Dynamics of a particle moving in a straight line or plane - Mathematics - Pearson A-Level

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Welcome to Dynamics! Understanding Motion and Force

Hello! This chapter, Dynamics of a particle, is where Mechanics truly comes alive. In the previous section (Kinematics), you learned how objects move (using SUVAT). Now, we learn why they move—by studying the forces acting on them!

Don't worry if applying forces feels tricky at first. We will break down Sir Isaac Newton’s famous laws step-by-step. By the end, you will be able to model and solve complex problems involving weight, friction, slopes, and pulleys!

Prerequisite Review: What is a Particle?

In M1, we almost always treat objects as a particle.

A particle is an object whose mass is concentrated at a single point.
This means we ignore its size, shape, and any rotational effects. We only care about its mass and its position.

Section 1: Newton’s Laws of Motion

1. Newton’s First Law (The Law of Inertia)

A particle will remain at rest or continue to move with constant velocity unless acted upon by a net external force.

In simple terms: If an object is in equilibrium (at rest or moving at a constant speed), the forces acting on it must be balanced. The resultant force is zero.

Mathematically, if velocity $\mathbf{v}$ is constant or zero, then the resultant force $\mathbf{F}_{net} = 0$.

2. Newton’s Second Law (The Core Equation)

This is the most important law for dynamics problems. It connects the cause (force) to the effect (acceleration).

The resultant force acting on a particle is equal to the rate of change of momentum, or more simply, it is equal to the mass of the particle multiplied by its acceleration.

The Formula:

$$ \mathbf{F} = m\mathbf{a} $$

Where:

$\mathbf{F}$ is the Resultant Force (measured in Newtons, N). This is the vector sum of ALL forces acting on the particle.
$m$ is the Mass (measured in kilograms, kg).
$\mathbf{a}$ is the Acceleration (measured in $\text{m s}^{-2}$).

🔥 Key Concept: The Resultant Force ($\mathbf{F}$)

When you write $\mathbf{F} = m\mathbf{a}$, remember that $\mathbf{F}$ is not just one force; it's the net force in the direction of motion (or acceleration).

Example: If a force $P$ pushes a box forward, and a frictional force $F$ resists it, the equation becomes:
$$ \text{(Force pushing forward)} - \text{(Force resisting)} = m\mathbf{a} $$ $$ P - F = m\mathbf{a} $$

3. Newton’s Third Law (Action and Reaction)

For every action, there is an equal and opposite reaction.

Crucial point: These paired forces act on different bodies. They never cancel each other out!

Example: When you stand on the floor:

The Earth pulls you down (Weight, $W$). (Force on you, by Earth).
You pull the Earth up. (Force on Earth, by you). These two form a N3 pair.

The force that balances your weight in a dynamics problem is usually the Normal Reaction ($R$), which is the force exerted by the floor on you. $W$ and $R$ are N1/N2 balancing forces, not N3 pairs.

Quick Review: Newton’s Laws

N1: Balanced forces = no acceleration ($F=0$).
N2: Unbalanced forces cause acceleration ($F=ma$).
N3: Action/Reaction forces act on separate bodies.

Section 2: The Cast of Characters (Forces)

Before solving problems, you must be able to identify and label the standard forces acting on a particle. Always include these on your force diagram! We use the standard acceleration due to gravity, $g \approx 9.8 \text{ m s}^{-2}$.

1. Weight ($W$ or $mg$)

Definition: The gravitational force exerted by the Earth on the particle.
Direction: Always acts vertically downwards.
Formula: $W = mg$

2. Normal Reaction ($R$ or $N$)

Definition: The force exerted by a surface (like a floor or slope) that prevents the object from falling through it.
Direction: Always acts perpendicular (normal) to the surface.
Analogy: When you push down on a table, the table pushes back up. If it didn't, you would break the table!

3. Tension ($T$)

Definition: The force transmitted through a string, rope, cable, or wire when it is pulled taut.
Direction: Acts along the string, pulling away from the particle.
In M1, strings are usually modeled as inextensible (fixed length) and light (zero mass). This means the tension is the same throughout the string.

