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Re: What is wring here?



Ludwik,
(1) The tension in the two strings is NOT the same
(2) The force on m2 to the left is NOT m1*g but is Tsubl, which can be
found by drawing the free body diagram for m1
(3) Similarly the force on m2 to the right is Tsubrm, which is found
by drawing the free body diagram for m2.
(4) When you do all this the expected expression for the magnitude of
the acceleration pops out.

(5) Summaru: you didn't draw enough free body diagrams and you nmade
an incorrect assumption about the tensions in the strings.

Brian McInnes
----------

I must be blinded by something; please help.

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