Why Tension Cancels When You Add Connected-Body Equations

Two blocks are connected by a light rope over a pulley. One block accelerates horizontally, while the other accelerates vertically. If you write Newton's second law for each block, both equations contain tension, \(T\). Yet, when you add the equations, the tension disappears.

Why?

Tension hasn't stopped acting, and it certainly isn't zero. It cancels because an ideal rope constrains the connected bodies to move together, while transmitting the same magnitude of tension throughout the rope. When the equations are written using consistent signs along each body's direction of motion, the tension contributions have opposite signs.

The important idea is that you can treat connected bodies as one combined system. This lets you find the acceleration without first calculating the tension. There is also a subtle distinction between cancelling tension in the equations and claiming that the tension forces are always equal and opposite vectors. We'll work through both.

## Why tension disappears when you add the equations

Consider a simple pulley system with two hanging masses. Mass \(m_1\) is heavier than mass \(m_2\), so \(m_1\) accelerates downwards and \(m_2\) accelerates upwards.

[IMAGE: An annotated Atwood machine with two unequal hanging blocks connected by one continuous light rope over a fixed pulley. Label the masses m1 and m2, with m1 heavier, show equal-magnitude tension T acting upwards on both blocks, gravitational forces m1g and m2g downwards, and acceleration arrows showing m1 down and m2 up. Include the pulley axle and rope arrangement clearly.]

Before writing any equations, make a prediction: if tension pulls upwards on both masses, why would its contributions cancel?

The answer comes from choosing a positive direction for each mass. Choose downwards as positive for \(m_1\), and upwards as positive for \(m_2\). These choices match the direction of acceleration.

For the heavier mass, gravity acts in the positive direction and tension acts in the negative direction:

\[
m_1g-T=m_1a
\]

For the lighter mass, tension acts in the positive direction and gravity acts in the negative direction:

\[
T-m_2g=m_2a
\]

Now add the equations:

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

The two tension terms cancel:

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

Therefore,

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

Notice what has happened. We have found the acceleration of the system without knowing the tension.

This works because the ideal rope has the same tension magnitude throughout, and both masses have the same acceleration magnitude. The tension terms enter the two equations with opposite signs because the masses accelerate in opposite directions.

First, the assumptions behind the cancellation

The cancellation depends on the rope and pulley behaving ideally. For the standard HSC fixed-pulley model, use these assumptions:

These assumptions give us two important relationships:

\[ T_1=T_2=T \]

and, for a simple fixed pulley,

\[ |a_1|=|a_2|=a \]

Here, \(T\) is the tension in newtons (N), and \(a\) is the common acceleration magnitude in metres per second squared (\(\mathrm{m\\,s^{-2}}\)).

Notice that these are two different conditions. Equal tension allows us to use the same \(T\) in both equations. The inextensible rope allows us to use the same acceleration magnitude.

You need both for the familiar cancellation.

The important distinction: tension does not have to cancel as a vector

Consider a slightly different arrangement. A trolley sits on a smooth horizontal track and connects to a hanging mass through a fixed pulley.

The trolley accelerates towards the pulley, while the hanging mass accelerates downwards.

Take right as positive for the trolley and down as positive for the hanging mass.

For the trolley:

\[ T=m_Aa \]

For the hanging mass:

\[ m_Bg-T=m_Ba \]

Adding:

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

Again, tension disappears.

But look carefully at the physical forces. Tension pulls the trolley horizontally towards the pulley and the hanging mass vertically upwards.

Those two tension vectors are perpendicular. They are certainly not equal and opposite vectors.

So why does the algebra still work?

Because we are adding scalar equations written along each object’s allowed direction of motion. We are not adding two force vectors in a single shared coordinate direction.

There is another way to understand this. Suppose the trolley moves right by a distance \(s\). The hanging mass must move down by the same distance.

• Tension does positive work \(+Ts\) on the trolley.
• Tension does negative work \(-Ts\) on the hanging mass.

The total work done by tension is therefore:

\[ W_T=Ts-Ts=0 \]

Tension transfers energy between the two moving objects, but does no net work on them together in this ideal 1:1 arrangement.

This is the deeper reason the tension contributions disappear.

Internal versus external forces: the system boundary matters

There is a subtle point in the language used to describe these problems.

An internal force acts between components inside a chosen system. An external force acts on that system from outside its boundary.

Whether tension is internal or external therefore depends on what you include in your system.

In HSC solutions, you will often hear that tension is an internal force that cancels when you consider the connected bodies together. This is useful shorthand.

