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[Phys-L] Re: Free body diagram misconception



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


By what means is F(pull) transmitted to
block 2, since the rope is connected to block 1?
Here is ONE problem with the formula you've
come up with. There is NO force on block 2
other than the static frictional force.

The second problem is that there is an assumption
made that the static friction force that acts between
block 1 and block 2 is equal to mu x m1 x g. This
is the maximum value only. The static friction
force varies according to the applied force - though
there is a maximal acceleration (and thus a maximal
applied force).

SO

The free body diagram for block 2 should
only have ONE horizontal force on it.
Block 1 will have the horizontal forces
of friction opposing the motion, to the
left, and then the force pulling to the right.

The equation for Newton's 2nd law for
the two blocks, then, should be

Block 2:
F(friction) = m2 x a

Block 1:
F(pull)-F(friction) = m1 x a

We don't know what the friction force is,
but by Newton's 3rd law we know that
F(friction) on Block 2 by Block 1 is equal and opposite
F(friction) on Block 1 by Block 2.

So we can substitute the first equation into the second:

F(pull) - m2 x a = m1 x a

Thus

a = F(pull)/(m1+m2)

Which is exactly what you'd expect if the two blocks
are accelerating together. Any force accelerates
the blocks. / There is no minimum. The blocks will
accelerate under any applied force./

Now let's figure out the maximum force that can be
applied without the blocks slipping relative to
each other:

In this case, when friction is maximal, F(friction) = mu x m1 x a.

So we have:

Block 2:
mu x m1 x g = m2 x a

Block 1:
F(pull)_max -mu x m1 x g = m1 x a

Here again each block has acceleration a, so we
can substitute equation 1 into equation 2 again after
solving it for a.

F(pull)_max - mu x m1 x g = m1 x (mu x m1 x g)/m2

F(pull)_max = (mu x g)(m1+m1^2/m2)

So, if the blocks are equal mass, the maximum pull
force that won't dislodge block 1 from block 2 would be
2xmu1xg.

Todd


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Todd K. Pedlar Assistant Professor of Physics

Luther College pedlto01@luther.edu

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/"Knowledge without integrity is dangerous and dreadful." --Samuel Johnson/

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