Atwood Machine

The Atwood machine consists of two masses (M₁ and M₂) connected via a rope running over a pulley (mass M₃ - 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 M₁> M₂. In this case the acceleration will be as shown in the diagram. (The formula derived below is correct for any case; if M₁< M₂, then the acceleration, a, will come out with a negative sign.) The total gravitational force is

(M₁- M₂) g

since the weight-force of M₁mass acts in the direction of the acceleration, a, but the weight of M₂ acts in the opposite direction. The total mass that has to be accelerated is (M₁+ M₂). So F = ma has the form:

F = m a =>
(M₁- M₂) g = (M₁+ M₂) a =>
\[ \rm \mathbf {a = \frac{g(M_{1} - M_{2})}{M_{1} + M_{2}}} \]

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.