However, don’t confuse it with Newton’s third law. The tension forces acting on two connected masses are not necessarily an equal-and-opposite action-reaction pair.

For example, in an Atwood machine, tension acts upwards on both masses. The forces point in the same physical direction, but have opposite signs in the two equations because the positive directions were chosen differently.

The cancellation comes from the rope constraint, not from blindly applying Newton’s third law.

01How can we test this experimentally?

The cancellation isn’t just an algebraic trick. We can investigate it using a trolley, a hanging mass, and a pulley.

Setting up the experiment

Start with a horizontal, low-friction track. Place a trolley on the track and attach one end of a light string to it.

Run the string towards a pulley at the edge of the track, over the pulley, and down to a hanging mass.

Attach a motion sensor behind the trolley to record its position as it moves towards the pulley. An inline force sensor between the trolley and string can measure the tension.

The mass of the trolley, including the force sensor, and the hanging mass are measured using a balance.

Release the system from rest.

The motion sensor provides position-time data. From this, we can determine the acceleration. The inline force sensor provides a measurement of \(T\).

What should the experiment show?

If the track and pulley were perfectly frictionless, the equations would be:

For the trolley:

\[ T=m_Aa \]

For the hanging mass:

\[ m_Bg-T=m_Ba \]

Adding them gives:

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

The acceleration should therefore depend on the hanging mass’s weight and the total moving mass, rather than on the tension.

A basic experimental design using a stopwatch to time a short fall would be less convincing. Human reaction time makes the acceleration difficult to measure accurately, and the experiment would not directly measure tension.

A better design uses the motion sensor and inline force sensor. A low-friction pulley reduces unwanted resistance, and including the force sensor’s mass in the trolley-side mass prevents an avoidable error.

An illustrative measurement

Suppose the following values are obtained in a hypothetical trial:

ILLUSTRATIVE DATA – NOT MEASURED EXPERIMENTAL RESULTS

The trolley experiences a small resistive force, \(f\), opposing its motion.

Its equation becomes:

\[ T-f=m_Aa \]

Rearrange to calculate the resistance:

\[ \begin{aligned} f&=T-m_Aa\\\ &=1.58-(0.80\times1.90)\\\ &=1.58-1.52\\\ &=0.06\ \mathrm{N} \end{aligned} \]

Now check the hanging mass:

\[ \begin{aligned} m_Bg-T &=(0.20\times9.8)-1.58\\\ &=1.96-1.58\\\ &=0.38\ \mathrm{N} \end{aligned} \]

And:

\[ m_Ba=0.20\times1.90=0.38\ \mathrm{N} \]

Both individual equations are satisfied.

Now add them:

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

The tension cancels, leaving:

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

Substitute:

\[ \begin{aligned} 1.96-0.06 &=(0.80+0.20)\times1.90\\\ 1.90&=1.90\ \mathrm{N} \end{aligned} \]

The result is consistent with the model.

Notice that the resistive force remains. It is an external force acting on the trolley, so it does not disappear when we combine the equations.

In a perfectly frictionless system, the predicted acceleration would be:

\[ a=\frac{0.20\times9.8}{0.80+0.20} =1.96\ \mathrm{m\\,s^{-2}} \]

The slightly smaller illustrative measurement is consistent with the small resistance included in our revised model.

What could prevent tension from cancelling?

The experiment also highlights the limitations of the ideal model.

Consider three situations:

The last two cases are important. Equal tension does not automatically mean you can add any two equations and expect the tension to disappear.

You must also account for the geometry of the rope and the motion it permits.

02A reliable method for cancelling tension

When you encounter a connected-body problem, use this sequence.

1. Choose your system. Identify which objects are connected and which external forces act on them.
2. Draw separate free-body diagrams. Include tension on every object pulled by the rope.
3. Choose positive directions. Ideally, choose them along the assumed acceleration of each object.
4. Write Newton’s second law separately. Use the correct forces and signs for each body.
5. Apply the rope constraint. For a simple fixed pulley, use the same tension magnitude and acceleration magnitude.
6. Add the equations. Check that the tension terms have opposite signs.
7. Inspect what remains. Gravity, friction, and applied forces generally remain because they are external driving or resisting forces.

If tension does not cancel, check four things:

• Have you used the same tension magnitude where the pulley model permits it?
• Have you chosen consistent positive directions?
• Does the rope actually give the objects equal acceleration magnitudes?
• Are there multiple rope segments pulling on a movable object?

