Force diagrams are a fundamental aspect of physics, providing a visual representation of the forces acting on an object. These diagrams are crucial in understanding static situations, where forces are in equilibrium, meaning there is no net force causing motion.

## Introduction to Force Diagrams

**Purpose: **Show force magnitude and direction on an object.

**Components:**

- Object shown as a dot or shape.
- Arrows for forces: length = magnitude, direction = force direction.

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## Creating and Interpreting Force Diagrams

**Steps:**

1. Represent object as a dot or shape.

2. Identify all forces on it.

3. Draw arrows from the object: length shows force size, direction shows force direction.

4. Label forces (e.g., F_gravity, F_tension).

**Analyzing Diagrams:** Check arrow balance. Balanced = static situation, net force = zero.

## Types of Forces in Static Situations

**Weight (Gravity):**Downward force, mass x gravity acceleration (F_gravity = m*g).**Tension:**Force in a stretched string/rope/wire.**Normal Reaction:**Force from a surface against an object, perpendicular to surface.**Friction:**Force opposing motion between two surfaces, parallel to surface.

## Analysis of Forces

**Weight and Gravity:**Force due to gravity, downward, equals mass x gravity $9.81 m/s^2$ on Earth.**Tension in Strings/Cables:**Force along a string/rope/cable when pulled.**Normal Reaction Force:**Surface force against an object, perpendicular to surface, equal and opposite to net pressing force.

## Example

**Problem:**

- Two boxes stacked: $m_1 = 10 kg, m_2 = 2 kg$.
- Find max horizontal force $F$ on bottom box to keep top box from sliding.
- Coefficient of static friction $μ_s$ between boxes = 0.2.
- Ignore ground friction.

**Solution:**

**1. Static Friction Force (F_R12):**

- $F_R12 = μ_s × m_2 × g$
- $F_R12 = 0.2 × 2 kg × 9.81 m/s^2$
- $F_R12 = 3.924 N$

**2. Max Force on Bottom Box (F_max):**

- $F_max = 3.924 N$
- $F_max = 3.924 N$

**3. Acceleration (a):**

- Total mass = $m_1 + m_2$
- $a = \frac{F}{m_1 + m_2}$
- $a = \frac{3.924 \, \text{N}}{10 \, \text{kg} + 2 \, \text{kg}} $</li><li>$a = 0.327 \, \text{m/s}^2 $</li></ul><p><strong>Results:</strong></p><ul><li>Max force without sliding =$3.924 N$</li><li>Acceleration of boxes =$0.327 \, \text{m/s}^2 $

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.