### Connected particles(Mechanics)

#### Problems involving connected particles

In a system that involves multiple particles in motion, we consider the particles separately.

If all particles are moving in the same straight line then we can treat the whole system as a single particle.

In problems involving particles which are connected by a string which passes over a pulley you cannot treat the whole system as a single particle because the particles are moving in different directions. Instead you resolve for each mass separately.

#### Worked Example: Two particles connected by a string

###### Finding the acceleration

Two particles of mass $m_1 = 4 \mathrm{kg}$ and $m_2 = 7 \mathrm{kg}$ are connected by a light inextensible string which is pulled taut. The heavier particle is pulled by a horizontal force $P = 50 \mathrm{N}$ along a rough horizontal plane. The coefficient of friction between each particle and the plane is $\mu = 0.3$. What is the acceleration of each particle? What is the tension in the string?

###### Solution

We can draw a diagram.

Note that each particle will have a different normal reaction, $R_1$ and $R_2$ and therefore a different friction (given by the corresponding $R \times \mu$). To find the acceleration we need to resolve $F = ma$ in the upwards direction, where acceleration is zero, to find $R_1$ and $R_2$. We do this for each particle separately. For the lighter particle of mass $m_1 = 4 \mathrm{kg}$ we have \begin{align} F & = ma, \\ R_1 - mg & = ma, \\ R_1 - \left( 4 \times 9.8 \right) & = 4 \times 0, \\ R_1 & = 4 \times 9.8, \\ & = 39.2 \mathrm{N}. \end{align} For the other particle of mass $m_2 = 7 \mathrm{kg}$ we have \begin{align} F & = ma, \\ R_2 - \left( 7 \times 9.8 \right) & = 0, \\ R_2 & = 68.6 \mathrm{N}. \end{align} As both particles are moving in the same straight line we can now use $R_1$ and $R_2$ when resolving $F=ma$ for the system as a whole. We resolve in the direction of acceleration to give \begin{align} F & = ma, \\ P - \mu R_2 - \mu R_1 & = ma, \\ 50 - \left(0.3 \times 68.6 \right) - \left(0.3 \times 39.2 \right) & = \left(4 + 7 \right) a, \\ \frac{17.66}{11} & = a, \\ a & = 1.605 \mathrm{ms^{-2} } \text{ (to 3d.p.).} \end{align} The acceleration of each particle is $1.605 \mathrm{ms^{-2} }$ since the string is inextensible. Note that we did not need to include the tension in the string, $T$, as we are treating the whole system as a single particle.

To find the tension in the string we only need to consider one of the particles. It will be easier to consider the particle of mass $4 \mathrm{kg}$ as this has less forces acting upon it. We then resolve $F = ma$ in the direction of acceleration. \begin{align} F & = ma, \\ T - \mu R_1 & = 4 \times 1.605, \\ T & = \left(0.3 \times 39.2 \right) + \left(4 \times 1.605\right), \\ & = 18.18 \mathrm{N}. \end{align} The tension in the string is $18.18 \mathrm{N}$ - this tension is the same throughout the length of the string and the mass of the string is negligible as we have modelled it as a light string.

#### More Support

You can get one-to-one support from Maths-Aid.