How to Find Rope Tension After Calculating Acceleration

A 2.0 kg mass and a 3.0 kg mass hang from opposite ends of a light rope over an ideal pulley. You’ve already calculated that the system accelerates at \(1.96\ \mathrm{m\,s^{-2}}\), with the 3.0 kg mass moving downwards.

What is the tension in the rope?

The answer is \(23.5\ \mathrm{N}\). To find it, isolate either mass and apply Newton’s second law:

\[
\sum F=ma
\]

For the 2.0 kg mass accelerating upwards:

\[
T-mg=ma
\]

Rearranging:

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

Therefore,

\[
T=2.0(9.8+1.96)
\]

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

That’s the essential method: once you know the system’s acceleration, draw a free-body diagram for one object and use Newton’s second law to find the tension.

The important part is choosing the correct forces and signs. Let’s work through why this method works, when to use each equation, and what changes when the system becomes more complicated.

01Why finding acceleration isn’t enough

When you calculate the acceleration of a connected system, you’re usually treating the objects together as one system.

This is useful because it allows you to eliminate the tension forces and find the acceleration without knowing the tension first.

But there’s a catch.

The equation for the whole system doesn’t normally tell you the tension. You need to look at what happens to an individual object.

Think about the two masses from our opening example.

• The 2.0 kg mass accelerates upwards.
• The 3.0 kg mass accelerates downwards.
• The rope pulls upwards on both masses.
• Gravity pulls downwards on both masses.

The rope tension must be strong enough to lift the lighter mass, but not strong enough to prevent the heavier mass from descending.

This gives us an immediate prediction:

\[
19.6\ \mathrm N<T<29.4\ \mathrm N
\]

Why?

The tension must exceed the lighter mass’s weight of \(19.6\ \mathrm N\) to accelerate it upwards. However, it must be less than the heavier mass’s weight of \(29.4\ \mathrm N\) to allow that mass to accelerate downwards.

Our calculated tension of \(23.5\ \mathrm N\) satisfies both conditions.

This is a useful way to check whether your answer makes physical sense.

02Internal and external forces: why tension disappears

Before calculating tension, you need to understand why it sometimes disappears from the equations.

An external force acts on a system from outside its chosen boundary. An internal force acts between parts of the system.

Consider two masses connected by an ideal rope.

When we treat both masses and their connecting rope as one system, the tension forces are internal. They act between parts of the system and cancel when we consider the system as a whole.

The remaining external forces determine the system’s acceleration.

For an Atwood machine, these include the two gravitational forces.

If the heavier mass is \(m_2\), and the lighter mass is \(m_1\), the net driving force is:

\[
F_{\text{net}}=m_2g-m_1g
\]

Therefore:

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

Rearranging:

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

Notice that tension doesn’t appear.

Now change your system boundary. Instead of considering both masses, isolate just the lighter mass.

The rope pulls upwards on this mass, and gravity pulls downwards. Both forces must now appear in its free-body diagram.

\[
T-m_1g=m_1a
\]

This equation contains tension, so we can rearrange it to find \(T\).

The important distinction is that internal and external describe forces relative to a chosen system boundary. If you choose only the two blocks as your system and exclude the rope, the rope forces are technically external to that boundary, but their equal and opposite contributions still cancel in the combined equation.

03Ideal rope and pulley assumptions

Most introductory HSC pulley questions assume an ideal rope and an ideal pulley.

These assumptions are important because they determine whether the tension is the same throughout the rope.

An ideal rope

An ideal rope is:

• Massless: it has negligible mass.
• Inextensible: it cannot stretch.
• Taut: it does not become loose during the motion.

For a simple fixed-pulley system, an inextensible rope means the connected masses have equal acceleration magnitudes in opposite directions.

For example, if one mass accelerates upwards at \(2.0\ \mathrm{m\,s^{-2}}\), the other accelerates downwards at \(2.0\ \mathrm{m\,s^{-2}}\).

An ideal pulley

An ideal pulley is:

• Massless, with negligible rotational inertia.
• Frictionless, so there is no frictional torque resisting the rope.

The pulley changes the direction of the rope without changing the magnitude of its tension.

