How to Decide Which Way a Pulley System Accelerates

Hang a \(0.60\ \mathrm{kg}\) mass from one end of a light string and a \(0.40\ \mathrm{kg}\) mass from the other. Pass the string over a fixed pulley and release both masses.

Which side accelerates downwards?

The \(0.60\ \mathrm{kg}\) mass accelerates down, while the \(0.40\ \mathrm{kg}\) mass accelerates up. That seems straightforward. But what happens when one mass is on a rough table? Or when the system is already moving in the opposite direction to its acceleration?

The reliable method is to choose a direction, calculate the net driving force, and check the sign of the acceleration.

For a simple pulley connecting two freely hanging masses, the heavier mass accelerates downwards. For more complicated arrangements, you must consider every external force acting along the direction the rope allows the objects to move.

01The quickest way to decide the direction

Start by imagining which way you think the system will move.

You don’t need to know the correct direction before beginning the calculation. You can assume a direction and let Newton’s second law tell you whether that assumption was correct.

For a simple Atwood machine, with two hanging masses connected by an ideal string over a fixed pulley:

• If the left mass is heavier, assume it accelerates downwards.
• If the right mass is heavier, assume it accelerates downwards.
• If both masses are equal, the system has zero acceleration when released from rest.

An Atwood machine is simply the standard arrangement of two hanging masses connected by a string over a pulley.

Why does the heavier mass accelerate downwards?

The gravitational force acting on an object is its weight:

\[
F_g=mg
\]

where:

• \(F_g\) is gravitational force in newtons (N),
• \(m\) is mass in kilograms (kg),
• \(g\) is gravitational acceleration, approximately \(9.8\ \mathrm{m\,s^{-2}}\).

The \(0.60\ \mathrm{kg}\) mass experiences a gravitational force of:

\[
F_{g1}=0.60\times9.8=5.88\ \mathrm{N}
\]

The \(0.40\ \mathrm{kg}\) mass experiences:

\[
F_{g2}=0.40\times9.8=3.92\ \mathrm{N}
\]

The heavier mass has the greater downward gravitational force.

However, both masses are connected by the same string. The lighter mass cannot simply remain stationary while the heavier one falls. It must accelerate upwards as the heavier mass accelerates downwards.

The difference between their weights provides the driving force for the system.

This is the first important idea:

Don’t compare the forces on just one object and forget about the other. Consider the entire connected system.

02Understand the ideal pulley assumptions

Before writing equations, identify what the question means by an ideal pulley system.

HSC physics problems commonly use three important assumptions.

We also usually neglect air resistance unless the question states otherwise.

What does an inextensible string tell us?

Imagine the heavier mass in our Atwood machine moves down by \(10\ \mathrm{cm}\).

Because the string cannot stretch, the lighter mass must move up by exactly \(10\ \mathrm{cm}\).

The two masses therefore have equal acceleration magnitudes:

\[
|a_1|=|a_2|
\]

Their accelerations point in opposite directions.

This relationship is extremely useful because it allows us to describe both objects using one acceleration variable.

Be careful, though. This equal-acceleration rule applies to the simple arrangement with a single fixed pulley. A movable pulley can produce a different acceleration relationship.

Why is the tension equal on both sides?

In an ideal system, the string is massless and the pulley has negligible rotational inertia.

If the tension differed on the two sides, there would be a net torque on the pulley. A massless pulley cannot have a finite net torque while having zero rotational inertia in this ideal model.

Therefore:

\[
T_1=T_2=T
\]

This lets us use one tension variable in both free-body equations.

A real pulley with significant rotational inertia is different. Its tensions can differ because some of the driving force produces rotational acceleration.

03Internal and external forces: what actually drives the system?

This is where many students lose marks.

Consider the two hanging masses connected by a string.

Each mass experiences:

• Its weight, acting downwards.
• Tension, acting upwards.

If you draw a free-body diagram for just one mass, tension is an external force acting on that mass. You must include it.

However, when you consider the connected pair as a system, tension is an internal interaction transmitted through the string. Its terms cancel when you add the two equations using the appropriate positive directions.

The gravitational forces do not cancel in this way. Their effects oppose each other because the two masses move in opposite directions.

Deriving the acceleration equation

Let:

• \(m_1\) be the heavier mass.
• \(m_2\) be the lighter mass.
• \(T\) be the tension in the string.
• \(a\) be the acceleration magnitude.

