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Explainer Video

Tutor: Meera

Summary

​

Connected particles

In a nutshell

You need to know how to solve problems involving particles that are connected together. Problems may involve particles suspended, or on a smooth or rough surface, connected by a string with pulleys involved. 


Equations

DESCRIPTION

EQUATION

Resultant force if there is acceleration in the plane.

​F=maF= m aF=ma​​

Resultant force if there is no motion in the plane. 

​∑F=0\sum F = 0∑F=0​​

Friction

​F=μRF = \mu RF=μR​​


Variable definitions

QUANTITY NAME

SYMBOL

UNIT NAME

UNIT

​Force of frictionForce \ of \ frictionForce of friction​​

​FFF​​

​NewtonsNewtonsNewtons​​

​NNN​​

​Reaction forceReaction \ forceReaction force​​

​RRR​​

​NewtonsNewtonsNewtons​​

​NNN​​

​WeightWeightWeight​​

​WWW​​

​NewtonsNewtonsNewtons​​

​NNN​​

​AccelerationAccelerationAcceleration​​

​aaa​​

​Meters per second squaredMeters \ per \ second \ squaredMeters per second squared​​

​m.s−2m.s^{-2}m.s−2​​

​Coefficient of frictionCoefficient \ of \ frictionCoefficient of friction​​

​μ\muμ​​

​​​UnitlessUnitlessUnitless​​

​−-−​​



Connected particles

procedure

1.

Draw a force diagram and resolve the forces into different components (parallel and perpendicular to the surface). 

2.

Form equations by considering each particle separately.

3.

Solve the equations.


Note: When the string connecting both particles is "inextensible", it means that the tension throughout the string is the same.


Example 1

​Note: In order to differentiate between the force of friction FFF in this problem, and the force from Newton's second law, F^\hat FF^ will be used in Newton's equation.

​

Two particles of mass 4kg4kg4kg and 6kg6 kg6kg are connected by an inextensible string and suspended from a pulley as shown. The system is released from rest. Find the tension in the string and the acceleration of the system:


Maths; Application of forces; KS5 Year 13; Connected particles


Consider the 4kg4kg4kg particle. It will accelerate upwards, so:

F^=maT−4g=4a\begin{aligned}\hat F &=ma \\T - 4g&= 4a\end{aligned}F^T−4g​=ma=4a​​​


Consider the 6kg6kg6kg particle. It will accelerate downwards, so:

​F^=ma6g−T=6a\begin{aligned}\hat F &=ma \\6g - T&= 6a\end{aligned}F^6g−T​=ma=6a​​​

​

Solve the equations simultaneously:

T−4g=4a1◯6g−T=6a2◯2g=10a1◯+2◯a=1.96 m.s−2‾T=4×1.96+4gT=47.04 N‾\begin{aligned}T - 4g&= 4a \quad &\textcircled{1} \\6g - T &= 6a \quad &\textcircled{2} \\\hline2g &=10a \quad &\textcircled{1}+\textcircled{2}\\\\a&= \underline{1.96 \ m.s^{-2}} \\\\T&= 4 \times 1.96 + 4g \\T&=\underline{47.04 \ N}\end{aligned}T−4g6g−T2gaTT​=4a=6a=10a=1.96 m.s−2​=4×1.96+4g=47.04 N​​1◯2◯1◯+2◯​​​


Note: The direction of the acceleration is taken to be the positive direction. This could be different for each particle involved. In this case, the 4kg4kg4kg particle will accelerate upwards, and the 6kg6kg6kg particle will accelerate downwards.


Example 2

​A particle AAA of mass mA=2 kgm_A =2\ kgmA​=2 kg is held on a rough slope inclined at θ=34∘\theta = 34^{\circ}θ=34∘ to the horizontal. The particle is connected by a light, inextensible string over a pulley to a particle BBB, with mass mB=5 kgm_B = 5 \ kgmB​=5 kg​, hanging vertically. The coefficient of friction is μ=0.5\mu = 0.5μ=0.5. Calculate the acceleration aaa when the system is released from rest.


First draw a forces diagram, in order to visualize the forces involved.


Resolve the forces into components, parallel and perpendicular to the inclined plane. You should add labels to your diagram as follows:



Consider particle BBB as it accelerates vertically downwards:

F^=ma5g−T=5a1◯\begin{aligned}\hat F&=ma \\5g - T&= 5a \quad \textcircled{1}\end{aligned}F^5g−T​=ma=5a1◯​​​​


Consider particle AAA, perpendicular to the plane:

R=2gcos⁡(34)R=2g \cos(34)R=2gcos(34)​​


Parallel to the plane:

F^=maT−2gsin⁡(34)−F=2a2◯\begin{aligned}\hat{F}&=ma \\T - 2g \sin(34) - F&= 2a \quad \textcircled{2}\end{aligned}F^T−2gsin(34)−F​=ma=2a2◯​​​


Use F=μRF= \mu RF=μR:

F=μRF=0.5×2gcos⁡(34)F = \mu R \\F = 0.5 \times 2g \cos(34)F=μRF=0.5×2gcos(34)​​


Substitute the friction into equation 2◯\textcircled{2}2◯ and solve simultaneously:

5g−T=5a1◯T−2gsin⁡(34)−0.5×2gcos⁡(34)=2a2◯5g−2gsin⁡(34)−0.5×2gcos⁡(34)=7a1◯+2◯a=4.27 m.s−2 (3 s.f.)‾\begin{aligned}5g - T&= 5a \qquad \textcircled{1} \\T - 2g \sin(34) - 0.5 \times 2g \cos(34)&= 2a \qquad \textcircled{2}\\ \hline5g - 2g \sin(34) - 0.5 \times 2g \cos(34) &=7a \qquad \textcircled{1}+\textcircled{2}\\\\a&= \underline{4.27 \ m.s^{-2} \ (3 \ s.f.)}\end{aligned}5g−TT−2gsin(34)−0.5×2gcos(34)5g−2gsin(34)−0.5×2gcos(34)a​=5a1◯=2a2◯=7a1◯+2◯=4.27 m.s−2 (3 s.f.)​​​​​



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FAQs - Frequently Asked Questions

What determines the positive direction of for the particles?

The direction of acceleration is usually taken as positive.

What does "inextensible string" mean?

That the tension throughout the string is the same.

Do all connected particles have the same mass?

No, connected particles can have different masses.

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