The physics behind engineering and the physical sciences: motion, forces, energy, momentum, torque, circuits, induction, waves and heat, with Python exercises and a home pendulum lab.

Physics for Engineering & Science

## Why this matters

Kinematics is the description of motion: where something is, how fast it is going and how quickly that speed changes. It comes before any question about *why* things move, and it shows up everywhere in engineering and science. A civil engineer checks how far a car travels while braking, a mechanical engineer works out how long a conveyor takes to bring a part up to speed, and a physicist reconstructs a particle's path from detector hits. All of them start with the same few quantities and four equations.

## The quantities

- **Position** x (metres, m): where the object is, measured from an origin you choose.
- **Displacement** Δx = x − x₀ (m): the change in position. It has a sign. Moving 5 m left and then 5 m right is a displacement of 0 m, even though you walked 10 m.
- **Velocity** v (m/s): the rate of change of position. Speed is its size; velocity also carries direction.
- **Acceleration** a (m/s²): the rate of change of velocity. An acceleration of 3 m/s² means the velocity grows by 3 m/s every second.
- **Time** t (s).

Pick a positive direction at the start of every problem and stick with it. Any quantity pointing the other way gets a minus sign. Most kinematics mistakes are sign mistakes.

## The constant-acceleration equations

When the acceleration stays constant, four equations connect the five quantities v₀ (starting velocity), v (final velocity), a, t and Δx:

```text
1.  v  = v₀ + a·t
2.  Δx = v₀·t + ½·a·t²
3.  v² = v₀² + 2·a·Δx
4.  Δx = ½·(v₀ + v)·t
```

Each equation leaves out one quantity. Equation 1 has no Δx, equation 2 has no v, equation 3 has no t and equation 4 has no a. The strategy is always the same: list the three values you know, name the one you want, and pick the equation that does not contain the fifth (the one you neither know nor need).

These equations are only valid while the acceleration is constant. If a problem has two phases (speeding up, then cruising), solve each phase separately and use the end of the first phase as the start of the second.

## Worked example 1: a car pulling away

A car starts from rest and accelerates at 3.0 m/s² for 6.0 s. How fast is it going, and how far has it travelled?

Known: v₀ = 0, a = 3.0 m/s², t = 6.0 s.

```text
v  = 0 + 3.0 × 6.0        = 18 m/s   (about 65 km/h)
Δx = 0 × 6.0 + ½ × 3.0 × 6.0² = 54 m
```

Check with equation 4: ½ × (0 + 18) × 6.0 = 54 m. Two equations agreeing is a quick way to catch arithmetic slips.

## Worked example 2: stopping distance

A car travels at 25 m/s (90 km/h) and brakes with a constant deceleration of 7.0 m/s². How far does it slide before stopping?

Take the direction of travel as positive, so a = −7.0 m/s². Known: v₀ = 25 m/s, v = 0, a = −7.0 m/s². We do not know or need t, so use equation 3:

```text
0 = 25² + 2 × (−7.0) × Δx
Δx = 625 / 14 = 44.6 m
```

In general the braking distance is v₀² / (2·|a|). Because speed is squared, doubling the speed makes the braking distance four times longer. That single fact explains a lot of road design.

Real stopping distance also includes the driver's reaction time. During a 1.5 s reaction the car still moves at 25 m/s, covering 37.5 m before the brakes even start, so the total is about 82 m. Road and vehicle standards use carefully chosen values for reaction time and friction; never take the numbers in a lesson as design values.

## Worked example 3: free fall

Near Earth's surface, anything falling freely (ignoring air resistance) accelerates downward at g ≈ 9.81 m/s². Drop a ball from a 20 m balcony. Take down as positive, so a = +9.81 m/s², v₀ = 0, Δx = 20 m.

```text
20 = ½ × 9.81 × t²   →   t = √(40 / 9.81) = 2.02 s
v  = 9.81 × 2.02 = 19.8 m/s
```

## The same idea in code

Engineers rarely solve these by hand more than once. A function makes the equations reusable and easy to test:

```python
def displacement(v0, a, t):
    """Displacement after t seconds of constant acceleration."""
    return v0 * t + 0.5 * a * t ** 2

print(displacement(0, 3.0, 6.0))   # 54.0
```

## Common mistakes

- **Mixing units.** Convert km/h to m/s first: divide by 3.6. 90 km/h is 25 m/s.
- **Forgetting signs.** A car that is slowing down while moving forward has a negative acceleration in your chosen frame.
- **Using the equations when a changes.** If the force (and so the acceleration) changes, split the motion into phases or use a numerical method.
- **Confusing distance and displacement.** The equations give displacement.

