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## Systems with Moveable Pulleys

Up to this point, we have only considered *fixed* pulleys, i.e. pulleys that are physically attached to either a ceiling or a table.

We can, however, also have **movable pulleys** (of a particular mass) for which a string simply passes around.

###### \[\,\]Figure 5.6.1

We can treat such a moveable pulley similarly to how we have treated objects attached to the ends of strings.

The main difference is that a movable pulley is, in a sense, “attached” to the string at *two* places rather than one.

This changes things in two different ways.

**1) There are two tension forces applied between a moveable pulley and a string**

As the pulley is “attached” to the string in two places, there will be two tension forces associated with it.

###### \[\,\]Figure 5.6.2

These tension forces have the same magnitude as each other as they are created by the same taut string. (They also typically have the same direction, as in this example, but that is not always the case.)

Also note that, if the moveable pulley is to accelerate downwards in the above figure, the moveable pulley’s weight force must be larger in magnitude than these two tension forces (of the same magnitude) combined.

In contrast, the object of mass \(m_1\) only needs a weight that is greater than one of these tension forces if it is to instead accelerate downwards.

As such, the moveable pulley will only accelerate downwards (and the object will only accelerate upwards) if \(m_2 > 2m_1\). Otherwise, the moveable pulley will instead accelerate upwards (and the object will accelerate downwards).

**2) The system no longer has a common acceleration**

Why? Well, let’s assume that \(m_2>2m_1\), i.e. that the object of mass \(m_1\) accelerates upwards and the movable pulley of mass \(m_2\) accelerates downwards.

If the object moves up a distance of \(|\vec{s_1}|\) metres, there is \(|\vec{s_1}|\) metres less string on the left hand side of the system.

Where has that \(|\vec{s_1}|\) metres of string gone? Well, it was used by the moveable pulley to move downwards!

However, half of that \(|\vec{s_1}|\) metres of string is on one side of the moveable pulley and the other half is on the other side.

###### \[\,\]Figure 5.6.3

Therefore, the pulley itself has only moved half of the distance that the object has in that same time, i.e. \(|\vec{s_2}|=\displaystyle\frac{|\vec{s_1}|}{2}\).

What does this imply in regards to acceleration? Well, consider the following 1D SUVAT equation

\begin{align}s=ut+\frac{1}{2}at^2\end{align}

Since we release the system from rest (\(u=0\)), we can simplify this to

\begin{align}s=\frac{1}{2}at^2\end{align}

and therefore the displacement coefficient is proportional to the acceleration coefficient \((s \propto a)\).

Hence, since \(|\vec{s_2}|=\displaystyle\frac{|\vec{s_1}|}{2}\), this implies that \(|\vec{a_2}|=\displaystyle\frac{|\vec{a_1}|}{2}\), i.e. the acceleration rate of the moveable pulley is also half the acceleration rate of the object!

###### \[\,\]Figure 5.6.4

Therefore, there is no longer a common acceleration in systems with movable pulleys. Instead, if a question asks you to find an acceleration for such a system, it will specifically state the object it wishes you to find that acceleration for.

\[\,\]

Besides these two changes, we treat a movable pulley in the same way that we treated objects in the previous lesson.

## Guided Example 5.6.1

An object of mass \(3\mbox{ kg}\) is attached by a light, inelastic string over a fixed pulley and over a moveable pulley of mass \(7\mbox{ kg}\), as shown below.

When the system is released:

**(a)** What is acceleration of the \(3\mbox{ kg}\) object?

**(b)** What is acceleration of the moveable pulley?

**(c)** What is the tension in the string?

**(a)** As the \(3\mbox{ kg}\) object has less forces acting on it, let us first apply Newton’s second law to this object.

**Step 1: Draw a free body diagram.**

As this object is less than half the mass of the moveable pulley, it will accelerate upwards. Therefore, the weight of the object must be less than the tension in the string.