For a fuller explanation of drawing separate free-body diagrams and setting up ordinary connected-mass problems, see Connected Masses and Pulleys: Forces, Tension, and Friction.

03Worked examples

04Questions and solutions

Question 1

Two masses of \(3.0\ \mathrm{kg}\) and \(2.0\ \mathrm{kg}\) are connected by a light, inextensible rope passing over an ideal fixed pulley. The system is released from rest.

Calculate:

a) The acceleration of the system.

b) The tension in the rope.

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

Solution 1

The \(3.0\ \mathrm{kg}\) mass accelerates downwards at \(1.96\ \mathrm{m\\,s^{-2}}\), and the rope tension is \(23.5\ \mathrm{N}\).

Step 1: Choose positive directions.

Take downwards as positive for the \(3.0\ \mathrm{kg}\) mass and upwards as positive for the \(2.0\ \mathrm{kg}\) mass.

Let the common acceleration magnitude be \(a\).

Step 2: Write Newton’s second law for each mass.

For the heavier mass:

\[ 3.0g-T=3.0a \]

For the lighter mass:

\[ T-2.0g=2.0a \]

Step 3: Add the equations.

\[ \begin{aligned} 3.0g-T+T-2.0g &=3.0a+2.0a\\\ g&=5.0a \end{aligned} \]

Therefore:

\[ a=\frac{9.8}{5.0} \]

\[ \boxed{a=1.96\ \mathrm{m\\,s^{-2}}} \]

The positive result confirms our assumed directions.

Step 4: Calculate tension.

Use the equation for the lighter mass:

\[ T-2.0g=2.0a \]

Rearrange:

\[ T=2.0(g+a) \]

Substitute:

\[ \begin{aligned} T&=2.0(9.8+1.96)\\\ &=23.52\ \mathrm{N} \end{aligned} \]

\[ \boxed{T=23.5\ \mathrm{N}} \]

The tension disappears when the equations are added because the ideal rope transmits the same tension magnitude to both masses, while their positive directions are opposite.

Question 2

Two masses of \(4.0\ \mathrm{kg}\) and \(3.0\ \mathrm{kg}\) are connected over an ideal fixed pulley.

The system is already moving. At the instant considered, the \(4.0\ \mathrm{kg}\) mass is moving upwards at \(1.5\ \mathrm{m\\,s^{-1}}\), while the \(3.0\ \mathrm{kg}\) mass moves downwards.

Calculate:

a) The acceleration of the system.

b) The tension in the rope.

c) The velocity and direction of the \(4.0\ \mathrm{kg}\) mass after \(1.5\ \mathrm{s}\), assuming the acceleration remains constant.

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

Solution 2

The \(4.0\ \mathrm{kg}\) mass accelerates downwards at \(1.4\ \mathrm{m\\,s^{-2}}\), the tension is \(33.6\ \mathrm{N}\), and after \(1.5\ \mathrm{s}\), it moves downwards at \(0.60\ \mathrm{m\\,s^{-1}}\).

The important distinction is that velocity and acceleration do not have to point in the same direction.

Step 1: Choose positive directions.

Take downwards as positive for the \(4.0\ \mathrm{kg}\) mass and upwards as positive for the \(3.0\ \mathrm{kg}\) mass.

The heavier mass has the greater weight, so the net acceleration is downwards for that mass.

Step 2: Write the two equations.

For the \(4.0\ \mathrm{kg}\) mass:

\[ 4.0g-T=4.0a \]

For the \(3.0\ \mathrm{kg}\) mass:

\[ T-3.0g=3.0a \]

Step 3: Add the equations to eliminate tension.

\[ \begin{aligned} 4.0g-3.0g&=7.0a\\\ g&=7.0a\\\ a&=\frac{9.8}{7.0} \end{aligned} \]

Therefore:

\[ \boxed{a=1.4\ \mathrm{m\\,s^{-2}}} \]

The \(4.0\ \mathrm{kg}\) mass accelerates downwards.

Step 4: Calculate tension.

Substitute into the equation for the heavier mass:

\[ 4.0g-T=4.0a \]

\[ \begin{aligned} T&=4.0(9.8-1.4)\\\ &=4.0(8.4)\\\ &=33.6\ \mathrm{N} \end{aligned} \]

\[ \boxed{T=33.6\ \mathrm{N}} \]

Step 5: Determine the velocity after \(1.5\ \mathrm{s}\).

This is where the initial direction of motion matters.

The \(4.0\ \mathrm{kg}\) mass initially moves upwards, but its acceleration is downwards.