Therefore:

\[
T_1=T_2=T
\]

This does not mean the forces point in the same direction. They act along different sections of the rope, but their magnitudes are equal.

In a real system, a pulley with rotational inertia can have different tensions on its two sides. Rope mass can also cause tension to vary along its length.

Don’t automatically assume equal tensions unless the question gives you the ideal conditions or makes them clear.

04The equations you actually need

Once you know the acceleration, your next step is to isolate one object.

The correct equation depends on the direction of its acceleration and the other forces acting on it.

Case A: A mass accelerating upwards

Suppose a mass \(m\) hangs vertically from a rope and accelerates upwards.

Choose upwards as positive.

The forces are:

• Tension \(T\), upwards.
• Weight \(mg\), downwards.

Newton’s second law gives:

\[
T-mg=ma
\]

Rearranging:

\[
\boxed{T=m(g+a)}
\]

Tension is greater than weight because the object accelerates upwards.

Case B: A mass accelerating downwards

Now suppose the same mass accelerates downwards.

Choose downwards as positive.

The forces are still tension upwards and weight downwards.

This time:

\[
mg-T=ma
\]

Rearranging:

\[
\boxed{T=m(g-a)}
\]

Tension is smaller than weight because gravity must provide the resultant downward force.

Notice that the equations are almost identical. The only difference is the direction of acceleration relative to the tension and weight forces.

Case C: A block on a horizontal surface

A block is pulled horizontally by a rope while friction opposes its motion.

If the block accelerates to the right, then:

\[
T-f=ma
\]

Rearranging:

\[
\boxed{T=ma+f}
\]

Here, \(f\) is the frictional force.

The tension must overcome friction and provide the resultant force needed to accelerate the block.

Case D: A block on an inclined plane

Suppose a block is pulled upwards along a frictionless inclined plane.

The component of its weight acting down the slope is:

\[
mg\sin\theta
\]

Taking upwards along the slope as positive:

\[
T-mg\sin\theta=ma
\]

Therefore:

\[
\boxed{T=ma+mg\sin\theta}
\]

If friction also acts down the slope, include it:

\[
T=ma+mg\sin\theta+f
\]

One common mistake is to use the direction of motion rather than the direction of acceleration to decide the signs.

An object can move upwards while accelerating downwards. In that situation, you must use the downward-acceleration equation.

05Testing the model: a simple pulley experiment

There’s a practical way to test whether tension really changes when a connected mass accelerates.

Imagine suspending a \(0.20\ \mathrm{kg}\) mass from a force sensor. If the mass is stationary, the sensor reads approximately:

\[
T=mg
\]

\[
T=0.20(9.8)
\]

\[
T=1.96\ \mathrm N
\]

A student might predict that the sensor will always read \(1.96\ \mathrm N\), even when the mass moves.

But the original setup has a limitation: the mass is stationary, so its acceleration is zero. It only tests equilibrium.

We need to revise the apparatus.

Revising the experimental setup

Attach a fixed, low-friction pulley to a bench or support. Pass a light cord over the pulley, with a \(0.20\ \mathrm{kg}\) mass hanger on one end and a \(0.30\ \mathrm{kg}\) mass hanger on the other.

Insert a low-mass force sensor into the rope near one hanger. Position a photogate or motion sensor so that the acceleration of the descending mass can be measured independently.

Release both masses from rest.

What should the experiment show?

First, calculate the acceleration of the system.

The heavier mass descends, so the net driving force is:

\[
F_{\text{net}}=(0.30-0.20)(9.8)
\]

The total mass is \(0.50\ \mathrm{kg}\).

Therefore:

\[
a=\frac{0.10(9.8)}{0.50}
\]

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

Now calculate the tension using the lighter mass, which accelerates upwards:

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

\[
T=0.20(9.8+1.96)
\]

\[
\boxed{T=2.35\ \mathrm N}
\]

This is the predicted force-sensor reading under the ideal model.

The expected values are:

Illustrative predictions, not recorded experimental measurements. Values assume an ideal rope and pulley and \(g=9.8\ \mathrm{m\,s^{-2}}\).

The key observation to look for is that the tension is greater than the lighter mass’s weight but smaller than the heavier mass’s weight.