**Step 2: Write the force and acceleration vectors in component form.**

The force vectors are given by

\begin{align} \vec{T}_1 = T_1 \vec{j} && \vec{W}_1=W_1\vec{j}\end{align}

and the acceleration vector is

\begin{align} \vec{a}_1 = a_1\vec{j} \end{align}

\[\,\]

**Step 3: Use \(\boldsymbol{\vec{F}_{\text{net}}=m\vec{a}}\).**

\[T_1\vec{j}+W_1\vec{j} = (3)(a_1\vec{j})\]

\[\,\]

**Step 4: Insert any known coefficients and replace any similar coefficients.**

We know that the magnitude of \(\vec{W}_1\) is \(mg = 3g\) and that it is in the *opposite* direction to \(\vec{j}\), i.e. \(W_1=-3g\).

Therefore, the equation becomes

\[T_1\vec{j}-3g\vec{j} = 3a_1\vec{j}\]

\[\,\]

**Step 5: Dot the equation with unit vectors.**

Dotting this equation with \(\vec{j}\), we obtain

\begin{align}T_1-3g=3a_1\end{align}

Let us next apply Newton’s second law to the \(5\mbox{ kg}\) moveable pulley.

**Step 1: Draw a free body diagram.**

As the moveable pulley is greater than twice the mass of the other object, it will accelerate downwards. Therefore, the weight of the object must be greater than sum of the tensions in the string.

**Step 2: Write the force and acceleration vectors in component form.**

The force vectors are given by

\begin{align} \vec{T}_{2l} = T_{2l} \vec{j} && \vec{T}_{2r} = T_{2r}\vec{j} && \vec{W}_2=W_2\vec{j}\end{align}

\[\,\]

**Step 3: Use \(\boldsymbol{\vec{F}_{\text{net}}=m\vec{a}}\).**

\[T_{2l}\vec{j}+T_{2r}\vec{j}+W_2\vec{j} = (7)(a_2\vec{j})\]

\[\,\]

**Step 4: Insert any known coefficients and replace any similar coefficients.**

We know that the magnitude of \(\vec{W}_2\) is \(mg = 7g\) and that it is in the *opposite* direction to \(\vec{j}\), i.e. \(W_2=-7g\).

We know that both \(\vec{T_{2l}}\) and \(\vec{T_{2r}}\) have the same magnitude as \(\vec{T_1}\) (as it is the same string for all) and all three are pointing in the same axis direction (positive \(y\) direction), i.e. \(T_{2l}=T_{2r}=T_1\).

Based on the logic above, the acceleration rates are related by \(|\vec{a_2}|=\displaystyle\frac{|\vec{a_1}|}{2}\). As they are moving in *opposite* axis directions (negative \(y\) and positive \(y\) directions respectively), this therefore implies that \(a_2=-\displaystyle\frac{a_1}{2}\).

Hence, the equation becomes

\[2T_1\vec{j}-7g\vec{j} = (7)\left(-\frac{a_1}{2}\vec{j}\right)\]

\[\,\]

**Step 5: Dot the equation with unit vectors.**

Dotting this equation with \(\vec{j}\), we obtain

\begin{align}2T_1-7g=-7\frac{a_1}{2}\end{align}

We now have the following two equations with two unknowns.

\begin{align}T_1-3g=3a_1\end{align}

\begin{align}2T_1-7g=-\frac{7a_1}{2}\end{align}

We can eliminate \(T_1\) easily by multiplying the first equation by \(2\)

\begin{align}2T_1-6g=6a_1\end{align}

\begin{align}2T_1-7g=-\frac{7a_1}{2}\end{align}

and then subtracting the second equation from the first

\begin{align}-6g – (-7g) = 6a_1- \left(- \frac{7a_1}{2}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}g = \frac{19a_1}{2}\end{align}

Solving for \(a_1\), this gives

\begin{align}a_1&=\frac{2g}{19}\\&\approx 1.03 \mbox { m/s}^2\end{align}

Hence, the acceleration of the \(3\mbox{ kg}\) object is \(\vec{a_1} = 1.03 \vec{j}\mbox{ m/s}^2\).