Taking downwards as positive:

\[ u=-1.5\ \mathrm{m\\,s^{-1}} \]

Use the constant-acceleration equation:

\[ v=u+at \]

Substitute:

\[ \begin{aligned} v&=-1.5+(1.4\times1.5)\\\ &=-1.5+2.1\\\ &=0.60\ \mathrm{m\\,s^{-1}} \end{aligned} \]

The positive result means the mass is now moving downwards.

It initially slows down, stops momentarily after approximately \(1.07\ \mathrm{s}\), and then reverses direction.

The cancellation of tension is unaffected by the initial velocity. Newton’s second law relates net force to acceleration, not velocity.

Question 3

A \(3.0\ \mathrm{kg}\) block rests on a rough horizontal table. It is connected over an ideal fixed pulley to a hanging mass of \(1.0\ \mathrm{kg}\).

The coefficients of friction are:

• Coefficient of static friction: \(\mu_s=0.40\)
• Coefficient of kinetic friction: \(\mu_k=0.25\)

Answer the following:

a) If the system is released from rest, will it begin moving?

b) A brief external push then starts the system sliding, with the hanging mass moving downwards. Once the push is removed, calculate the acceleration and tension while the system continues sliding.

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

Solution 3

The system remains stationary when released from rest. However, once it is pushed into motion with the hanging mass descending, it accelerates downwards at \(0.613\ \mathrm{m\\,s^{-2}}\), with a tension of \(9.19\ \mathrm{N}\).

Step 1: Calculate the normal force.

For the block on the horizontal table:

\[ N=m_Ag \]

\[ N=3.0\times9.8 \]

\[ N=29.4\ \mathrm{N} \]

Step 2: Determine whether static friction can hold the system.

The hanging mass has a weight of:

\[ F_g=m_Bg \]

\[ F_g=1.0\times9.8=9.8\ \mathrm{N} \]

If the system remains stationary, the tension must balance the hanging mass’s weight.

Therefore, the static friction required to hold the system is also \(9.8\ \mathrm{N}\).

However, the maximum available static friction is:

\[ f_{s,\max}=\mu_sN \]

\[ f_{s,\max}=0.40\times29.4 \]

\[ f_{s,\max}=11.76\ \mathrm{N} \]

Since:

\[ 9.8<11.76 \]

static friction is sufficient to prevent motion.

The system remains stationary.

Notice that static friction does not automatically equal its maximum value. It adjusts to the force required to prevent slipping.

Step 3: Calculate kinetic friction once the system is sliding.

After the external push starts the motion, kinetic friction acts on the block.

\[ f_k=\mu_kN \]

\[ \begin{aligned} f_k&=0.25\times29.4\\\ &=7.35\ \mathrm{N} \end{aligned} \]

Step 4: Write Newton’s second law for both objects.

Take right as positive for the block and downwards as positive for the hanging mass.

For the block:

\[ T-7.35=3.0a \]

For the hanging mass:

\[ 9.8-T=1.0a \]

Step 5: Add the equations.

\[ \begin{aligned} T-7.35+9.8-T&=3.0a+1.0a\\\ 2.45&=4.0a\\\ a&=\frac{2.45}{4.0} \end{aligned} \]

Therefore:

\[ \boxed{a=0.6125\ \mathrm{m\\,s^{-2}}} \]

Step 6: Calculate tension.

Use the hanging mass equation:

\[ 9.8-T=1.0(0.6125) \]

Rearrange:

\[ \begin{aligned} T&=9.8-0.6125\\\ &=9.1875\ \mathrm{N} \end{aligned} \]

\[ \boxed{T=9.19\ \mathrm{N}} \]

The key lesson is that friction remains in the combined equation because it is an external force. Tension cancels because it is transmitted through the ideal rope and has opposite signed contributions in the two equations.

The change from static to kinetic friction also explains why the system can remain stationary when released but accelerate once it has been pushed into motion.

Question 4

Two blocks, A and B, have masses of \(2.0\ \mathrm{kg}\) and \(3.0\ \mathrm{kg}\), respectively. They are connected by a light, inextensible rope on a rough horizontal surface.

A horizontal force of \(20\ \mathrm{N}\) pulls block A to the left. Both blocks slide to the left.

The kinetic friction forces are:

• Block A: \(2.0\ \mathrm{N}\)
• Block B: \(3.0\ \mathrm{N}\)

Calculate:

a) The acceleration of both blocks.

b) The tension in the rope.

c) Explain why tension cancels even though the blocks are accelerating in the same physical direction.

