Forces and newtons laws of motion

Free-Body Force Diagrams and Centre of Gravity

A free-body force diagram shows a single object isolated from its surroundings with every external force acting on it represented by an arrow. Each arrow starts at the point where the force acts. The arrow’s direction shows the direction of the force, and its length represents the magnitude.

For an extended rigid body [an object whose size cannot be ignored], weight is drawn as a single arrow acting from the centre of gravity. The centre of gravity is the single point at which the entire weight of the object appears to act. For a uniform symmetric object, this point lies at its geometric centre.

MisconceptionForces are not “stored” inside an object. A free-body diagram only shows external forces acting on the chosen body. Never include forces the object exerts on something else.

Exam TipLabel each arrow with the force name and the body exerting it.

Newton’s First and Second Laws

Newton’s second law states that the resultant force on an object of constant mass equals the product of its mass and acceleration. Newton’s first law is the special case where the resultant force is zero, so the object remains at rest or continues at constant velocity.

Key Equations

Newton’s second law:

$$F=ma$$

Variables:

• ΣF = resultant force on the object, in newtons (N)
• m = mass of the object, in kilograms (kg)
• a = acceleration of the object, in metres per second squared (m s⁻²)

SI unit: newton (N), where 1 N = 1 kg m s⁻²

Rearrangements:

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

ProportionalityFor constant mass, acceleration is directly proportional to resultant force. For constant resultant force, acceleration is inversely proportional to mass. Doubling the resultant force doubles the acceleration; doubling the mass halves the acceleration.

The resultant force ΣF is the vector sum of every force acting on the body. When ΣF = 0, the acceleration is zero, so the object stays at rest or moves at constant velocity. This is Newton’s first law of motion.

Terminal velocity is reached by a falling object when the upward drag force grows until it equals the weight. The resultant force then becomes zero and therefore the acceleration is zero, so the object falls at constant velocity.

Examiner InsightWhen asked about terminal velocity, name the two opposing forces, state that they are equal in magnitude, then state the consequence: zero resultant force and therefore zero acceleration.

Exam TipAlways link “balanced forces” → “zero resultant” → “zero acceleration”.

Gravitational Field Strength and Weight

Gravitational field strength at a point is the gravitational force exerted per unit mass on a small test mass placed at that point. Weight is the gravitational force on a body and acts vertically downwards at the centre of gravity.

Key Equations

Gravitational field strength:

$$g=\frac{F}{m}$$

Weight from gravitational field strength:

$$W=mg$$

Variables:

• g = gravitational field strength, in newtons per kilogram (N kg⁻¹) or m s⁻²
• F = gravitational force on the test mass, in newtons (N)
• m = mass of the object, in kilograms (kg)
• W = weight of the object, in newtons (N)

SI unit of g: N kg⁻¹ (numerically equal to m s⁻²) SI unit of W: newton (N)

Rearrangements:

$$F=mg$$ $$m=\frac{W}{g}$$

ProportionalityWeight is directly proportional to mass at a given location. Doubling the mass doubles the weight. At the same point in a field, all objects experience the same gravitational field strength.

Near the surface of the Earth, g ≈ 9.81 N kg⁻¹. The numerical equivalence between N kg⁻¹ and m s⁻² explains why a freely-falling object near Earth’s surface accelerates at approximately 9.81 m s⁻². The unit N kg⁻¹ emphasises the field interpretation, while m s⁻² emphasises the acceleration interpretation.

Mass is a scalar measure of the quantity of matter in an object and stays constant everywhere. Weight is a vector and depends on the local gravitational field strength. An astronaut on the Moon has the same mass as on Earth but a smaller weight because Moon’s g is smaller.

Core Practical 1: Acceleration of Free Fall

This Core Practical determines the acceleration due to gravity, g, by timing how long an object takes to fall a measured distance under gravity alone. The key idea is that drag on a small dense object is negligible, so the only significant force on it is its weight, giving acceleration g.

Aim

To determine the acceleration of free fall, g, by measuring the time taken for a small steel ball-bearing to fall through a known vertical distance.

Method

1. Set up the apparatus with the electromagnet directly above the trapdoor and the metre rule vertical.
2. Measure the fall distance s from the bottom of the ball-bearing to the trapdoor using the metre rule.
3. Switch on the electromagnet to hold the ball-bearing in place and reset the timer to zero.
4. Open the switch to release the ball-bearing; this simultaneously starts the timer.
5. The ball-bearing strikes the trapdoor, opening it and stopping the timer. Record the time t.
6. Repeat the timing three times for the same distance and calculate a mean time.
7. Vary s over a suitable range (for example 0.20 m to 1.00 m in 0.10 m steps) and repeat steps 3–6 for each new distance.
8. Plot a graph of s on the y-axis against ½t² on the x-axis; the gradient equals g.

Variables

• Independent variable (IV): fall distance s, in metres, varied between 0.20 m and 1.00 m
• Dependent variable (DV): time of fall t, measured in seconds using an electronic timer (resolution 0.001 s)
• Control variables (CV): the same ball-bearing throughout (constant mass and shape), the same release height of the electromagnet, the apparatus kept perfectly vertical

Expected results

Using s = ½gt² with initial velocity zero, a graph of s against ½t² is a straight line through the origin with gradient g. The expected gradient is approximately 9.81 m s⁻².

Precaution

A systematic error arises because the ball-bearing takes a short time to leave the electromagnet after the current is switched off. This makes recorded times slightly too long and reduces the calculated value of g. Using a low-residual-magnetism electromagnet and clean iron core minimises this delay.

Measurement technique

Time the fall electronically rather than by hand to remove human reaction time. The timer resolution of 0.001 s gives a small percentage uncertainty in t. Repeat each timing three times and use the mean to reduce random error.

SafetyMechanical hazard: place a soft pad or tray below the trapdoor so the falling ball-bearing does not strike feet or floor. Keep fingers clear of the release point.

Newton’s Third Law of Motion

Newton’s third law states that when body A exerts a force on body B, body B exerts an equal and opposite force on body A. The two forces in such a pair are called an interaction pair or Newton’s third law pair.

The two forces in an interaction pair share five essential properties:

• They are equal in magnitude.
• They act in opposite directions.
• They are of the same type (for example, both gravitational, both contact, both electrostatic).
• They act along the same line.
• They act on two different bodies.

A common test scenario is a book resting on a table. The book’s weight (Earth pulling the book down) pairs with the gravitational pull of the book on the Earth, not with the normal contact force from the table. The normal contact force from the table on the book pairs with the contact force from the book pushing down on the table.

MisconceptionA book on a table has weight (down) and normal contact force (up), but these are NOT a third-law pair. They act on the same body and they are different types of force.

Exam TipThird-law pairs always act on two different bodies and are the same type of force.