Assume \(m_1\) accelerates downwards and \(m_2\) accelerates upwards.

For the heavier mass, take downwards as positive:

\[
m_1g-T=m_1a
\]

For the lighter mass, take upwards as positive:

\[
T-m_2g=m_2a
\]

Add the equations:

\[
m_1g-T+T-m_2g=m_1a+m_2a
\]

The tension terms cancel:

\[
(m_1-m_2)g=(m_1+m_2)a
\]

Rearranging:

\[
\boxed{a=\frac{(m_1-m_2)g}{m_1+m_2}}
\]

This equation gives the acceleration magnitude when \(m_1\) is the heavier mass.

Notice what determines the result:

• The difference in masses determines the driving force.
• The total mass determines how difficult the system is to accelerate.

If the masses are equal, the numerator is zero, giving zero acceleration.

If the masses are unequal, the heavier side accelerates downwards.

The difference in weights is the effective driving force along the allowed motion of the connected system. It is not the ordinary vector sum of the gravitational forces on both masses.

What changes when you choose a different system?

For a block on a table connected to a hanging mass, friction becomes important.

The block’s friction is an external force that can oppose the hanging mass’s weight. The block’s normal force and weight act vertically, so they do not contribute directly along the horizontal string direction.

The general principle remains the same:

Include every external force that acts along the permitted direction of motion.

04A reliable decision-making process

Use this sequence whenever you encounter a pulley problem.

Step 1: Identify the connected objects.

Work out which objects are attached to the same string and how their movements are related.

Step 2: Choose an assumed direction.

For a simple Atwood machine, assume the heavier mass accelerates downwards.

For a more complicated arrangement, choose whichever direction seems reasonable. It does not have to be correct.

Step 3: Draw a free-body diagram for each object.

Include all relevant forces:

• Weight.
• Tension.
• Normal reaction.
• Friction.
• Any applied forces.

Step 4: Choose positive directions.

Make your positive directions consistent with your assumed motion.

For example, if you assume the left mass moves downwards, take downwards as positive for that mass and upwards as positive for the other.

Step 5: Apply Newton’s second law.

For each object, use:

\[
\sum F=ma
\]

Only include force components along the object’s direction of acceleration.

Step 6: Combine the equations.

The tension terms will often cancel, leaving an equation involving the external driving forces and the total mass.

Step 7: Interpret the sign.

• Positive acceleration: your assumed direction was correct.
• Negative acceleration: the system accelerates in the opposite direction.
• Zero acceleration: there is no acceleration under the stated conditions.

One final detail: acceleration and velocity are not necessarily in the same direction. A mass can be moving upwards while accelerating downwards. In that situation, it is slowing down.

This is the same distinction used when analysing an object after a string breaks in When a String Breaks: Which Way Does the Object Move?.

05Check the prediction experimentally: an Atwood machine

The ideal pulley model makes a prediction. We can investigate whether a real apparatus behaves approximately as expected.

Consider a school experiment using two mass hangers connected by a light string over a fixed pulley.

The pulley is attached to a clamp stand. One mass hangs on each side, and the slightly heavier mass is allowed to descend.

The prediction is straightforward: the heavier mass should accelerate downwards, while the lighter mass accelerates upwards.

What should the experiment measure?

A basic first design could use a stopwatch to measure how long the heavier mass takes to fall a known distance.

This can demonstrate the direction of motion, but manual timing introduces reaction-time errors. A small accidental push during release can also affect the result.

A better design uses two light gates.

Attach a card of known length \(L\) to the descending mass. As the card passes through each light gate, the gate measures how long the card blocks the light beam.

The instantaneous speed at each gate can be estimated using:

\[
v=\frac{L}{t}
\]

where \(t\) is the time taken for the card to pass through the gate.

If the two gates are separated by a known distance \(s\), we can calculate the acceleration using:

\[
v^2=u^2+2as
\]

Here:

• \(u\) is the speed at the first gate, in \(\mathrm{m\,s^{-1}}\).
• \(v\) is the speed at the second gate, in \(\mathrm{m\,s^{-1}}\).
• \(s\) is the distance between the gates, in metres.
• \(a\) is the acceleration, in \(\mathrm{m\,s^{-2}}\).

Comparing the theoretical and experimental results

Suppose the two hanging masses are:

• Heavier mass: \(0.600\ \mathrm{kg}\).
• Lighter mass: \(0.450\ \mathrm{kg}\).

