 ### 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 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.

### 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 The

Things get interesting when the masses are unequal. Now if we let go of one, 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 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 The

The

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The The net force in the system is The net force in the system is ### Tension in the system The

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:

### Tleft = M1g + M1a = M1(g + a)

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

### Tright = M2g + M2a = M2(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 ### You can help

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