**Atwood"s machine** used to offer me fits. It have the right to be confounding. However I"ve found that to understand all of the troubles that can be make by thinking about this very basic system is to have actually a deep understanding about forces and acceleration. So I market my best understanding in the hope that it will help you.

You are watching: Atwood machine special cases

Atwood"s maker is illustrated in the computer animation on the right. It couldn"t it is in simpler. It"s just a pulley, with which operation a string or rope fastened to two masses. We generally make the simplification that the string/rope and the wheel wheel are of negligible mass.

In this section, we"ll go v a number of Atwood"s scenarios, starting with the really simplest, two equal masses in **static equilibrium** (not moving). Very first a definition:

**Tension** is the sum of pressures *pulling* top top either finish of a string, rope, wire, cable, &c. If forces pull at both ends, they room additive. In certain problems, the force at one and also of a moving string is no a pulling force, but works in the contrary direction. In that case, it need to be subtracted native the pressure pulling indigenous the other end. (A string through two "pushing" pressures is no under tension).

The illustration shows an **equilibrium** situation. The 2 masses (**M**) are equivalent, for this reason the pressure of gravity on every is equal. The upward pressure opposing gravity is the **tension** (**T**) in the string.

For the mechanism to be in equilibrium,** T = Fg**. The net pressure is **2Fg - 2T = 0**, so over there is no acceleration.

The tension in the cable is **2T** or **2Fg**. The string supports both masses, so we would mean the stress in this case to it is in the sum of the two downward forces.

**Note**: In the most basic of Atwood"s an equipment problems, we typically make 2 simplifications, the the massive of the pulley is zero and that pulley/rope device is frictionless. We have the right to relax those restrictions later when we get good at the straightforward problems.

**Constant velocity **means **no acceleration**, therefore the net force in the system must still it is in zero. So this instance is a lot prefer the revolution one above.

In this case, the increase velocity has the exact same magnitude, however opposite direction as the downward velocity, so there is no network velocity, thus no acceleration.

You may have encountered difficulties like this. For example, if a consistent force exactly equal come the opposing force of friction is used to a slide object, the object move at continuous velocity — no acceleration.

Of course, once the mass on the left access time the pulley, the system stops, and also that"s acceleration, a adjust in velocity, and that"s a different scenario...

Things get exciting when the masses room unequal. Currently if we let the mechanism go, we get acceleration in the downward direction the the more heavier mass.

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

The complete acceleration the the mechanism is the very same for both masses; M1 speeds up upward in ~ the same price as the bottom acceleration that M2 because they room tied together. We deserve to treat the totality system together a solitary mass, M = M1 + M2.

The net pressure in the mechanism (see the diagram above) is just the sum of the pressures at work-related in one direction, to speak the downward direction of massive 2,

$$ \beginalign F+net = F_2 - F_1 &= m_2g - m_1g \\ &= (m_2 - m_1)g \endalign$$

**F2** and **F1** room the gravitational pressures on the masses. Using **F = mg** and combining the state (factoring out **g**) provides us the result.

Now the net force in the system can likewise be represented by Newton"s second law as the sum of the masses multiply by their acceleration:

$$F_net = (m_1 + m_2)a$$

Rearranging, we find that the acceleration the the mechanism is the network force, **Fnet**, split by the full mass (we are assuming a massless pully and also rope):

$$a = \frac(m_2 - m_1) gm_1 + m_2$$

The **tension** in the wire of an Atwood"s device is the same everywhere when the system is in ~ equilibrium, yet it is different for every mass in an speeding up system.

To find the tension, treat every mass independently and also use the usual acceleration.

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

$$T_left = M_1g + M_1a = M_1(g + a)$$

The anxiety in the ideal side that the string is the pressure of gravity on M2 *minus* the downward acceleration force:

$$T_right = M_2 g - M_2 a = M_2 (g - a)$$

Think that the right side this way: anet walk not work-related to further stretch the ideal side that the string. It functions in the direction the compression, which quantities to reduce the equilibrium tension

### Example 1 – acceleration & tension

Consider the Atwood machine below. Calculation the acceleration the the system and also the stress in every rope after the system is released.

**Solution**: The very first thing we should always do is brand the numbers with the appropriate vectors:

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

$$a = \frac(m_2 - m_1) gm_1 + m_2$$

where us take **m2** as the bigger mass and **m1** as the smaller (it doesn"t really matter as lengthy as we know that hopeful acceleration is in the direction that the bigger 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 devices cancel come give

$$a = \frac2 \; Kg \cdot 9.8 \; m/s^26$$

and finally, ours accleration is

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

tensionWe showed above that the tension on the hefty side (**Th**) is

$$T_heavy = M_2 g + M_2 a = M_2(g - a).$$

where **a** is the acceleration the the system we calculated above. Plugging in what we understand gives

$$ \beginalign T_heavy &= M_h g - M_h a = M_h (g - a) \\ &= 4 \; Kg (9.8 - 3.27) \; m/s^2 \\ &= 26.12 \; N \endalign$$

The

$$ \beginalign T_light &= M_L g + M_L a = M_L (g + a) \\ &= 2 \; Kg (9.8 + 3.27) \; m/s^2 \\ &= 26.14 \; N \endalign$$

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