The ideal model predicts:

\[
\begin{aligned}
a&=\frac{(0.600-0.450)9.8}
{0.600+0.450}\\
&=1.40\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

Now consider the following illustrative measurements. These are example values, not results from an actual experiment.

Calculate the experimental acceleration:

\[
\begin{aligned}
a&=\frac{v^2-u^2}{2s}\\
&=\frac{1.08^2-0.60^2}{2(0.300)}\\
&=\frac{1.1664-0.3600}{0.600}\\
&=1.34\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

The measured acceleration is approximately \(1.34\ \mathrm{m\,s^{-2}}\), compared with the theoretical prediction of \(1.40\ \mathrm{m\,s^{-2}}\).

The difference is small. It could be explained by friction in the pulley bearings, the pulley having some rotational inertia, or measurement uncertainty.

The revised light-gate design improves the timing measurement, but it does not remove every non-ideal feature of the apparatus.

The experiment supports the ideal model as a useful approximation. It does not prove that a real pulley is perfectly frictionless or massless.

08Questions and solutions

The following questions build from a standard Atwood machine to systems involving initial velocity, inclined planes, friction, and a movable pulley.

Question 1

Two masses of \(0.50\ \mathrm{kg}\) and \(0.30\ \mathrm{kg}\) are connected by a light, inextensible string passing over a fixed, ideal pulley.

The system is released from rest.

Calculate:

a) Which mass accelerates downwards.

b) The magnitude of the acceleration.

Take \(g=9.8\ \mathrm{m\,s^{-2}}\).

Solution 1

The \(0.50\ \mathrm{kg}\) mass accelerates downwards at \(2.45\ \mathrm{m\,s^{-2}}\), while the \(0.30\ \mathrm{kg}\) mass accelerates upwards.

The heavier mass experiences the greater gravitational force, so assume it accelerates downwards.

Let \(T\) be the tension and \(a\) the acceleration magnitude.

For the heavier mass:

\[
0.50g-T=0.50a
\]

For the lighter mass:

\[
T-0.30g=0.30a
\]

Add the equations:

\[
(0.50-0.30)g=(0.50+0.30)a
\]

Rearrange:

\[
a=\frac{(0.50-0.30)9.8}{0.80}
\]

\[
\boxed{a=2.45\ \mathrm{m\,s^{-2}}}
\]

The result is positive, confirming the assumed direction.

The heavier mass accelerates downwards, and the lighter mass accelerates upwards.

Question 2

Two masses of \(0.40\ \mathrm{kg}\) and \(0.55\ \mathrm{kg}\) are connected by a light, inextensible string over a fixed, ideal pulley.

At a particular instant, the \(0.40\ \mathrm{kg}\) mass is moving downwards at \(0.80\ \mathrm{m\,s^{-1}}\). The \(0.55\ \mathrm{kg}\) mass is therefore moving upwards.

Ignoring air resistance, determine:

a) The direction of acceleration of each mass.

b) The time taken for the system to momentarily stop.

c) The distance travelled by the \(0.40\ \mathrm{kg}\) mass before it stops.

Solution 2

The \(0.40\ \mathrm{kg}\) mass accelerates upwards, opposite to its current motion. It slows down and momentarily stops after approximately \(0.52\ \mathrm{s}\), having travelled \(0.21\ \mathrm{m}\) downwards.

The key is to distinguish velocity from acceleration. The heavier mass determines the direction of acceleration, not necessarily the direction in which the masses are currently travelling.

Part a: Determine the acceleration.

The \(0.55\ \mathrm{kg}\) mass is heavier, so it accelerates downwards.

The \(0.40\ \mathrm{kg}\) mass therefore accelerates upwards.

Calculate the acceleration:

\[
\begin{aligned}
a&=\frac{(0.55-0.40)g}{0.55+0.40}\\
&=\frac{0.15\times9.8}{0.95}\\
&=1.55\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

Part b: Calculate the stopping time.

The \(0.40\ \mathrm{kg}\) mass is initially moving downwards, but its acceleration is upwards.

Choose downwards as positive for this mass:

• Initial velocity: \(u=+0.80\ \mathrm{m\,s^{-1}}\).
• Acceleration: \(a=-1.55\ \mathrm{m\,s^{-2}}\).
• Final velocity: \(v=0\).

Use:

\[
v=u+at
\]

Substitute:

\[
0=0.80-1.55t
\]

Rearrange:

\[
t=\frac{0.80}{1.55}
\]

\[
\boxed{t=0.52\ \mathrm{s}}
\]

Part c: Calculate the distance travelled.

