Consider a modified Atwood machine, a horizontal
frictionless table with two massless and frictionless
pulleys. The vertically suspended mass on the left, m1,
is smaller than the vertically suspended mass on the
right, m3. What is the acceleration? The correct answer is
a=[(m3-m1)/(m1+m2+m3)]*g
But what is wrong with the following reasoning?
I am interested in what happens to m2. Therefore I
draw the free body diagram for m2 with four forces.
Vertical forces cancel. The remaining two forces are
F1=m1*g (to the leftt) and F3=m3*g (to the right).
The net force on m2 must thus be F=(m3-m1)*g; it is
directed to the right. Thus the expected acceleration is
a=[(m3-m1)/m2]*g.
But this contradicts the first answer. The acceleration
of m2 must be the same as the acceleration of two
other masses, or as the acceleration of the entire
system, (m1+m2+m3). Furthermore the second
answer allows for a>g, which is silly. Clearly
something is wrong with my free-body-diagram
approach. What is it?
Ludwik Kowalski