Connected particles and pulleys

Connected particles are objects whose motions are linked by a string, rope, rod or contact force. The key idea is to combine two kinds of statement:

1. a constraint links the particles’ accelerations;
2. Newton’s second law, $F=ma$, applies to each particle separately.

For two particles joined by a light, inextensible string that remains taut, both particles have the same magnitude of acceleration. If the string passes over a smooth pulley, its direction may change, but its tension is the same on both sides.

Prerequisites

You should be able to:

• draw forces on one object using free body diagrams;
• apply Newton’s laws in a chosen positive direction;
• resolve weight parallel and perpendicular to a slope using resolving forces;
• use $F\leq \mu R$ and $F=\mu R$ in limiting cases from friction;
• use constant acceleration formulae after finding a constant acceleration.

Throughout, $g$ denotes gravitational acceleration. Use the value stated in the question, commonly $9.8\ \mathrm{m\,s^{-2}}$.

The standard string model

Exam questions usually state some or all of these modelling assumptions.

Assumption	Consequence
String is light	Its mass is ignored, so it does not require a resultant force to accelerate it
String is inextensible	Its length is fixed
String is taut	Connected particles cannot move independently
Pulley is smooth	There is no friction between string and pulley
Pulley is light	Its rotational inertia is ignored
Particles are point masses	Their size and rotation are ignored

For the usual light string and smooth pulley model:

$$\boxed{a_A=a_B\text{ in magnitude}}, \qquad \boxed{T_A=T_B=T}.$$

These conclusions have different causes. Equal acceleration follows from the fixed string length. Equal tension follows from the ideal light string and pulley model.

Do not combine the directions blindly

The particles may accelerate in different spatial directions. If one particle moves down, the other might move up or horizontally. Their acceleration magnitudes are equal. Choose a positive direction for each particle that follows the expected motion, then use the same symbol $a$.

Why an inextensible string gives equal acceleration

Suppose two straight sections of a taut string have variable lengths $x$ and $y$. The rest of the string has constant length $C$, so

$$x+y+C=L,$$

where $L$ is the fixed total length. Differentiating with respect to time gives

$$\dot x+\dot y=0,$$

and differentiating again gives

$$\ddot x+\ddot y=0.$$

Thus the displacements, velocities and accelerations have equal magnitudes and opposite signs when measured in the same geometric sense.

A reliable solution method

For most connected particle problems:

1. Decide which way the system is likely to move.
2. Draw a separate free body diagram for each particle.
3. Choose the positive direction for each particle along the expected motion.
4. Write $F=ma$ for each particle in its positive direction.
5. Use the common acceleration magnitude $a$ and common tension $T$ only when the model justifies them.
6. Solve the simultaneous equations.
7. Interpret negative answers and check units.

The equation should always have the form

$$\text{forces forwards}-\text{forces backwards}=ma.$$

This phrasing prevents many sign errors.

Two particles on a smooth horizontal surface

Consider particles $A$ and $B$, of masses $m_A$ and $m_B$, joined by a light taut string on a smooth horizontal surface. A horizontal force $P$ pulls $B$.

For the entire system, tension is internal, so

$$P=(m_A+m_B)a.$$

Hence

$$a=\frac{P}{m_A+m_B}.$$

For particle $A$, tension is its only horizontal force:

$$T=m_Aa.$$

Worked example 1: force, acceleration and tension

Particles $A$ and $B$ have masses $3\ \mathrm{kg}$ and $5\ \mathrm{kg}$. They are joined by a light taut string on a smooth horizontal surface. A horizontal force of $24\ \mathrm N$ acts on $B$. Find the acceleration and tension.

Treat both particles as one system:

$$24=(3+5)a.$$

Therefore

$$a=3\ \mathrm{m\,s^{-2}}.$$

Now isolate $A$. Its only horizontal force is the tension:

$$T=3a=3(3)=\boxed{9\ \mathrm N}.$$

For a check, isolate $B$:

$$24-T=5a.$$

Substitution gives $24-9=5(3)$, as required.