Use:

\[
v^2=u^2+2as
\]

Substitute:

\[
0=(0.80)^2+2(-1.55)s
\]

Rearrange:

\[
s=\frac{0.80^2}{2(1.55)}
\]

\[
\boxed{s=0.21\ \mathrm{m}}
\]

The \(0.40\ \mathrm{kg}\) mass travels \(0.21\ \mathrm{m}\) downwards before momentarily stopping.

After stopping, it reverses direction because the \(0.55\ \mathrm{kg}\) mass continues to accelerate downwards.

This question demonstrates why the direction of velocity cannot be used to determine the direction of acceleration.

Question 3

A \(2.0\ \mathrm{kg}\) block rests on a smooth inclined plane at an angle of \(30^\circ\) to the horizontal.

The block is connected by a light, inextensible string over an ideal pulley to a hanging mass of \(1.2\ \mathrm{kg}\).

The system is released from rest.

Calculate:

a) Which object accelerates downwards.

b) The magnitude of the acceleration.

c) The tension in the string.

Take \(g=9.8\ \mathrm{m\,s^{-2}}\).

Solution 3

The \(1.2\ \mathrm{kg}\) hanging mass accelerates downwards, while the \(2.0\ \mathrm{kg}\) block accelerates up the incline. The acceleration magnitude is \(0.613\ \mathrm{m\,s^{-2}}\), and the tension is \(11.0\ \mathrm{N}\).

The tempting mistake is to compare the total masses and conclude that the \(2.0\ \mathrm{kg}\) block must move down the incline.

However, only the component of its weight parallel to the incline contributes to motion along the string.

Part a: Determine the direction.

The gravitational force acting down the incline is:

\[
F_{\parallel}=m_Ag\sin\theta
\]

where:

• \(m_A=2.0\ \mathrm{kg}\) is the mass on the incline.
• \(\theta=30^\circ\) is the angle of the incline.

Substitute:

\[
\begin{aligned}
F_{\parallel}
&=2.0\times9.8\times\sin30^\circ\\
&=9.8\ \mathrm{N}
\end{aligned}
\]

The hanging mass has a weight of:

\[
F_g=1.2\times9.8=11.76\ \mathrm{N}
\]

Since \(11.76\ \mathrm{N}\) is greater than \(9.8\ \mathrm{N}\), the hanging mass accelerates downwards.

Part b: Calculate the acceleration.

Assume the hanging mass accelerates downwards and the block accelerates up the incline.

For the block on the incline:

\[
T-m_Ag\sin\theta=m_Aa
\]

For the hanging mass:

\[
m_Bg-T=m_Ba
\]

Add the equations:

\[
m_Bg-m_Ag\sin\theta
=(m_A+m_B)a
\]

Rearrange:

\[
a=\frac{m_Bg-m_Ag\sin\theta}
{m_A+m_B}
\]

Substitute:

\[
\begin{aligned}
a&=\frac{1.2(9.8)-2.0(9.8)\sin30^\circ}
{2.0+1.2}\\
&=\frac{11.76-9.80}{3.20}\\
&=0.6125\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

Therefore:

\[
\boxed{a=0.613\ \mathrm{m\,s^{-2}}}
\]

Part c: Calculate the tension.

Use the equation for the hanging mass:

\[
m_Bg-T=m_Ba
\]

Rearrange:

\[
T=m_B(g-a)
\]

Substitute:

\[
\begin{aligned}
T&=1.2(9.8-0.6125)\\
&=11.025\ \mathrm{N}
\end{aligned}
\]

Therefore:

\[
\boxed{T=11.0\ \mathrm{N}}
\]

The important lesson is that you must compare force components along the direction of motion, not simply compare the masses.

Question 4

A \(2.0\ \mathrm{kg}\) block, labelled A, rests on a rough horizontal table. It is connected over an ideal pulley to a hanging mass B of \(0.40\ \mathrm{kg}\).

The coefficients of friction between block A and the table are:

\[
\mu_s=0.35
\]

\[
\mu_k=0.25
\]

At a particular instant, block A is moving to the left, away from the pulley, at \(0.80\ \mathrm{m\,s^{-1}}\). Mass B is therefore moving upwards at the same speed.

Determine:

a) The direction and magnitude of acceleration.

b) The time taken for the system to momentarily stop.

c) The distance travelled by block A before stopping.

d) Whether the system starts moving again after stopping or remains stationary.

