Mathematics (9660) | Newton’s Law of Motion

Welcome to M1: Newton’s Law of Motion!

Hello! This chapter is the absolute heart of Mechanics. If kinematics (motion equations) tells us *how* things move, Newton’s Laws tell us *why* they move—it’s all about Forces.

Don't worry if this seems tricky at first; we will break down the concepts into clear, easy steps. Mastering this chapter means mastering the fundamental math needed to solve almost any problem involving motion and interaction in the physical world. Let's dive in!

1. Understanding the Forces at Play (M1.3)

Before we apply Newton's Laws, we must be able to identify and label all the forces acting on an object. This starts with drawing a clear Force Diagram (or Free Body Diagram).

1.1 Standard Forces

• Weight (\(W\)) or Force of Gravity: This is the force exerted on an object due to gravity. It always acts vertically downwards.

  \(W = mg\)

  where \(m\) is the mass (in kg) and \(g\) is the acceleration due to gravity. In OxfordAQA A Level Mechanics, \(g\) is taken as \(9.8 \text{ ms}^{-2}\).

  Did you know? Mass is constant, but Weight changes depending on the gravitational field (e.g., you weigh less on the Moon!).



• Normal Reaction (\(R\)): When an object rests on a surface, the surface pushes back. This force is always perpendicular (or 'normal') to the surface.

  Analogy: If you push down on a table, the table pushes back up with a Normal Reaction force.



• Tension (\(T\)): The pulling force transmitted axially by a string, cable, or chain. It always pulls away from the object along the direction of the string.


• Thrust (Compression): The pushing force transmitted by a rigid rod (opposite of tension).


• Resistive Forces (e.g., Air Resistance, Friction): These forces oppose the direction of motion. If a particle is moving right, air resistance and friction act left.

---

2. Newton’s Three Laws of Motion (M1.3)

These three laws are the foundation of classical mechanics. You must know what each law means and, crucially, how the Second Law is used mathematically.

2.1 Newton's First Law (The Law of Inertia)

An object will remain at rest, or continue to move at a constant velocity (constant speed in a straight line), unless acted upon by a resultant force.

• If an object is at rest or moving at constant velocity, the forces acting on it are balanced.
• The Resultant Force (\(F_{\text{net}}\)) is zero.
• When the resultant force is zero, we say the object is in equilibrium.

2.2 Newton's Second Law (The Calculation Law)

The resultant force acting on an object is equal to the rate of change of momentum, and it is proportional to the mass and the acceleration.

For a constant mass, this simplifies to the most important formula in this chapter:

Resultant Force = Mass × Acceleration

$$\mathbf{F} = \mathbf{ma}$$

• \(F\) is the Net Force or Resultant Force (in Newtons, N) acting on the object.
• \(m\) is the mass (in kilograms, kg).
• \(a\) is the acceleration (in \(\text{ms}^{-2}\)).

Remember: \(F\) in this equation is not just *any* force; it is the total force causing the motion! If you have a driving force \(D\) and a resistance \(R\), then \(F = D - R\).

2.3 Newton's Third Law (Action and Reaction)

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

• The two forces are equal in magnitude.
• The two forces are opposite in direction.
• Crucially: They act on different objects.

Common Misconception: If a box is resting on the floor, the weight \(W\) pulling it down and the Normal Reaction \(R\) pushing it up are NOT a third law pair, as they both act on the box. A third law pair would be: (1) Earth pulls box (Weight) and (2) Box pulls Earth (Equal and opposite gravitational force).

---

3. Friction and Resistive Forces (M1.3)

Friction is a key resistive force. It always acts parallel to the surface and opposes motion (or impending motion).

3.1 Dynamic Friction

When an object is actually moving (sliding), the friction acting is called dynamic friction. Its maximum value (limiting friction) is proportional to the Normal Reaction force.

$$\mathbf{F} = \mathbf{\mu R}$$

• \(F\) is the magnitude of the friction force (N).
• \(\mu\) is the coefficient of friction (a dimensionless number, usually between 0 and 1).
• \(R\) is the Normal Reaction force (N).

