2 a                              [4]
                             m2
            = F_applied  ---------              [5]
                          m1 + m2

subject to the restriction that

         F2 <= mu m1 g                          [6]

If [6] is not satisfied, you are in the sliding friction regime,
and a whole new analysis is required.  But it is even simpler
than the above.

I emphasize there are two cases (static versus sliding) that must
be analyzed separately.