Solution 4

Both blocks accelerate leftwards at \(3.0\ \mathrm{m\\,s^{-2}}\), and the rope tension is \(12\ \mathrm{N}\).

Step 1: Choose a positive direction.

Take left as positive for both blocks.

The applied force acts leftwards, while friction opposes the motion and acts rightwards.

Tension pulls A towards B and B towards A.

Step 2: Write Newton’s second law for block A.

The forces on A are:

• Applied force: \(20\ \mathrm{N}\) left.
• Friction: \(2.0\ \mathrm{N}\) right.
• Tension: \(T\) right.

Therefore:

\[ 20-2.0-T=2.0a \]

Step 3: Write Newton’s second law for block B.

Tension pulls B leftwards, while friction acts rightwards:

\[ T-3.0=3.0a \]

Step 4: Add the equations.

\[ \begin{aligned} 20-2.0-T+T-3.0 &=2.0a+3.0a\\\ 15&=5.0a\\\ a&=\frac{15}{5.0} \end{aligned} \]

Therefore:

\[ \boxed{a=3.0\ \mathrm{m\\,s^{-2}}} \]

Step 5: Calculate tension.

Use the equation for block B:

\[ T-3.0=3.0(3.0) \]

\[ \begin{aligned} T&=9.0+3.0\\\ &=12\ \mathrm{N} \end{aligned} \]

\[ \boxed{T=12\ \mathrm{N}} \]

Step 6: Explain the cancellation.

Both blocks accelerate in the same physical direction, but tension acts in opposite directions on them.

It pulls A to the right and B to the left.

Because we chose left as positive for both blocks, tension has a negative sign in A’s equation and a positive sign in B’s equation.

When the equations are added, these contributions cancel.

The applied force and both friction forces remain because they are external forces acting on the two-block system.

This example shows that you should never decide the sign of tension simply by looking at whether the connected objects move in the same or opposite directions. Look at the force direction relative to your chosen positive direction for each object.

Question 5

Extension challenge

A \(1.0\ \mathrm{kg}\) block A sits on a smooth horizontal table. It is connected to a \(2.0\ \mathrm{kg}\) hanging load B using a light, inextensible rope.

The rope is arranged as follows:

• One end is fixed to the ceiling.
• The rope runs down around a movable pulley attached to B.
• It then runs upwards over a fixed pulley at the edge of the table.
• Finally, the rope runs horizontally from the fixed pulley to block A.

Both pulleys are ideal and massless. The rope remains taut.

The load B is released from rest and moves downwards.

Answer the following:

a) Determine the relationship between the acceleration magnitudes of A and B.

b) Calculate the acceleration of each mass.

c) Calculate the tension in the rope.

d) Explain why simply adding the two unmodified Newton’s second law equations does not eliminate tension.

e) If B descends \(0.10\ \mathrm{m}\), calculate the total work done by tension on A and B.

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

Solution 5

The hanging load B accelerates downwards at \(3.27\ \mathrm{m\\,s^{-2}}\), while block A accelerates rightwards at \(6.53\ \mathrm{m\\,s^{-2}}\). The rope tension is \(6.53\ \mathrm{N}\), and the total work done by tension on the two moving bodies is zero.

This problem exposes an important limitation of the simple cancellation rule: equal tension does not guarantee equal acceleration magnitudes.

Step 1: Establish the relationship between the accelerations.

When B moves down by a distance \(s\), both vertical rope segments supporting its movable pulley become longer by \(s\).

The total increase in vertical rope length is:

\[ s+s=2s \]

Because the rope has a fixed total length, its horizontal section must shorten by \(2s\).

Therefore, A moves towards the pulley by twice the distance that B moves down.

The displacement relationship is:

\[ s_A=2s_B \]

Differentiating with respect to time gives the velocity relationship:

\[ v_A=2v_B \]

Differentiating again gives:

\[ \boxed{a_A=2a_B} \]

This is the crucial constraint.

Step 2: Identify the forces on each object.

Block A experiences:

• Tension \(T\) horizontally towards the pulley.
• Normal force upwards.
• Weight downwards.

The vertical forces balance because A has no vertical acceleration.

For the hanging load B, the movable pulley is pulled upwards by two sections of rope.

Each section has tension \(T\), so the total upward force is:

\[ F_T=2T \]

The load also experiences its weight \(m_Bg\) downwards.

Step 3: Write Newton’s second law for block A.

