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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

Variable definitions

Connected particles

procedure

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 $$F$$ in this problem, and the force from Newton's second law, $$\hat F$$ will be used in Newton's equation.

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Two particles of mass $$4kg$$ and $$6 kg$$ 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:

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

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

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

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

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Solve the equations simultaneously:

$$\begin{aligned}T - 4g&= 4a \quad &\textcircled{1} \\6g - T &= 6a \quad &\textcircled{2} \\\hline2g &=10a \quad &\textcircled{1}+\textcircled{2}\\\\a&= \underline{1.96 \ m.s^{-2}} \\\\T&= 4 \times 1.96 + 4g \\T&=\underline{47.04 \ N}\end{aligned}$$​​

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 $$4kg$$ particle will accelerate upwards, and the $$6kg$$ particle will accelerate downwards.

Example 2

​A particle $$A$$ of mass $$m_A =2\ kg$$ is held on a rough slope inclined at $$\theta = 34^{\circ}$$ to the horizontal. The particle is connected by a light, inextensible string over a pulley to a particle $$B$$, with mass $$m_B = 5 \ kg$$​, hanging vertically. The coefficient of friction is $$\mu = 0.5$$. Calculate the acceleration $$a$$ 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 $$B$$ as it accelerates vertically downwards:

$$\begin{aligned}\hat F&=ma \\5g - T&= 5a \quad \textcircled{1}\end{aligned}$$​​​

Consider particle $$A$$, perpendicular to the plane:

$$R=2g \cos(34)$$​​

Parallel to the plane:

$$\begin{aligned}\hat{F}&=ma \\T - 2g \sin(34) - F&= 2a \quad \textcircled{2}\end{aligned}$$​​

Use $$F= \mu R$$:

$$F = \mu R \\F = 0.5 \times 2g \cos(34)$$​​

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

$$\begin{aligned}5g - T&= 5a \qquad \textcircled{1} \\T - 2g \sin(34) - 0.5 \times 2g \cos(34)&= 2a \qquad \textcircled{2}\\ \hline5g - 2g \sin(34) - 0.5 \times 2g \cos(34) &=7a \qquad \textcircled{1}+\textcircled{2}\\\\a&= \underline{4.27 \ m.s^{-2} \ (3 \ s.f.)}\end{aligned}$$​​