The illustration shows an equilibrium situation. The two masses (M) are equivalent, thus the force of gravity on each is equal. The upward force opposing gravity is the tension (T) in the string.

For the system to be in equilibrium, T = F_g. The net force is 2F_g - 2T = 0, so there is no acceleration.

The tension in the string is 2T or 2F_g. The string supports both masses, so we would expect the tension in this case to be the sum of the two downward forces.

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Note: In the simplest of Atwood's machine problems, we usually make two simplifications, that the mass of the pulley is zero and that pulley/rope system is frictionless. We can relax those restrictions later when we get good at the easy problems.

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Things get interesting when the masses are unequal. Now if we let the system go, we get acceleration in the downward direction of the heavier mass.

The gravitational forces on each mass are now unequal, and the net force, the vector sum of the two gravitational forces, is in the direction of the heavier mass. That means the net acceleration vector is in the direction of M₂, as shown (upper right).

The total acceleration of the system is the same for both masses; M₁ accelerates upward at the same rate as the downward acceleration of M₂ because they are tied together. We can treat the whole system as a single mass, M = M₁ + M₂.

Now the net force in the system can also be represented by Newton's second law as the sum of the masses multiplied by their acceleration:

$$F_{net} = (m_1 + m_2)a$$

Rearranging, we find that the acceleration of the system is the net force, F_net, divided by the total mass (we are assuming a massless pully and rope):

$$a = \frac{(m_2 - m_1) g}{m_1 + m_2}$$

The tension in the string of an Atwood's machine is the same everywhere when the system is at equilibrium, but it is different for each mass in an accelerating system.

To find the tension, treat each mass independently and use the common acceleration.

The net force (tension) on the left-side of the string is the sum of the downward gravitational force and the upward accelerating force of the system these forces both stretch the string:

$$T_{left} = M_1g + M_1a = M_1(g + a)$$

The tension in the right side of the string is the force of gravity on M₂ minus the downward acceleration force:

$$T_{right} = M_2 g - M_2 a = M_2 (g - a)$$

Think of the right side this way: a_net does not work to further stretch the right side of the string. It works in the direction of compression, which amounts to reducing the equilibrium tension

$$a = \frac{[(4 - 2) \; Kg] 9.8 \; m/s^2}{(2 + 4) \; Kg}$$

The Kg units cancel to give

$$a = \frac{2 \; Kg \cdot 9.8 \; m/s^2}{6}$$

and finally, our accleration is

$$a = 3.27 \; m/s^2$$

Tension

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We showed above that the tension on the heavy side (T_h) is

$$T_{heavy} = M_2 g + M_2 a = M_2(g - a).$$

where a is the acceleration of the system we calculated above. Plugging in what we know gives

$$ \begin{align} T_{heavy} &= M_h g - M_h a = M_h (g - a) \\[5pt] &= 4 \; Kg (9.8 - 3.27) \; m/s^2 \\[5pt] &= 26.12 \; N \end{align}$$

The

$$ \begin{align} T_{light} &= M_L g + M_L a = M_L (g + a) \\[5pt] &= 2 \; Kg (9.8 + 3.27) \; m/s^2 \\[5pt] &= 26.14 \; N \end{align}$$