If a problem states the surface is smooth, you assume \(\mu = 0\), so \(F = 0\). If the surface is rough, friction is present.

---

4. Solving Problems Using F = ma

When applying Newton's Second Law, we restrict ourselves to straight-line motion (horizontal or vertical, including motion up or down an inclined plane). Since resolution of forces is not required in this unit, forces will typically be co-linear with the acceleration.

Step-by-Step Guide for \(F=ma\) Problems

1. Draw the Diagram: Sketch the particle and show all forces acting on it (W, R, T, F, etc.).
2. Choose the Direction of Motion: Determine the direction of acceleration (\(a\)). This direction is the positive direction for your calculation.
3. Apply \(F = ma\): The resultant force (\(F_{\text{net}}\)) is the sum of forces in the positive direction minus the sum of forces in the negative direction.

   $$(\text{Forces in direction of } a) - (\text{Forces opposing } a) = ma$$

4. Solve: Substitute known values and solve for the unknown (mass, acceleration, or force).

Example: A particle of mass 5 kg is pulled horizontally by a force of 20 N across a rough surface where friction is 5 N.

Step 1 & 2: Diagram shows 20 N forward, 5 N backward. \(a\) is forward.
Step 3: \(F_{\text{net}} = ma\)
$$(20) - (5) = 5a$$
$$15 = 5a$$
Step 4: \(a = 3 \text{ ms}^{-2}\).

---

5. Connected Particle Problems (M1.3)

These involve two or more objects linked together, often by a light inextensible string or rod. The two main scenarios are cars/trailers and pulleys.

5.1 Car and Trailer/Connected Boxes

Imagine a car (mass \(M_C\)) pulling a trailer (mass \(M_T\)) with a driving force \(D\). The trailer is connected by a tow bar (exerting Tension \(T\) or Thrust).

Key Concept: Since the objects are connected by an inextensible string or rigid rod, they must have the same acceleration (\(a\)).

Method: Treating the System as a Whole

If you consider the whole system (Car + Trailer), the internal forces (Tension \(T\) or Thrust) cancel out, as they are action/reaction pairs acting between the objects.

$$F_{\text{net (System)}} = m_{\text{total}} a$$

The forces to consider are ONLY the external forces (Driving force, total Resistance).

Method: Treating Particles Separately

To find the internal force (\(T\)), you must apply \(F=ma\) to each body individually:

• Car: \(D - R_{\text{car}} - T = M_C a\)
• Trailer: \(T - R_{\text{trailer}} = M_T a\)

You now have simultaneous equations that you can solve for \(a\) and \(T\).

5.2 Pulley Systems (Atwood Machines)

These involve two masses connected by a light inextensible string passing over a smooth fixed peg or pulley.

Assumptions:

• The pulley/peg is smooth (no friction).
• The string is light (massless, so we ignore its weight).
• The string is inextensible (it doesn't stretch, so both masses have the same speed and acceleration \(a\)).

Step-by-Step for Pulleys

Consider two masses, \(M_1\) and \(M_2\), connected over a smooth pulley, with \(M_1 > M_2\).

1. Determine Motion: \(M_1\) will accelerate down, and \(M_2\) will accelerate up. Use the same Tension \(T\) for both masses (since the string is light).
2. Apply \(F=ma\) to \(M_1\) (Downward is positive): $$W_1 - T = M_1 a$$ $$\text{or } M_1 g - T = M_1 a \hspace{1cm} (\text{Equation 1})$$
3. Apply \(F=ma\) to \(M_2\) (Upward is positive): $$T - W_2 = M_2 a$$ $$\text{or } T - M_2 g = M_2 a \hspace{1cm} (\text{Equation 2})$$
4. Solve Simultaneously: Add Equation 1 and Equation 2 to eliminate \(T\) and find \(a\). $$(M_1 g - T) + (T - M_2 g) = M_1 a + M_2 a$$ $$g(M_1 - M_2) = a(M_1 + M_2)$$ $$a = \frac{g(M_1 - M_2)}{(M_1 + M_2)}$$

This method is robust! Always define your positive direction based on the expected motion of the individual mass.