4. Resistance or Thrust

Resistance ($R_E$): Forces that oppose motion, such as air resistance or drag. Unless otherwise stated, we often ignore air resistance in M1.
Thrust (or Driving Force $P$): The force generated by an engine or a push, acting in the direction of intended motion.

Section 3: Applying $F = ma$ in a Straight Line

Solving dynamics problems requires a structured approach. Use the following steps every time!

Step-by-Step Problem Solving

Step 1: Draw a Diagram and Define Direction

Draw a clear diagram of the particle. Label all forces acting on it ($mg$, $R$, $T$, Friction, etc.).

Crucially, indicate the direction of acceleration ($\mathbf{a}$) with an arrow. This tells you which forces are positive and which are negative in your equation.

Step 2: Resolve Perpendicular to Motion (Find R)

In the direction perpendicular to the line of motion, the acceleration is zero ($a=0$). The forces must balance (N1). This step is often used to find the value of the Normal Reaction $R$.

$$ \sum F_{perpendicular} = 0 $$

Step 3: Resolve Parallel to Motion (Apply N2)

In the direction of motion, use Newton's Second Law ($F=ma$).

$$ \sum F_{motion} = m\mathbf{a} $$

Remember: Forces acting in the direction of acceleration are positive; forces acting against the direction of acceleration are negative.

Example: Box being Pulled Horizontally

A box of mass 5 kg is pulled along a smooth horizontal surface by a tension $T$ of 20 N. Find the acceleration.

Diagram: Forces are $T$ (Right), $mg$ (Down = 5g), $R$ (Up). We define acceleration ($a$) as Right.
Perpendicular (Vertical): $R - 5g = 0 \implies R = 5g$ (We don't need $R$ here, but it's good practice).
Parallel (Horizontal): The only force horizontally is $T$.
$$ \sum F = ma $$ $$ T = ma $$ $$ 20 = 5a $$ $$ a = 4 \text{ m s}^{-2} $$

💡 Common Mistake to Avoid

Using $a=g$ when the motion is horizontal: When an object is moving horizontally, its acceleration ($a$) is caused by the net horizontal force. The acceleration due to gravity ($g$) is only used if the object is in free fall (like a vertical projection) or for calculating the weight ($mg$).

Section 4: Understanding and Calculating Friction

Friction ($F$) is a force that opposes motion between two surfaces in contact.

Limiting Friction (Maximum Friction)

Friction has a maximum possible value, called Limiting Friction. Once the pushing force exceeds this limit, the object starts moving, and friction remains constant at this maximum value.

The Friction Formula

The magnitude of the limiting friction is directly proportional to the magnitude of the Normal Reaction ($R$).

$$ F_{max} = \mu R $$

Where:

$\mu$ (mu) is the Coefficient of Friction (a dimensionless number, usually between 0 and 1).
$R$ is the Normal Reaction force.

Two States of Motion

Case 1: Particle is Moving or about to Move (Limiting Equilibrium)

If the question states the particle is moving, or is at the point of slipping (limiting equilibrium), we use the maximum friction value:

$$ F = \mu R $$

Case 2: Particle is at Rest (Static Equilibrium)

If the particle is at rest, the friction force is only as large as it needs to be to prevent motion.

$$ F \le \mu R $$

In this case, we use N1 ($\sum F = 0$) to find the friction force $F$. We then check if this calculated $F$ is less than or equal to $\mu R$.

Did you know?

The coefficient of static friction ($\mu_s$, used for an object at rest) is often slightly higher than the coefficient of kinetic friction ($\mu_k$, used when the object is sliding). In M1, we usually assume $\mu_s = \mu_k = \mu$.

Section 5: Dynamics on Inclined Planes (2D Resolution)

This is where students often struggle, but if you master resolving the weight force, it becomes straightforward.

The Strategy: Tilting the Axes

When a particle rests on a slope (inclined at angle $\alpha$ to the horizontal), we resolve forces along axes that are parallel and perpendicular to the slope.