Take right as positive:

\[ T=m_Aa_A \]

Substitute \(m_A=1.0\ \mathrm{kg}\):

\[ T=1.0a_A \]

Using the acceleration constraint:

\[ \boxed{T=2.0a_B} \]

Step 4: Write Newton’s second law for load B.

Take downwards as positive.

Gravity acts downwards, while the two tension forces act upwards:

\[ m_Bg-2T=m_Ba_B \]

Substitute \(m_B=2.0\ \mathrm{kg}\):

\[ 19.6-2T=2.0a_B \]

Step 5: Eliminate tension using the acceleration constraint.

From block A:

\[ T=2.0a_B \]

Substitute into the equation for B:

\[ 19.6-2(2.0a_B)=2.0a_B \]

Expand:

\[ 19.6-4.0a_B=2.0a_B \]

Rearrange:

\[ 19.6=6.0a_B \]

Therefore:

\[ \boxed{a_B=3.27\ \mathrm{m\\,s^{-2}}} \]

Step 6: Calculate the acceleration of A.

Since:

\[ a_A=2a_B \]

\[ a_A=2(3.2667) \]

Therefore:

\[ \boxed{a_A=6.53\ \mathrm{m\\,s^{-2}}} \]

Step 7: Calculate tension.

Use the equation for A:

\[ T=m_Aa_A \]

\[ \begin{aligned} T&=1.0\times6.5333\\\ &=6.5333\ \mathrm{N} \end{aligned} \]

Therefore:

\[ \boxed{T=6.53\ \mathrm{N}} \]

Step 8: Explain why ordinary addition does not cancel tension.

The two original equations are:

\[ T=m_Aa_A \]

\[ m_Bg-2T=m_Ba_B \]

If you simply add these equations, you obtain:

\[ m_Bg-T=m_Aa_A+m_Ba_B \]

Tension has not disappeared.

A tempting conclusion is that the internal-force cancellation principle has failed.

It hasn’t.

The problem is that the two equations use acceleration variables with different magnitudes. A moves twice as far, and accelerates twice as much, as B.

We must first express both equations using the same acceleration variable.

From:

\[ T=2m_Aa_B \]

multiply the entire equation by two:

\[ 2T=4m_Aa_B \]

Now add this equation to B’s equation:

\[ 2T=4m_Aa_B \]

\[ m_Bg-2T=m_Ba_B \]

Adding:

\[ m_Bg=(4m_A+m_B)a_B \]

The tension now cancels correctly.

The acceleration of B is therefore:

\[ \boxed{a_B=\frac{m_Bg}{m_B+4m_A}} \]

This gives the same result as our earlier calculation.

Step 9: Check the work done by tension.

Suppose B descends \(0.10\ \mathrm{m}\).

Because of the rope constraint, A moves:

\[ s_A=2(0.10)=0.20\ \mathrm{m} \]

Tension acts rightwards on A, in the same direction as its displacement.

The work done by tension on A is:

\[ W_A=Ts_A \]

\[ W_A=6.5333\times0.20 \]

\[ W_A=1.31\ \mathrm{J} \]

For B, the total upward tension force is \(2T\), while its displacement is downwards.

Therefore:

\[ W_B=-2Ts_B \]

\[ W_B=-2(6.5333)(0.10) \]

\[ W_B=-1.31\ \mathrm{J} \]

Add the two contributions:

\[ \begin{aligned} W_{\text{total}}&=W_A+W_B\\\ &=1.31-1.31\\\ &=0\ \mathrm{J} \end{aligned} \]

\[ \boxed{W_{\text{total}}=0\ \mathrm{J}} \]

This confirms that tension does no net work on the connected moving bodies, even though their acceleration magnitudes are different.

The number of rope segments supporting a movable pulley changes the relationship between the bodies’ motion. It does not invalidate the underlying principle.

05The key idea to remember

For a simple ideal fixed-pulley system, adding the equations eliminates tension because the rope transmits equal tension magnitudes and constrains the connected bodies to have equal acceleration magnitudes.

But the cancellation is more subtle than simply saying that two internal forces are equal and opposite.

You must account for:

• The system boundary you have chosen.
• The direction of each tension force relative to the positive direction.
• The relationship between the accelerations of the connected bodies.
• The number of rope segments supporting each moving object.

Once you understand these relationships, you can find the acceleration of a connected system without calculating tension first.

The next step is to isolate one object and use its individual Newton’s second law equation to calculate the tension. That is where the combined-system method and separate free-body diagrams work together.“`