## Quick recap

- Choose a positive direction and give every vector quantity a sign.
- With constant acceleration, four equations link v₀, v, a, t and Δx. Pick the one that leaves out the quantity you do not need.
- Braking distance grows with the square of speed.
- Free fall near Earth: a = 9.81 m/s² downward.

## Try it

Open the **Do** tab and write the stopping-distance function, then use it to see how braking distance changes with speed.


Describe motion with displacement, velocity and acceleration, and solve constant-acceleration problems such as braking and free fall.

Kinematics

Motion in a Straight Line: Crash Course Physics #1

Kinematics Part 1: Horizontal Motion

## Why this matters

Kinematics describes motion; Newton's laws explain it. Every structural load calculation, every machine part sizing and every orbit calculation is built on three short statements Isaac Newton published in 1687. The tool that makes them usable is the **free-body diagram**: a sketch of one object with every force acting on it. Engineers draw these before they write a single equation, because a missing or invented force makes every number after it wrong.

## The three laws

1. **First law (inertia).** An object keeps moving at a constant velocity (which includes staying at rest) unless a net force acts on it. A hockey puck slides a long way on ice because there is very little friction to change its velocity.
2. **Second law.** The net force on an object equals its mass times its acceleration:

```text
ΣF = m·a
```

Force is measured in **newtons** (N). One newton is the force that gives 1 kg an acceleration of 1 m/s², so 1 N = 1 kg·m/s². The Σ means "add up every force, as vectors".

3. **Third law.** When object A pushes on object B, B pushes back on A with a force of equal size in the opposite direction. The two forces act on *different* objects, so they never cancel each other in a single free-body diagram.

## Mass and weight

Mass (kg) is how much matter an object has and how strongly it resists acceleration. **Weight** is the gravitational force on it: W = m·g, with g ≈ 9.81 m/s² near Earth's surface. A 2.0 kg box weighs 2.0 × 9.81 = 19.6 N. On the Moon the same box has the same mass but weighs only about a sixth as much.

## Drawing a free-body diagram

1. Pick **one** object and draw it as a dot or a box.
2. Draw every force acting **on** that object, starting at the object and pointing in the direction of the force. Common ones:
   - **Weight** W, straight down.
   - **Normal force** N, perpendicular to any surface it rests on, pushing away from the surface.
   - **Tension** T, along a rope or cable, pulling away from the object.
   - **Friction** f, along the surface, opposing sliding (next lesson).
   - **Applied forces**, such as a push.
3. Choose axes (usually x along the motion, y perpendicular to it).
4. Write ΣF = m·a separately for each axis.

Only forces from things touching the object (plus gravity, which acts at a distance) belong on the diagram. "The force of motion" is not a force; velocity is not a force.

## Worked example 1: pushing a box

You push a 2.0 kg box across a floor with a horizontal force of 10 N. Friction opposes the motion with 4.0 N. What is the box's acceleration?

Free-body diagram: push 10 N to the right, friction 4.0 N to the left, weight 19.6 N down, normal force up.

```text
y-axis:  N − 19.6 = m·0      →  N = 19.6 N   (the box does not leave the floor)
x-axis:  10 − 4.0 = 2.0·a    →  a = 3.0 m/s² to the right
```

## Worked example 2: an elevator cable

A 500 kg elevator car hangs from a cable. What is the cable tension when the car (a) sits still, (b) accelerates upward at 2.0 m/s², and (c) accelerates downward at 2.0 m/s²?

Take up as positive. Two forces act on the car: tension T up and weight m·g down.

```text
ΣF = T − m·g = m·a     →     T = m·(g + a)

(a) a = 0:      T = 500 × 9.81        = 4905 N
(b) a = +2.0:   T = 500 × (9.81 + 2.0) = 5905 N
(c) a = −2.0:   T = 500 × (9.81 − 2.0) = 3905 N
```

This is why you feel heavier when an elevator starts upward: the floor has to push harder on you to accelerate you. If the car were in free fall (a = −9.81 m/s²), the tension would be zero.