Why the system equation is useful

When equations for all particles are added, internal tensions cancel. This often finds $a$ quickly. You must still isolate one particle to find $T$.

Self-check 1

Masses of $2\ \mathrm{kg}$ and $6\ \mathrm{kg}$ are connected on a smooth horizontal surface. A force of $32\ \mathrm N$ pulls the $2\ \mathrm{kg}$ mass away from the other mass. Find $a$ and $T$.

Answer

For the whole system,

$$32=8a,$$

so $a=4\ \mathrm{m\,s^{-2}}$.

The tension accelerates the $6\ \mathrm{kg}$ mass, so

$$T=6a=\boxed{24\ \mathrm N}.$$

The fact that the force is applied to the smaller mass matters for tension, but not for the system acceleration.

A particle hanging over a pulley

Suppose $A$, of mass $m_A$, lies on a smooth horizontal table and is connected over a smooth pulley to hanging particle $B$, of mass $m_B$. If $B$ descends:

$$T=m_Aa$$

for $A$, and

$$m_Bg-T=m_Ba$$

for $B$.

Adding eliminates $T$:

$$m_Bg=(m_A+m_B)a,$$

so

$$\boxed{a=\frac{m_Bg}{m_A+m_B}}.$$

The driving force is the hanging weight $m_Bg$, while the total inertial mass is $m_A+m_B$.

Worked example 2: table and hanging particle

A particle of mass $4\ \mathrm{kg}$ lies on a smooth horizontal table. It is connected by a light inextensible string over a smooth pulley to a hanging particle of mass $2\ \mathrm{kg}$. The system is released from rest. Find its acceleration and the tension, taking $g=9.8\ \mathrm{m\,s^{-2}}$.

Let the $2\ \mathrm{kg}$ particle move down and the $4\ \mathrm{kg}$ particle move towards the pulley.

For the $4\ \mathrm{kg}$ particle:

$$T=4a. \tag{1}$$

For the $2\ \mathrm{kg}$ particle:

$$2g-T=2a. \tag{2}$$

Substitute (1) into (2):

$$19.6-4a=2a,$$

so

$$a=\frac{19.6}{6}=\boxed{\frac{49}{15}\ \mathrm{m\,s^{-2}}}.$$

Then

$$T=4a=\boxed{\frac{196}{15}\ \mathrm N}.$$

The tension is less than the hanging weight $19.6\ \mathrm N$, which is consistent with downward acceleration.

Self-check 2

A $3\ \mathrm{kg}$ particle on a smooth table is connected to a hanging $1\ \mathrm{kg}$ particle. Find $a$ and $T$ in terms of $g$.

Answer

For the system,

$$g=(3+1)a,$$

so

$$a=\frac g4.$$

For the particle on the table,

$$T=3a=\boxed{\frac{3g}{4}}.$$

Two hanging particles

Two particles hanging on opposite sides of a pulley form an idealised Atwood machine. Let $m_2>m_1$, so $m_2$ descends and $m_1$ rises.

For $m_1$:

$$T-m_1g=m_1a.$$

For $m_2$:

$$m_2g-T=m_2a.$$

Adding gives

$$(m_2-m_1)g=(m_1+m_2)a,$$

so

$$\boxed{a=\frac{(m_2-m_1)g}{m_1+m_2}}.$$

Substitution gives

$$\boxed{T=\frac{2m_1m_2g}{m_1+m_2}}.$$

The acceleration is less than $g$ because $0<m_2-m_1<m_1+m_2$.

Worked example 3: two hanging masses

Particles of masses $3\ \mathrm{kg}$ and $7\ \mathrm{kg}$ hang on opposite sides of a smooth pulley. Find the acceleration and tension, taking $g=9.8\ \mathrm{m\,s^{-2}}$.

The $7\ \mathrm{kg}$ particle descends. For the whole system, the net driving force is the difference in weights:

$$7g-3g=(7+3)a.$$

Therefore

$$a=\frac{4g}{10}=0.4g=\boxed{3.92\ \mathrm{m\,s^{-2}}}.$$

Use the rising $3\ \mathrm{kg}$ particle to find tension:

$$T-3g=3a.$$

Thus

$$T=3(g+a)=3(9.8+3.92)=\boxed{41.16\ \mathrm N}.$$

Check the physical bounds:

$$3g<T<7g.$$

Indeed, $29.4<41.16<68.6$.