A sensor reading near \(2.35\ \mathrm N\), alongside an acceleration near \(1.96\ \mathrm{m\,s^{-2}}\), would be consistent with the model.

If the measurements differ systematically, pulley friction, rotational inertia, sensor mass, and air resistance may be responsible. A more refined investigation could use lower-inertia equipment or measure the tension on both sides of the pulley.

The experiment demonstrates why a stationary spring-balance reading cannot automatically be applied to an accelerating object.

06Worked examples

Let’s put the method into practice with two common HSC-style setups.

07A quick decision process for HSC questions

When you have already calculated the system acceleration, use this sequence to find tension.

Keep these common equations handy:

These are not equations to memorise blindly. They’re rearrangements of Newton’s second law for particular situations.

If the force directions change, your equation must change too.

08Questions and solutions

Try these five questions in order. The first checks the basic method, while the later questions introduce more complicated force systems and rope constraints.

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

Question 1

A \(1.5\ \mathrm{kg}\) load is suspended from a light vertical rope. It accelerates upwards at \(1.2\ \mathrm{m\,s^{-2}}\).

Calculate the tension in the rope.

Solution 1

The tension is \(16.5\ \mathrm N\).

Since the load accelerates upwards, tension must be greater than its weight.

Choose upwards as positive:

\[
T-mg=ma
\]

Rearrange:

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

Substitute:

\[
T=1.5(9.8+1.2)
\]

\[
T=1.5(11.0)
\]

\[
\boxed{T=16.5\ \mathrm N}
\]

The answer is greater than the load’s weight:

\[
mg=1.5(9.8)=14.7\ \mathrm N
\]

This makes physical sense because the resultant force must act upwards.

Question 2

A \(4.0\ \mathrm{kg}\) block is placed on a rough inclined plane at an angle of \(30^\circ\) to the horizontal.

It is connected by a light rope over a frictionless pulley to a hanging \(3.0\ \mathrm{kg}\) mass.

The hanging mass accelerates downwards at \(0.673\ \mathrm{m\,s^{-2}}\), pulling the block upwards along the incline.

The coefficient of kinetic friction between the block and the incline is \(0.15\).

Calculate the rope tension. Then verify that your answer is consistent with the motion of the block on the incline.

Solution 2

The rope tension is approximately \(27.4\ \mathrm N\).

The easiest approach is to isolate the hanging mass because its acceleration is already known.

Step 1: Apply Newton’s second law to the hanging mass.

Choose downwards as positive.

The forces are weight downwards and tension upwards:

\[
m_hg-T=m_ha
\]

Rearrange:

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

Substitute:

\[
T=3.0(9.8-0.673)
\]

\[
T=3.0(9.127)
\]

\[
\boxed{T=27.38\ \mathrm N}
\]

Step 2: Verify the result using the inclined block.

The block moves upwards along the incline, so kinetic friction acts down the incline.

First, calculate the normal force:

\[
N=mg\cos\theta
\]

\[
N=4.0(9.8)\cos30^\circ
\]

\[
N=33.95\ \mathrm N
\]

Calculate the frictional force:

\[
f_k=\mu_kN
\]

\[
f_k=0.15(33.95)
\]

\[
f_k=5.09\ \mathrm N
\]

The component of weight acting down the incline is:

\[
F_{g,\parallel}=mg\sin\theta
\]

\[
F_{g,\parallel}=4.0(9.8)\sin30^\circ
\]

\[
F_{g,\parallel}=19.6\ \mathrm N
\]

Step 3: Check Newton’s second law along the incline.

Choose upwards along the incline as positive:

\[
T-mg\sin\theta-f_k=ma
\]

Substitute:

\[
27.38-19.6-5.09
\approx 4.0(0.673)
\]

\[
2.69\approx2.69\ \mathrm N
\]

The equation is satisfied, allowing for rounding.

The tempting mistake here is to use \(T=m_hg\), giving \(29.4\ \mathrm N\). But the hanging mass accelerates downwards, so gravity must provide a resultant force after tension has been subtracted.

Question 3

Three blocks, A, B, and C, are connected in a straight line by two separate light strings on a rough horizontal surface.

Their masses are:

• Block A: \(2.0\ \mathrm{kg}\)
• Block B: \(3.0\ \mathrm{kg}\)
• Block C: \(1.0\ \mathrm{kg}\)

A horizontal force of \(15\ \mathrm N\) pulls block A to the right.

