The
Atwood machine consists of two masses (M_{1} and
M_{2}) connected via a rope running over a pulley (mass
M_{3} - only comes into play in the case with friction). In
the friction-free case, the pulley does not move, and the rope just
glides over it. In the presence of friction, the pulley is set into
rotation, and the energy of rotation must be taken into account - as
described in a later chapter.

Let's consider the friction-free case with M_{1 }>
M_{2}. In this case the acceleration will be as shown in the
diagram. (The formula derived below is correct for any case; if
M_{1 }< M_{2}, then the acceleration, a, will come
out with a negative sign.) The total gravitational force is

since the weight-force of M_{1 }mass acts in the direction
of the acceleration, a, but the weight of M_{2} acts in the
opposite direction. The total mass that has to be accelerated is
(M_{1 }+ M_{2}). So F = ma has the form:

(M

From this equation you can see that the acceleration, a, in this case is always smaller than g. If both masses are equal, then we obtain the expected result of 0 acceleration. By selecting the proper combination of masses, we can generate any value of the acceleration between zero and g that we desire.

© *MultiMedia
Physics*, 1999