Connected particles on an inclined plane

For a particle of mass $m$ on a plane inclined at angle $\theta$ to the horizontal, the component of weight down the plane is

$$mg\sin\theta.$$

If no other force has a perpendicular component, the normal reaction is

$$R=mg\cos\theta.$$

Write the equation of motion along the plane. The reaction does not appear in that equation because it is perpendicular to the motion.

Worked example 4: smooth slope and hanging mass

Particle $A$, of mass $5\ \mathrm{kg}$, lies on a smooth plane inclined at $30^\circ$ to the horizontal. It is connected over a smooth pulley to a hanging particle $B$ of mass $4\ \mathrm{kg}$. Find the acceleration and tension, taking $g=9.8\ \mathrm{m\,s^{-2}}$.

Compare the competing forces:

$$\text{weight of }B=4g=39.2\ \mathrm N,$$ $$\text{component of }A\text{'s weight down the plane} =5g\sin30^\circ=24.5\ \mathrm N.$$

Therefore $B$ moves down and $A$ moves up the plane.

For $A$, positive up the plane:

$$T-5g\sin30^\circ=5a. \tag{1}$$

For $B$, positive downwards:

$$4g-T=4a. \tag{2}$$

Add (1) and (2):

$$4g-5g\sin30^\circ=9a.$$

Hence

$$a=\frac{39.2-24.5}{9}=\boxed{\frac{49}{30}\ \mathrm{m\,s^{-2}}}.$$

From (2),

$$T=4g-4a =39.2-\frac{98}{15} =\boxed{\frac{98}{3}\ \mathrm N}.$$

What if the assumed direction is wrong?

You may choose either direction as positive. If the equations give $a<0$, the true acceleration is opposite to your chosen direction. Do not discard the result or make tension negative.

Self-check 3

A $6\ \mathrm{kg}$ particle on a smooth $30^\circ$ slope is connected to a hanging $2\ \mathrm{kg}$ particle. State the direction of motion and find the acceleration in terms of $g$.

Answer

The component down the slope is

$$6g\sin30^\circ=3g,$$

which exceeds the hanging weight $2g$. Therefore the $6\ \mathrm{kg}$ particle moves down the slope.

For the whole system,

$$3g-2g=(6+2)a,$$

so

$$\boxed{a=\frac g8}.$$

Adding friction

Friction acts opposite to the actual motion or the tendency to move. On a rough plane,

$$F\leq \mu R.$$

Use $F=\mu R$ only when the particle is moving, or when it is in limiting equilibrium and about to move. In many A level connected particle questions, motion is stated, so kinetic friction has magnitude $\mu R$ under the model used.

Worked example 5: rough horizontal table

A particle $A$ of mass $6\ \mathrm{kg}$ lies on a rough horizontal table with coefficient of friction $0.25$. It is connected over a smooth pulley to a hanging particle $B$ of mass $4\ \mathrm{kg}$. The system is released and $B$ moves down. Find $a$ and $T$, taking $g=9.8\ \mathrm{m\,s^{-2}}$.

For $A$, vertical equilibrium gives

$$R=6g.$$

Therefore the friction magnitude is

$$F=\mu R=0.25(6g)=1.5g=14.7\ \mathrm N.$$

Since $A$ moves towards the pulley, friction acts away from the pulley.

For $A$:

$$T-14.7=6a. \tag{1}$$

For $B$:

$$4g-T=4a. \tag{2}$$

Adding:

$$39.2-14.7=10a,$$

so

$$a=\boxed{2.45\ \mathrm{m\,s^{-2}}}.$$

Then, from (2),

$$T=39.2-4(2.45)=\boxed{29.4\ \mathrm N}.$$

Friction is not always $\mu R$

If the question asks whether the system moves from rest, first find the friction needed for equilibrium. The system remains at rest if that required value is at most $\mu R$.