**(b) **We can now easily find this by using the following relation

\begin{align}a_2&=-\frac{a_1}{2}\\&=-\frac{1.03}{2}\\&\approx -0.52 \mbox{ m/s}^2\end{align}

Hence, the acceleration of the \(7\mbox{ kg}\) moveable pulley is \(\vec{a_2} = -0.52 \vec{j}\mbox{ m/s}^2\).

**(c)** Again, this can now be easily found by using one of our above equations for \(T_1\), e.g.

\begin{align}T_1-3g=3a_1\end{align}

Rearranging, this gives

\begin{align}T_1&=3g+3a_1\\&=3(9.8)+3(1.03)\\&=32.49 \mbox{ N}\end{align}

Hence, the string’s tension has a magnitude of \(32.49\mbox{ N}\) and a direction as shown in the free body diagrams above.

We can apply the same logic for other systems involving moveable pulleys, including the following.

###### \[\,\]Figure 5.6.5

As this system also contains a moveable pulley, it is again treated in the same manner as the previous system.

Once again, the tension force is applied twice between the moveable pulley and the string.

###### \[\,\]Figure 5.6.6

We now have three objects, each with their won acceleration rates. Before we jump into discussing how to find these rates, it should be noted that even the *direction* of these acceleration vectors is not as obvious as before.

In the previous system, we stated that the moveable pulley will accelerate downwards if \(m_2>2m_1\). However, the similar relation in this case is not as straightforward to state. (In fact, finding that relation is an exam question in itself!)

Even with such a relation, it is not guaranteed that the two remaining objects will move in the opposite direction. Depending on the masses, it is entirely possible that one of the two objects could accelerate in the same direction as the moveable pulley!

How do we therefore proceed?

Well, specifying the direction of the acceleration vectors is not actually necessary and is only done for our own benefit so that we may draw them on free body diagrams. Newton’s second law will tell us what the correct sign of the acceleration coefficient is, regardless as to whether we initially drew it in the right direction or not.

### Exam Tip

If you are unsure about the direction of motion of an object (or a system), you can randomly choose a direction when drawing the free body diagram.

Once you find that object’s acceleration coefficient, the sign of that coefficient will tell you if that guess was correct or not. If it wasn’t, go back and redraw the correct acceleration vector.

Therefore, unless we are told otherwise, we are free to assume for example that the pulley accelerates downwards and the two objects accelerate upwards. If that guess was wrong, i.e. if the acceleration coefficients we find indicates that we drew at least one of the acceleration vectors incorrectly, we simply go back, cross out that vector and draw the correct vector!

With this assumption, if the object of mass \(m_1\) moves up a distance of \(|\vec{s_1}|\) metres and the object of mass \(m_2\) moves up a distance of \(|\vec{s_2}|\) metres, there is \(|\vec{s_1}|+|\vec{s_2}|\) metres in total less string on the left and right hand sides of the system.

This quantity of string has again gone to the pulley. Half of those \(|\vec{s_1}|+|\vec{s_2}|\) metres of string is on one side of the moveable pulley and the other half is on the other side.

###### \[\,\]Figure 5.6.7

As before, the pulley itself has therefore only moved half of this distance, i.e. \(|\vec{s_3}|=\displaystyle\frac{|\vec{s_1}|+|\vec{s_2}|}{2}\) metres.

Hence, using the previous logic, the acceleration rate of the moveable pulley must be \(|\vec{a_3}|=\displaystyle\frac{|\vec{a_1}|+|\vec{a_2}|}{2}\).

###### \[\,\]Figure 5.6.8

## Guided Example 5.6.2

Objects of masses \(3\mbox{ kg}\) and \(5\mbox{ kg}\) are attached to each other by a light, inelastic string over a moveable pulley of mass \(10\mbox{ kg}\), as shown below.

When the system is released:

**(a)** What is acceleration of the moveable pulley?

**(b)** What is the tension in the string?

**(a) **As the direction of the objects is not obvious, we shall randomly assume that the moveable pulley accelerates downwards and that the two remaining objects accelerate upwards.

(If that assumption is correct, we should obtain a negative acceleration coefficient for the moveable pulley and a positive acceleration for the two objects. If we don’t get those signs in any of the three cases, we go back and redraw the correct acceleration vector in the corresponding free body diagram.)

