When I say difficult, I mean difficult in comparison to the usual standard of Higher Level Leaving Cert Applied Maths Connected Particles Questions

## Question (a)

Consider the following:

A rectangular block moves across a stationary horizontal surface with acceleration $g/5$ (the question had $g/5$ m/s${}^2$ but the m/s${}^2$ is repeated as $g=9.8$ m/s${}^2$ and so includes the unit).

There is a serious problem with this question and that is that the asymmetry in the problem means that there is an ambiguity: is the block moving left to right or right to left? I am going to assume the block moves from left to right. One would hope not to see such ambiguity in an official exam paper.

Two particles of mass $M$ placed on the block, are connected by a taut inextensible string. A second string passes over a light, smooth, fixed pulley to a third particle of mass $2M$ which presses against the block as shown in the diagram.

### (i)

If contact between the particles and the block is smooth, find the magnitude and direction of the resultant forces acting on the particles.

#### Solution

Note firstly that there are two accelerations at play. The acceleration of the block relative to the horizontal surface, $g/5$, and the accelerations of the particles relative to the block, say $a$:

We draw all the forces (I lazily didn’t add arrows to the force vectors):

We know that the normal forces for the particles on top of the block because their vertical acceleration is zero and so the sum of the forces in that direction must be zero, and as the down forces are equal for both, necessarily the up forces must be equal too.

The accelerations of the particles on top are $a+g/5$ (it would be $a-g/5$ if the block was moving right-to-left), while the acceleration of the $2M$ particle is $g/5$ in the horizontal direction, and $a$ in the vertical direction. Thus we can form four equations via Newton’s Second Law $ma_i=F_i$:

$M(a+g/5)=T$   (A)

$M(a+g/5)=S-T$   (B)

$2M(g/5)=N_2$   (C)

$2M(a)=2Mg-S$   (D)

So we have $N_2=2Mg/5$. If we add (A)+(B)+(D) we get

$2M(a+g/5)+2Ma=2Mg$

$\displaystyle\Rightarrow 4Ma+\frac{2Mg}{5}=2Mg\Rightarrow a=\frac{2g}{5}$.

We can thus find $T$ using (A):

$\displaystyle T=M\left(\frac{2g}{g}+\frac{g}{5}\right)=\frac{3Mg}{5}$.

Let us impose the following axis on the system:

Therefore the magnitude of the force on the first particle is $\displaystyle \frac{3Mg}{5}$ and the direction is in the $x$-direction. As the acceleration and mass of the second particle is the same as the first, by Newton’s second law, the resultant force must be acting on the second particle. Therefore $\displaystyle S=\frac{6Mg}{5}$ (in order that $\displaystyle S-T=\frac{3Mg}{5}$.

Now turning our attention to the $2M$ particle. The force in the negative $y$ direction is

$\displaystyle 2Mg-\frac{6Mg}{5}=\frac{4Mg}{5}$,

and so the resultant force on this particle is $\displaystyle N_2\mathbf{i}-\frac{4Mg}{5} \mathbf{j}=\frac{2Mg}{5}\mathbf{i}-\frac{4Mg}{5}\mathbf{j}$:

Call this vector $\vec{F}_3$. We find the magnitude using Pythagoras Theorem:

$\displaystyle |\vec{F}_3|=\sqrt{\left(\frac{2Mg}{5}\right)^2+\left(-\frac{4Mg}{5}\right)^2}=\frac{2Mg}{\sqrt{5}}$.

The angle of the resultant force is, below the positive $x$-axis:

$\displaystyle \tan^{-1}\left(\frac{\frac{4Mg}{5}}{\frac{2Mg}{5}}\right)=\tan^{-1}(2)$.

### (ii)

If contact between the particles and the block is rough, for what same value of the coefficient of friction will the particles remain at rest relative to the block?

#### Solution

Let the coefficient of friction be given by $\mu$. We have the same acceleration (labeling) but now additional friction forces:

The normal forces are the same, and so the frictions are $\mu Mg$, $\mu Mg$, and $\displaystyle \frac{2\mu Mg}{5}$. This gives three equations:

$M(a+g/5)=T-\mu Mg$

$M(a+g/5)=S-T-\mu Mg$

$2M(a)=2Mg-S-\frac{2\mu Mg}{5}$

If the particles are at rest relative to the block $a\underset{!}{=}0$:

$M(g/5)=T-\mu Mg$

$M(g/5)=S-T-\mu Mg$

$\displaystyle 0=2Mg-S-\frac{2\mu Mg}{5}$

$\displaystyle \frac{2Mg}{5}=2Mg-\frac{12\mu Mg}{5}$

$\underset{\div_{Mg}\circ \times_5}{\Rightarrow} 2=10-12\mu$

$\Rightarrow \mu =\frac23$.

## Question (b)

A bucket with mass $m_2$ and a block with mass $m_1$ are hung on a pulley system, as shown.

The pulley with the mass effectively has a mass of $m_1$.

### (i)

Find the magnitude of the acceleration with which the bucket and the block are moving.

#### Solution

First, the accelerations. We will assume that the $m_2$ moves ‘down’ (if it doesn’t its acceleration will come out as negative).

If the $m_2$ mass moves down then the $m_1$ mass-pulley system goes ‘up’. Let $a$ be the acceleration that the $m_1$ mass-pulley system has.

Problem: Convince yourself that the $m_2$ mass has acceleration $2a$. We draw the forces:

Now we write down two equations via Newton’s Second Law (the masses $m_i$ are ‘known’).

For the mass-pulley system:

$m_1a=2T-m_1g$,   (A)

and for the other mass, the bucket:

$m_2(2a)=m_2g-T$

Multiply both sides by two:

$\underset{\times 2}{\Rightarrow}4m_2a=2m_2g-2T$.   (B)

$\Rightarrow m_1a+4m_2a=2m_2g-m_1g$

$\Rightarrow a(m_1+4m_2)=2m_2g-m_1g$

$\displaystyle \underset{\div_{m_1+4m_2}}{\Rightarrow} a=\frac{2m_2-m_1}{m_1+4m_2}g$.

This is the acceleration of the block. The acceleration of the block is twice this:

$\displaystyle 2a=\frac{4m_2-2m_1}{m_1+4m_2}g$.

### (ii)

Find the magnitude of the tension force $T$ by which the rope is stressed.

#### Solution

i.e. find the tension. From Newton’s Second Law for the bucket, we have

$T=m_2g-m_2(2a)$

$\displaystyle =m_2g-m_2\cdot \frac{4m_2-2m_1}{m_1+4m_2}g$

Exercise

Show that this simplifies to

$\displaystyle T=\frac{3 m_1 m_2 }{m_1 +4m_2 }\cdot g.$