The kinetic frictional forces acting on A, B, and C have magnitudes of \(2.0\ \mathrm N\), \(3.0\ \mathrm N\), and \(1.0\ \mathrm N\), respectively.

All three blocks accelerate to the right at \(1.5\ \mathrm{m\,s^{-2}}\).

Calculate the tension in each string.

Explain why the tensions are different even though both strings are ideal.

Solution 3

The tension between A and B is \(10.0\ \mathrm N\), while the tension between B and C is \(2.5\ \mathrm N\).

The key is to recognise that the two strings are separate and each must accelerate a different combination of masses.

Step 1: Find the tension between B and C.

Isolate block C.

The tension in the right-hand string acts to the right, while friction acts to the left.

Newton’s second law gives:

\[
T_{BC}-f_C=m_Ca
\]

Rearrange:

\[
T_{BC}=m_Ca+f_C
\]

Substitute:

\[
T_{BC}=1.0(1.5)+1.0
\]

\[
\boxed{T_{BC}=2.5\ \mathrm N}
\]

Step 2: Find the tension between A and B.

Now consider blocks B and C together as one subsystem.

The tension between A and B must accelerate both blocks and overcome their combined friction.

Therefore:

\[
T_{AB}-(f_B+f_C)=(m_B+m_C)a
\]

Rearrange:

\[
T_{AB}=(m_B+m_C)a+f_B+f_C
\]

Substitute:

\[
T_{AB}=(3.0+1.0)(1.5)+3.0+1.0
\]

\[
T_{AB}=6.0+4.0
\]

\[
\boxed{T_{AB}=10.0\ \mathrm N}
\]

Step 3: Explain why the tensions differ.

Both strings are ideal, but they are two separate strings.

The right-hand string only needs to accelerate block C and overcome its friction.

The left-hand string must accelerate both B and C and overcome the friction acting on both blocks.

Therefore, the left-hand string requires a greater tension.

The misconception is that ideal strings always have the same tension. They don’t.

An individual ideal string has uniform tension throughout its length. Two different strings can have completely different tensions.

Question 4

A \(600\ \mathrm{kg}\) service lift is travelling upwards at \(2.5\ \mathrm{m\,s^{-1}}\).

It begins slowing down at a constant rate of \(1.0\ \mathrm{m\,s^{-2}}\). A single vertical cable supports the lift.

Calculate the cable tension.

Then determine how long it takes for the lift to momentarily stop, and explain whether the cable tension changes at that instant.

Assume the downward acceleration remains constant throughout this interval.

Solution 4

The cable tension is \(5280\ \mathrm N\), and the lift takes \(2.5\ \mathrm s\) to momentarily stop.

The important distinction is that the lift is moving upwards but accelerating downwards.

Step 1: Determine the direction of acceleration.

Since the lift is slowing down while travelling upwards, its acceleration must act downwards.

Choose upwards as positive.

Therefore:

\[
a=-1.0\ \mathrm{m\,s^{-2}}
\]

Step 2: Apply Newton’s second law.

The cable tension acts upwards, and the lift’s weight acts downwards.

\[
T-mg=ma
\]

Substitute:

\[
T-600(9.8)=600(-1.0)
\]

Rearrange:

\[
T=5880-600
\]

\[
\boxed{T=5280\ \mathrm N}
\]

Notice that the tension is smaller than the lift’s weight:

\[
mg=5880\ \mathrm N
\]

This is because the resultant force must act downwards.

Step 3: Calculate the stopping time.

Use the kinematics equation:

\[
v=u+at
\]

At the instant the lift stops, its velocity is zero.

Therefore:

\[
0=2.5-1.0t
\]

\[
t=\frac{2.5}{1.0}
\]

\[
\boxed{t=2.5\ \mathrm s}
\]

Step 4: Does tension change when the lift stops?

No. Under the stated constant-acceleration assumption, the tension remains \(5280\ \mathrm N\).

At the instant the velocity reaches zero, the acceleration is still downwards. Newton’s second law depends on acceleration, not velocity.

This is an important exam distinction: an object can have zero instantaneous velocity and still experience a non-zero resultant force.