Take \(g=9.8\ \mathrm{m\,s^{-2}}\).

Solution 4

Block A accelerates to the right, towards the pulley, at \(3.68\ \mathrm{m\,s^{-2}}\). It initially continues moving left but slows down. It stops after approximately \(0.218\ \mathrm{s}\), having travelled \(0.087\ \mathrm{m}\) to the left. The system then remains stationary because static friction is sufficient to prevent it from reversing.

This problem requires careful treatment of friction because the system is already moving.

Part a: Determine the acceleration.

Choose the initial direction of motion as positive:

• Block A: left is positive.
• Mass B: upwards is positive.

Because block A is sliding left, kinetic friction acts to the right.

The tension also pulls block A towards the pulley, so it acts to the right.

For block A:

\[
-f_k-T=m_Aa
\]

For mass B:

\[
T-m_Bg=m_Ba
\]

First calculate the kinetic friction.

The normal reaction is:

\[
N=m_Ag
\]

Therefore:

\[
\begin{aligned}
f_k&=\mu_kN\\
&=0.25(2.0\times9.8)\\
&=4.90\ \mathrm{N}
\end{aligned}
\]

Add the two equations:

\[
-f_k-m_Bg=(m_A+m_B)a
\]

The tension cancels.

Substitute:

\[
\begin{aligned}
a&=\frac{-(4.90)-(0.40\times9.8)}
{2.0+0.40}\\
&=\frac{-4.90-3.92}{2.40}\\
&=-3.675\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

Therefore:

\[
\boxed{a=-3.68\ \mathrm{m\,s^{-2}}}
\]

The negative sign means the acceleration is opposite to the chosen positive direction.

Block A accelerates right, towards the pulley, while mass B accelerates downwards.

Part b: Calculate the stopping time.

For block A, the initial velocity is:

\[
u=0.80\ \mathrm{m\,s^{-1}}
\]

Its acceleration is:

\[
a=-3.675\ \mathrm{m\,s^{-2}}
\]

At the instant it stops:

\[
v=0
\]

Using:

\[
v=u+at
\]

Substitute:

\[
0=0.80-3.675t
\]

Rearrange:

\[
t=\frac{0.80}{3.675}
\]

\[
\boxed{t=0.218\ \mathrm{s}}
\]

Part c: Calculate the stopping distance.

Use:

\[
v^2=u^2+2as
\]

Substitute:

\[
0=(0.80)^2+2(-3.675)s
\]

Rearrange:

\[
s=\frac{0.80^2}{2(3.675)}
\]

\[
\boxed{s=0.087\ \mathrm{m}}
\]

Block A travels approximately \(8.7\ \mathrm{cm}\) to the left before stopping.

Part d: Determine what happens after stopping.

This is the part that requires the most care.

Once block A stops, kinetic friction no longer applies. We must determine whether static friction can hold the system stationary.

If the system is to remain at rest, the tension must balance the hanging mass’s weight:

\[
T=m_Bg
\]

Therefore:

\[
T=0.40\times9.8=3.92\ \mathrm{N}
\]

Static friction must balance this tension on block A.

The maximum available static friction is:

\[
\begin{aligned}
f_{s,\max}
&=\mu_sN\\
&=0.35(2.0\times9.8)\\
&=6.86\ \mathrm{N}
\end{aligned}
\]

Since the required static friction of \(3.92\ \mathrm{N}\) is less than the maximum available \(6.86\ \mathrm{N}\), static friction can hold the system.

Therefore, the system remains stationary after stopping.

Notice that friction changes direction here. While block A slides left, kinetic friction acts right. Once it stops, static friction acts left to prevent the block from being pulled towards the pulley.

This is why you must reconsider the forces whenever the motion changes state.

Question 5

A pulley system is arranged as follows:

• One end of a light, inextensible string is fixed to the ceiling.
• The string passes downwards around a light movable pulley attached to a \(1.5\ \mathrm{kg}\) mass, labelled M.
• The string then travels upwards over a fixed, ideal pulley.
• The free end of the string supports a hanging \(1.0\ \mathrm{kg}\) mass, labelled m.

Both pulleys are ideal, the movable pulley has negligible mass, and the string remains taut.

The system is released from rest.

Determine:

a) Which mass accelerates downwards.

b) The acceleration magnitude of each mass.

c) Why comparing the two hanging masses directly gives the wrong prediction.