As the moveable pulley has the most forces acting on it, we shall start with one of the other objects. Of these, we shall randomly choose the \(5\mbox{ kg}\) object to start with.

**Step 1: Draw a free body diagram.**

As stated, we are assuming that this object is accelerating upwards.

**Step 2: Write the force and acceleration vectors in component form.**

The force vectors are given by

\begin{align} \vec{T}_1 = T_1\vec{j} && \vec{W}_1=W_1\vec{j}\end{align}

and the acceleration vector is

\begin{align}\vec{a}_1 = a_1\vec{j} \end{align}

\[\,\]

**Step 3: Use \(\boldsymbol{\vec{F}_{\text{net}}=m\vec{a}}\).**

\[T_1\vec{j}+W_1\vec{j} = (5)(a_1\vec{j})\]

\[\,\]

**Step 4: Insert any known coefficients and replace any similar coefficients.**

We know that the magnitude of \(\vec{W}\) is \(mg = 5g\) and that it is in the *opposite* direction to \(\vec{j}\), i.e. \(W=-5g\).

Therefore, the equation becomes

\[T_1\vec{j}-5g\vec{j} = 5a_1\vec{j}\]

\[\,\]

**Step 5: Dot the equation with unit vectors.**

Dotting this equation with \(\vec{i}\), we obtain

\begin{align}T_1-5g=5a_1\end{align}

Let us next apply Newton’s second law to the \(3\mbox{ kg}\) hanging object on the right.

**Step 1: Draw a free body diagram.**

As stated, we are assuming that this object is accelerating upwards.

**Step 2: Write the force and acceleration vectors in component form.**

The force vectors are given by

\begin{align} \vec{T}_2 = T_2\vec{j} && \vec{W}_2=W_2\vec{j}\end{align}

and the acceleration vector is

\begin{align}\vec{a}_2 = a_2\vec{j} \end{align}

\[\,\]

**Step 3: Use \(\boldsymbol{\vec{F}_{\text{net}}=m\vec{a}}\).**

\[T_2\vec{j}+W_2\vec{j} = (3)(a_2\vec{j})\]

\[\,\]

**Step 4: Insert any known coefficients and replace any similar coefficients.**

We know that the magnitude of \(\vec{W}\) is \(mg = 3g\) and that it is in the *opposite* direction to \(\vec{j}\), i.e. \(W=-3g\).

We know that \(\vec{T_2}\) has the same magnitude as \(\vec{T_1}\) (as it is the same string for both) and both are pointing in the same axis direction (positive \(y\) direction), i.e. \(T_2=T_1\).

Therefore, the equation becomes

\[T_1\vec{j}-3g\vec{j} = 3a_2\vec{j}\]

\[\,\]

**Step 5: Dot the equation with unit vectors.**

Dotting this equation with \(\vec{i}\), we obtain

\begin{align}T_2-3g=3a_2\end{align}

Finally, let us apply Newton’s second law to the \(10\mbox{ kg}\) moveable pulley.

**Step 1: Draw a free body diagram.**

As stated, we are assuming that the moveable pulley is accelerating downwards.

**Step 2: Write the force and acceleration vectors in component form.**

The force vectors are given by

\begin{align} \vec{T}_{3l} = T_{3l} \vec{j} && \vec{T}_{3r} = T_{3r}\vec{j} && \vec{W}_3=W_3\vec{j}\end{align}

\[\,\]

**Step 3: Use \(\boldsymbol{\vec{F}_{\text{net}}=m\vec{a}}\).**

\[T_{3l}\vec{j}+T_{3r}\vec{j}+W_3\vec{j} = (10)(a_3\vec{j})\]

\[\,\]

**Step 4: Insert any known coefficients and replace any similar coefficients.**

We know that the magnitude of \(\vec{W}_3\) is \(mg = 10g\) and that it is in the *opposite* direction to \(\vec{j}\), i.e. \(W_3=-10g\).