Question 5

Extension challenge

A \(2.0\ \mathrm{kg}\) load is attached to a movable pulley.

A light, inextensible rope has one end fixed to the ceiling. It passes down around the movable pulley, then upwards over a second fixed pulley, before hanging vertically downwards. A separate \(2.0\ \mathrm{kg}\) mass is attached to the free end of the rope.

Both pulleys are massless and frictionless.

When the system is released, the load attached to the movable pulley accelerates upwards at \(1.96\ \mathrm{m\,s^{-2}}\).

Calculate:

(a) The acceleration of the freely hanging mass.

(b) The tension in the rope.

Explain why treating the movable pulley as if it were supported by only one rope segment produces an incorrect result.

Solution 5

(a) The freely hanging mass accelerates downwards at \(3.92\ \mathrm{m\,s^{-2}}\).

(b) The rope tension is \(11.76\ \mathrm N\).

The important difference from a standard Atwood machine is that the movable pulley is supported by two sections of rope.

This changes both the acceleration relationship and the total upward force on the load.

Step 1: Establish the acceleration relationship.

Imagine raising the movable pulley by a distance \(x\).

Each of the two rope sections supporting the movable pulley becomes shorter by \(x\).

The total rope length in those two sections therefore decreases by:

\[
2x
\]

Since the rope has a fixed total length, the freely hanging end must move down by \(2x\).

The same relationship applies to velocity and acceleration.

Therefore:

\[
a_{\text{hanging}}=2a_{\text{load}}
\]

Substitute:

\[
a_{\text{hanging}}=2(1.96)
\]

\[
\boxed{a_{\text{hanging}}=3.92\ \mathrm{m\,s^{-2}}}
\]

Step 2: Isolate the freely hanging mass.

The hanging mass accelerates downwards.

Choose downwards as positive.

Newton’s second law gives:

\[
mg-T=ma
\]

Rearrange:

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

Substitute:

\[
T=2.0(9.8-3.92)
\]

\[
T=2.0(5.88)
\]

\[
\boxed{T=11.76\ \mathrm N}
\]

Step 3: Verify the result using the movable pulley and load.

This is where the most important detail appears.

The movable pulley is supported by two sections of the same ideal rope.

Each section provides an upward force of magnitude \(T\).

Therefore, the total upward force on the load is:

\[
F_{\text{up}}=2T
\]

The load has a weight of \(Mg\), acting downwards.

Taking upwards as positive:

\[
2T-Mg=Ma_{\text{load}}
\]

Substitute:

\[
2(11.76)-2.0(9.8)=2.0(1.96)
\]

\[
23.52-19.60=3.92
\]

\[
3.92=3.92\ \mathrm N
\]

The result is consistent.

Why the tempting approach fails

A student might isolate the \(2.0\ \mathrm{kg}\) load and write:

\[
T-Mg=Ma
\]

This incorrectly assumes that only one section of rope supports the load.

The movable pulley is actually pulled upwards by two rope segments, so the correct equation is:

\[
2T-Mg=Ma
\]

There is also a second trap: the hanging mass does not have the same acceleration magnitude as the load.

The geometry of the rope makes its acceleration twice as large.

This question combines two ideas that are easy to overlook when working separately:

• The number of rope segments supporting an object determines the total tension force acting on it.
• The rope’s geometry determines the relationship between connected objects’ accelerations.

The tension is identical in both supporting sections because they belong to the same ideal rope. However, the total upward force on the movable pulley is \(2T\), not \(T\).

08What to remember when finding tension

Once you’ve calculated the acceleration of a connected system, don’t return to the whole-system equation and expect tension to appear.

Instead, isolate one object.

Draw its free-body diagram, identify the direction of acceleration, and apply Newton’s second law. Include weight, friction, normal forces, and any other forces acting along your chosen axis.

For a standard ideal Atwood machine, either mass will give you the same rope tension. For more complicated systems, check how many rope segments act on each object and whether the pulley assumptions still apply.

The next useful step is becoming confident with free-body diagrams involving multiple connected objects. Once you can choose the correct system boundary and account for every force, finding tension becomes a straightforward application of Newton’s second law rather than a formula-guessing exercise.