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[Phys-L] Re: Free body diagram misconception



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

Now all of us agreed on this, but we also saw a problem. This means
that there is some value for the force of the pull that Block 2 will not
move. However it is on a frictionless surface and IF they were
connected the Frictional component would disappear - thus the object
would accelerate. What is wrong with our reasoning? Did we make a bad
assumption somewhere?

I love these problems but it is driving me bonkers since I know that I
am forgetting something fundamental --- Block 2 should accelerate for
ANY force pulling on it if it is on a frictionless surface.

There is only one horizontal force applied to Block 2--the frictional
force between it and Block 1. Thus, the equation for Block 2 is

force of friction = m2 * a2

So your intuition is certainly correct: Block 2 accelerates for any
pulling force.

The interesting question is whether the surface between the two
blocks slips or not. You can show that there will be no slipping as
long as F(pull) remains less than (1 + m1/m2) * the maximum
attainable static frictional force.

--
John "Slo" Mallinckrodt

Professor of Physics, Cal Poly Pomona
<http://www.csupomona.edu/~ajm>

and

Lead Guitarist, Out-Laws of Physics
<http://www.csupomona.edu/~hsleff/OoPs.html>