Worked example 6: deciding whether motion begins

A $5\ \mathrm{kg}$ particle rests on a rough horizontal table and is connected to a hanging $1\ \mathrm{kg}$ particle. The coefficient of friction is $0.3$. Determine whether the system moves when released.

If the system remains at rest, the hanging particle has vertical equilibrium, so

$$T=g.$$

The table particle therefore needs friction $F=g$ to remain at rest. Its maximum available friction is

$$\mu R=0.3(5g)=1.5g.$$

Since

$$g<1.5g,$$

static friction can supply the required force. The system does not move. The actual friction is $g$, not $1.5g$.

Finding unknown masses or coefficients

Connected particle questions often give the acceleration and ask for a parameter. The mechanics is unchanged: write one equation for each particle, then solve for the unknowns.

Worked example 7: finding an unknown mass

A particle of mass $m\ \mathrm{kg}$ lies on a smooth horizontal table and is connected to a hanging particle of mass $3\ \mathrm{kg}$. The hanging particle accelerates downwards at $2\ \mathrm{m\,s^{-2}}$. Find $m$ and the tension, taking $g=9.8\ \mathrm{m\,s^{-2}}$.

For the hanging particle:

$$3g-T=3(2).$$

Thus

$$T=29.4-6=23.4\ \mathrm N.$$

For the particle on the table:

$$T=ma.$$

Therefore

$$m=\frac{23.4}{2}=\boxed{11.7\ \mathrm{kg}}.$$

Worked example 8: finding a coefficient of friction

A $4\ \mathrm{kg}$ particle on a rough horizontal table is connected to a hanging $2\ \mathrm{kg}$ particle. The hanging particle descends with acceleration $1.4\ \mathrm{m\,s^{-2}}$. Find the coefficient of friction, taking $g=9.8\ \mathrm{m\,s^{-2}}$.

For the hanging particle:

$$2g-T=2(1.4),$$

so

$$T=19.6-2.8=16.8\ \mathrm N.$$

For the table particle:

$$T-F=4(1.4).$$

Hence

$$F=16.8-5.6=11.2\ \mathrm N.$$

Also $R=4g=39.2\ \mathrm N$. Since the particle is moving,

$$F=\mu R,$$

so

$$\mu=\frac{11.2}{39.2}=\boxed{\frac27}.$$

The coefficient of friction is dimensionless.

When the string becomes slack or a particle hits the ground

The common acceleration constraint holds only while the string is taut and the original arrangement persists. If a hanging particle reaches the floor, the tension usually becomes zero immediately and the other particle begins a new phase of motion.

Worked example 9: motion in two phases

A $3\ \mathrm{kg}$ particle on a smooth horizontal table is connected to a hanging $1\ \mathrm{kg}$ particle. The system starts from rest with the hanging particle $0.8\ \mathrm m$ above the floor. Find the speed of the table particle $1.5\ \mathrm s$ after release, taking $g=9.8\ \mathrm{m\,s^{-2}}$.

While the string is taut,

$$a=\frac{g}{3+1}=2.45\ \mathrm{m\,s^{-2}}.$$

Let $t_1$ be the time until the hanging particle reaches the floor. Using $s=ut+\frac12at^2$:

$$0.8=\frac12(2.45)t_1^2.$$

Thus

$$t_1=\sqrt{\frac{1.6}{2.45}}\approx0.808\ \mathrm s.$$

The common speed at that instant satisfies

$$v^2=u^2+2as=2(2.45)(0.8)=3.92,$$

so

$$v=\sqrt{3.92}\approx1.98\ \mathrm{m\,s^{-1}}.$$

After the hanging particle hits the floor, the string is slack. The table is smooth, so the table particle has zero resultant horizontal force and continues at constant speed. Therefore its speed at $1.5\ \mathrm s$ is

$$\boxed{1.98\ \mathrm{m\,s^{-1}}\text{, approximately}}.$$

Do not continue using the original connected system acceleration after the impact.