Why? Because the acceleration ($a$) happens parallel to the slope, and the Normal Reaction ($R$) happens perpendicular to the slope.

The Crucial Step: Resolving Weight ($mg$)

The Weight ($mg$) always acts vertically down, so it must be resolved into components along these new axes.

If the slope angle is $\alpha$:

Component perpendicular to the plane: This component balances the Normal Reaction $R$. $$ mg \cos \alpha $$
Component parallel to the plane: This component acts down the slope, trying to pull the particle down. $$ mg \sin \alpha $$

Step-by-Step for Slopes

Assume a particle of mass $m$ is sliding down a rough slope inclined at $\alpha$.

Step 1: Resolve Perpendicular (Find R)

The forces perpendicular to the slope must balance.

$$ R = mg \cos \alpha $$

This is vital because $R$ is needed to calculate friction ($F = \mu R$).

Step 2: Resolve Parallel (Apply $F=ma$)

Forces acting down the slope are positive (direction of acceleration):

$$ \text{(Weight component down)} - \text{(Friction up)} = ma $$ $$ mg \sin \alpha - F = ma $$

If the slope were smooth ($\mu=0$), then $F=0$, and the acceleration down the slope would be $a = g \sin \alpha$.

Memorization Trick

Remembering which component is sine and which is cosine:

$R$ is cosy (close) to the angle $\alpha$. So, the force balancing $R$ is $mg \mathbf{cos} \alpha$.
The force pulling the object sliding down is $mg \mathbf{sin} \alpha$.

Section 6: Advanced Scenarios – Connected Particles and Pulleys

Connected particles are objects linked by a light, inextensible string, often involving a pulley (a smooth, fixed wheel used to change the direction of the string).

These problems are solved by writing a separate $F=ma$ equation for each particle, and then solving the resulting simultaneous equations.

Key Assumptions for Connected Systems

Inextensible String: Both particles must have the same magnitude of acceleration ($a$).
Light String: The tension ($T$) is the same throughout the entire string.
Smooth Pulley: The pulley does not affect the tension in the string.

Solving a Pulley System (Vertical Motion)

Consider two masses, $m_1$ and $m_2$, hanging vertically over a pulley, with $m_1 > m_2$.

The system will accelerate downwards for $m_1$ and upwards for $m_2$.

Equation for Particle 1 ($m_1$, moving down)

Weight is greater than Tension, so $W_1$ is positive:

$$ m_1 g - T = m_1 a \quad \text{ (Equation 1)} $$

Equation for Particle 2 ($m_2$, moving up)

Tension is greater than Weight, so $T$ is positive:

$$ T - m_2 g = m_2 a \quad \text{ (Equation 2)} $$

Solving Simultaneously

Add (1) and (2) together to eliminate $T$:

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

This yields acceleration $a$. You can then substitute $a$ back into either equation to find the Tension $T$.

The 'Whole System' Trick

If you only need the acceleration ($a$) and not the tension ($T$), you can sometimes treat the connected particles as one system.

Resultant Force = Total Mass $\times$ Acceleration

The internal forces (like Tension $T$) cancel out when viewing the whole system. The resultant force is just the net external driving force (often the difference in weights).

Example: Box on Table Connected to Hanging Mass (Smooth Table)

Mass $m_A$ rests on a smooth table, connected by a string over a pulley to a hanging mass $m_B$.

External force driving the system: Only the weight of the hanging mass, $m_B g$.
Total mass moving: $m_A + m_B$.

$$ \text{Total } F = \text{Total } m \times a $$ $$ m_B g = (m_A + m_B) a $$

This provides a very fast way to find $a$. If the table were rough, the frictional force on $m_A$ would also need to be included on the left side of the equation.

🚀 Key Takeaway for Dynamics

Master the diagram! Once your force diagram is perfect, and you have correctly identified the direction of acceleration, the problem becomes a simple substitution into $F=ma$ and basic algebra. Always resolve forces perpendicular to motion first!