Atwood's machine

### Atwood's machine

A simple device for studying the laws of motion & forces

Atwood's machine used to give me fits. It can be confounding. Yet I've found that to understand all of the problems that can be posed by thinking about this very simple system is to have a deep understanding about forces and acceleration. So I offer my best understanding in the hope that it will help you.

Atwood's machine is illustrated in the animation on the right. It couldn't be simpler. It's just a pulley, through which runs a string or rope attached to two masses. We generally make the simplification that the string/rope and the pulley wheel are of negligible mass.

In this section, we'll go through a number of Atwood's scenarios, starting with the very simplest, two equal masses in static equilibrium (not moving). First a definition:

Tension is the sum of forces pulling on either end of a string, rope, wire, cable, &c. If forces pull at both ends, they are additive. In certain problems, the force at one and of a moving string is not a pulling force, but works in the opposite direction. In that case, it must be subtracted from the force pulling from the other end. (A string with two "pushing" forces is not under tension).

### 1. Equal masses, no acceleration 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 = Fg. The net force is 2Fg - 2T = 0, so there is no acceleration.

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

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.

### 2. Equal masses; Constant velocity

Constant velocity means no acceleration, so the net force in the system must still be zero. So this situation is a lot like the static one above.

In this case, the upward velocity has the same magnitude, but opposite direction as the downward velocity, so there is no net velocity, therefore no acceleration.

You may have encountered problems like this. For example, if a constant force exactly equal to the opposing force of friction is applied to a sliding object, the object moves at constant velocity — no acceleration.

Of course, when the mass on the left hits the pulley, the system stops, and that's acceleration, a change in velocity, and that's a different scenario... ### 3. Unequal masses; Acceleration 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.

### Net force in the system

The net force in the system (see the diagram above) is just the sum of the forces at work in one direction, say the downward direction of mass 2,

\begin{align} F+{net} = F_2 - F_1 &= m_2g - m_1g \\[5pt] &= (m_2 - m_1)g \end{align}

F2 and F1 are the gravitational forces on the masses. Using F = mg and combining the terms (factoring out g) gives us the result.

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, Fnet, divided by the total mass (we are assuming a massless pully and rope):

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

### Tension in the system 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: anet 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

### Example 1 – acceleration & tension

Consider the Atwood machine below. Calculate the acceleration of the system and the tension in each rope after the system is released. Solution: The first thing we should always do is label the figures with the appropriate vectors: #### Acceleration

We showed above that the acceleration of such a system is:

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

where we take m2 as the larger mass and m1 as the smaller (it doesn't really matter as long as we know that positive acceleration is in the direction of the larger mass). Plugging in our masses and g = 9.8 m/s2 gives

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

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

We showed above that the tension on the heavy side (Th) 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}   Dr. Cruzan's Math and Science Web by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012,2022, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.