Common misconceptions

Tension equals weight

$T=mg$ only when a hanging particle has zero vertical acceleration. If it accelerates downwards, $mg-T=ma$; if it accelerates upwards, $T-mg=ma$.

Tension is the driving force on the whole system

Tension is internal to a system containing both connected particles. It cancels when the particle equations are added. External forces determine the system acceleration.

Every string has the same tension

Tension is constant throughout one ideal light string passing over smooth pulleys. Two distinct strings may have different tensions, so label them $T_1$ and $T_2$.

Equal acceleration means equal resultant force

Connected particles can have the same acceleration but different masses. Their resultant forces are $m_Aa$ and $m_Ba$, which are generally different.

The heavier particle must move down

Compare forces along the possible motion, not masses alone. On a slope, use $mg\sin\theta$ along the plane. Friction may also change or prevent the motion.

A negative acceleration is impossible

A negative result means the true acceleration is opposite to the chosen positive direction. The algebra has supplied useful information.

The normal reaction is always $mg$

On a slope, $R=mg\cos\theta$ if no other force has a perpendicular component. On a horizontal surface, an angled pull can also change $R$.

Mixed self-check

Question 1

Particles of masses $2\ \mathrm{kg}$ and $5\ \mathrm{kg}$ hang on opposite sides of a smooth pulley. Find $a$ and $T$ in terms of $g$.

Answer

The $5\ \mathrm{kg}$ particle descends. For the system,

$$5g-2g=7a,$$

so

$$a=\frac{3g}{7}.$$

For the $2\ \mathrm{kg}$ particle,

$$T-2g=2a,$$

giving

$$T=2g+\frac{6g}{7}=\boxed{\frac{20g}{7}}.$$

Question 2

A $4\ \mathrm{kg}$ particle on a smooth plane inclined at $30^\circ$ is connected to a hanging $3\ \mathrm{kg}$ particle. Find the acceleration and tension in terms of $g$.

Answer

The hanging weight is $3g$ and the component down the plane is

$$4g\sin30^\circ=2g.$$

The $3\ \mathrm{kg}$ particle moves down. For the system,

$$3g-2g=7a,$$

so

$$a=\frac g7.$$

For the hanging particle,

$$3g-T=3a,$$

therefore

$$T=3g-\frac{3g}{7}=\boxed{\frac{18g}{7}}.$$

Question 3

A $5\ \mathrm{kg}$ particle lies on a rough horizontal table where $\mu=0.2$. It is connected to a hanging $2\ \mathrm{kg}$ particle. Assuming motion occurs, find $a$ in terms of $g$ and verify that the assumed motion is consistent.

Answer

The friction magnitude is

$$F=\mu R=0.2(5g)=g.$$

If the hanging particle descends, the net driving force is

$$2g-g=g>0.$$

Thus that direction is consistent. For the complete system,

$$g=7a,$$

so

$$\boxed{a=\frac g7}.$$

Question 4

Explain why replacing a light pulley by a massive pulley can make the tensions on its two sides unequal.

Answer

A massive pulley requires a turning effect to produce angular acceleration. A difference between the two tensions supplies this resultant moment. In the ideal light pulley model, rotational inertia is ignored, so no tension difference is required.

Exam checklist

Before finishing a connected particles problem, check that you have:

• drawn a separate force diagram for each particle;
• stated or clearly used a positive direction;
• resolved weight as $mg\sin\theta$ along a slope;
• put friction opposite the motion or tendency to move;
• used equal acceleration magnitudes only while the string is taut and inextensible;
• used equal tension only where the ideal string and pulley model permits it;
• written one valid $F=ma$ equation for each isolated particle;
• cancelled tension only when treating the particles as one system;
• checked whether a negative value reverses your assumed direction;
• started a new phase when a string becomes slack or a particle hits an obstacle;
• included units and an appropriate value of $g$.

Next steps

• Review how to identify every force in forces and free body diagrams.
• Strengthen the underlying equations in Newton’s laws.
• Practise rough surfaces and limiting equilibrium in friction.
• Combine a calculated acceleration with constant acceleration and SUVAT.
• Extend force equations to several directions in dynamics in a plane.