Take \(g=9.8\ \mathrm{m\,s^{-2}}\).

Solution 5

The \(1.0\ \mathrm{kg}\) mass accelerates downwards at approximately \(1.78\ \mathrm{m\,s^{-2}}\), while mass M accelerates upwards at approximately \(0.89\ \mathrm{m\,s^{-2}}\).

The tempting prediction is that the \(1.5\ \mathrm{kg}\) mass should descend because it is heavier than the \(1.0\ \mathrm{kg}\) mass.

That prediction is wrong because the movable pulley is supported by two sections of string, not one. The force and acceleration relationships are therefore different from those in a simple Atwood machine.

Part a: Establish the relationship between the movements.

Let mass M move downwards by a distance \(x\).

The two string sections supporting the movable pulley each become longer by \(x\).

Therefore, the final hanging section must become shorter by \(2x\) to maintain the total string length.

This means mass m must move upwards by \(2x\).

Consequently, their acceleration magnitudes satisfy:

\[
a_m=2a_M
\]

where:

• \(a_M\) is the acceleration magnitude of the \(1.5\ \mathrm{kg}\) mass.
• \(a_m\) is the acceleration magnitude of the \(1.0\ \mathrm{kg}\) mass.

Their acceleration directions are opposite.

Part b: Calculate the accelerations.

Assume mass M accelerates downwards.

Take downwards as positive for M and upwards as positive for m.

Let \(T\) be the tension in the string.

The movable pulley is supported by two sections of string, each exerting an upward tension \(T\).

Therefore, the total upward force on the movable pulley is:

\[
F_T=2T
\]

For mass M:

\[
1.5g-2T=1.5a_M
\]

For mass m:

\[
T-1.0g=1.0(2a_M)
\]

Rearrange the second equation:

\[
T=g+2a_M
\]

Substitute this into the first equation:

\[
1.5g-2(g+2a_M)=1.5a_M
\]

Expand:

\[
1.5g-2g-4a_M=1.5a_M
\]

Collect terms:

\[
-0.5g=5.5a_M
\]

Therefore:

\[
a_M=\frac{-0.5\times9.8}{5.5}
\]

\[
a_M=-0.891\ \mathrm{m\,s^{-2}}
\]

The negative result means our assumed direction for M was incorrect.

Mass M actually accelerates upwards at:

\[
\boxed{a_M=0.89\ \mathrm{m\,s^{-2}}}
\]

Since the other mass has twice the acceleration magnitude:

\[
a_m=2(0.891)
\]

\[
\boxed{a_m=1.78\ \mathrm{m\,s^{-2}}}
\]

Mass m accelerates downwards.

Part c: Explain why the simple heavier-mass rule fails.

For a fixed pulley connecting two hanging masses, the heavier mass wins because each mass is supported by one section of string.

Here, mass M is supported by two sections of string.

The two upward tensions combine to produce a total upward force of \(2T\).

The movable pulley also changes the displacement relationship: mass M moves half the distance travelled by mass m.

A useful way to check the direction is through gravitational potential energy.

If M moves downwards by a distance \(x\):

• M loses gravitational potential energy of \(1.5gx\).
• Mass m rises by \(2x\), gaining gravitational potential energy of \(2.0gx\).

The net change is:

\[
\Delta E_p=-1.5gx+2.0gx
\]

\[
\Delta E_p=0.5gx
\]

The system would gain gravitational potential energy if M moved downwards. With no external energy input, that cannot be the direction of spontaneous acceleration from rest.

Instead, M accelerates upwards and m accelerates downwards.

The deeper lesson is that you cannot decide the direction of acceleration by comparing masses alone. You must first understand how the string constrains their movements and how many string sections support each movable pulley.

09The key decision to remember

For a simple fixed-pulley system, the heavier hanging mass accelerates downwards.

For other pulley arrangements, use a more general method:

1. Identify the forces acting on each object.
2. Establish the relationship between the connected objects’ accelerations.
3. Choose an assumed direction.
4. Apply Newton’s second law to each object.
5. Combine the equations and interpret the sign of the acceleration.

Pay particular attention to friction, inclined-plane components, and movable pulleys. These are the situations where the simple rule of comparing masses can lead you astray.

Once you can establish the acceleration direction correctly, calculating tension becomes much more straightforward. The next step is to use the same free-body diagrams to investigate how the tension changes when the pulley has rotational inertia or when the string is no longer taut.