We know that both \(\vec{T_{3l}}\) and \(\vec{T_{3r}}\) have the same magnitude as \(\vec{T_1}\) (as it is the same string for all) and all three are pointing in the same axis direction (positive \(y\) direction), i.e. \(T_{3l}=T_{3r}=T_1\).

Based on the logic above, the acceleration rates are related by \(|\vec{a_3}|=\displaystyle\frac{|\vec{a_1}|+|\vec{a_2}|}{2}\). As the pulley is moving in the *opposite* axis direction to the \(5\mbox{ kg}\) object (negative \(y\) and positive \(y\) directions respectively) and as the pulley is moving in the *opposite* axis direction to the \(3\mbox{ kg}\) object (negative \(y\) and positive \(y\) directions respectively), this therefore implies that \(a_3=\displaystyle\frac{-a_1-a_2}{2}\).

Therefore, the equation becomes

\[2T_1\vec{j}-10g\vec{j} = (10)\left(\frac{-a_1-a_2}{2}\vec{j}\right)\]

\[\,\]

**Step 5: Dot the equation with unit vectors.**

Dotting this equation with \(\vec{j}\), we obtain

\begin{align}2T_1-10g=-5(a_1+a_2)\end{align}

We now have the following three equations with three unknowns.

\begin{align}T_1-5g=5a_1\end{align}

\begin{align}T_1-3g=3a_2\end{align}

\begin{align}2T_1-10g=-5(a_1+a_2)\end{align}

Let us rearrange the first two equations for the acceleration rates

\begin{align}a_1 = \frac{T_1-5g}{5}\end{align}

\begin{align}a_2 = \frac{T_1-3g}{3}\end{align}

\begin{align}2T_1-10g=-5(a_1+a_2)\end{align}

and insert those rates into the third equation

\begin{align}2T_1-10g=-5\left(\frac{T_1-5g}{5}+\frac{T_1-3g}{3}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2T_1-10g=-(T_1-5g)-\frac{5}{3}(T_1-3g)\end{align}

Multiplying both sides by \(3\)

\begin{align}6T_1-60g = -6T_1+30g-5T_1+15g\end{align}

\begin{align}\downarrow\end{align}

\begin{align}17T_1=105g\end{align}

\begin{align}\downarrow\end{align}

\begin{align}T_1 &=\frac{105g}{17}\\&\approx 60.53 \mbox{ N}\end{align}

We can now calculate both \(a_1\) and \(a_2\) using

\begin{align}a_1 = \frac{T_1-5g}{5}\end{align}

\begin{align}a_2 = \frac{T_1-3g}{3}\end{align}

First, let us calculate \(a_1\)

\begin{align}a_1 &= \frac{T_1-5g}{5}\\&=\frac{60.53-5(9.8)}{5}\\&\approx 2.31 \mbox{ m/s}^2\end{align}

(This coefficient is associated with the \(5\mbox{ kg}\) object, which we assumed was accelerating in the positive \(y\) direction. As it is positive, i.e. as \(\vec{a}_1 = 2.31\vec{j}\mbox{ m/s}^2\), our assumption was correct.)

For \(a_2\), we obtain

\begin{align}a_2 &= \frac{T_1-3g}{3}\\&=\frac{60.53-3(9.8)}{3}\\&\approx 10.38 \mbox{ m/s}^2\end{align}

(This coefficient is associated with the \(3\mbox{ kg}\) object, which we assumed was accelerating in the positive \(y\) direction. As it is positive, i.e. as \(\vec{a}_2 = 10.38\vec{j}\mbox{ m/s}^2\), our assumption was correct.)

Finally, for the moveable pulley, we obtain

\begin{align}a_3 &= -\frac{a_1+a_2}{2}\\&=-\frac{2.31+10.38}{2}\\&\approx -6.35\mbox{ m/s}^2\end{align}

and it therefore has an acceleration of \(-6.35\vec{j}\mbox{ m/s}^2\) (which, again, agrees with our assumption about its direction).

**(b)** As we calculated above, the tension in the string has a magnitude of \(60.53\mbox{ N}\) and a direction as shown in the free body diagrams above.