Real elevator cables, cranes and lifting gear are designed with large safety factors and to published standards, and are signed off by qualified engineers and inspectors. This example shows the physics, not a design method.

## Writing it in code

Treat each force as an (x, y) pair. Adding forces is then adding pairs:

```python
forces = [(10, 0), (-4, 0), (0, -19.62), (0, 19.62)]
fx = sum(f[0] for f in forces)
fy = sum(f[1] for f in forces)
print(fx / 2.0, fy / 2.0)   # 3.0 0.0  (acceleration in m/s²)
```

## Common mistakes

- **Putting both third-law forces on one diagram.** The floor pushes on the box (on the box's diagram); the box pushes on the floor (on the floor's diagram).
- **Assuming the normal force always equals the weight.** It does on a flat floor with no vertical acceleration. On a ramp, in an elevator, or with someone pressing down, it does not.
- **Adding a "force of motion".** A moving object with no net force keeps moving; it does not need a force to sustain its velocity.
- **Mixing up mass and weight.** Mass in kg, weight in N.

## Quick recap

- ΣF = m·a, applied separately along each axis.
- Weight is m·g; 1 N = 1 kg·m/s².
- Third-law pairs act on different objects.
- Draw the free-body diagram first, every time.

## Try it

In the **Do** tab, finish the elevator-tension function and test it for an elevator that is still, speeding up and in free fall.


Draw free-body diagrams and use Newton's three laws to solve for forces and accelerations.

Newton's laws

Newton's Laws: Crash Course Physics #5

Free Body Diagrams - Tension, Friction, Inclined Planes, & Net Force

## Why this matters

Friction is the force that lets you walk, lets tyres grip a road and keeps bolts from shaking loose. It is also the force that wears out bearings and wastes energy in machines. Inclines (ramps, roads on hills, conveyor slopes, soil on a hillside) are where friction and gravity compete most visibly, and the inclined-plane problem is the standard way to practise splitting forces into components. The method you learn here, choosing smart axes and resolving forces along them, is the same one used later in statics and structural analysis.

## Two kinds of friction

When two dry surfaces press together, friction acts along the surface and opposes sliding. The simple model (good to roughly 10–20% for many dry materials) is:

```text
static friction:   f_s ≤ μ_s · N     (surfaces not sliding; f_s is only as big as needed)
kinetic friction:  f_k = μ_k · N     (surfaces sliding)
```

- N is the **normal force**, the push of the surface perpendicular to itself.
- μ_s and μ_k are the **coefficients** of static and kinetic friction. They have no units and depend on both surfaces. Usually μ_k < μ_s, which is why it is harder to start a heavy box moving than to keep it moving.
- Rough values: rubber on dry concrete about 0.6–0.8, wood on wood about 0.25–0.5, ice on ice about 0.03. Real values vary with surface finish, moisture, temperature and contamination, so engineers use measured or published values with margins, not textbook numbers.

Static friction is a "whatever it takes, up to a limit" force. A book resting on a level table feels zero friction, because nothing is trying to slide it.

## Forces on an incline

Put a block of mass m on a ramp tilted at angle θ. Three forces act on it: weight m·g (straight down), the normal force N (perpendicular to the ramp) and friction f (along the ramp).

Choose the x-axis **along the ramp** (down-slope positive) and the y-axis **perpendicular** to it. Now only the weight needs splitting:

```text
component of weight along the ramp:          m·g·sin θ
component of weight into the ramp:           m·g·cos θ
```

Perpendicular to the ramp nothing accelerates, so N = m·g·cos θ. The steeper the ramp, the smaller the normal force, and so the smaller the maximum friction.

## Will it slide?

The block stays put as long as static friction can balance the down-slope pull:

```text
m·g·sin θ ≤ μ_s · m·g·cos θ    →    tan θ ≤ μ_s
```

The mass cancels. The block starts to slide at the angle where tan θ = μ_s. That gives a neat household measurement: rest a coin on a hardcover book, tilt the book slowly until the coin just slips, measure the angle, and μ_s ≈ tan θ.

## Acceleration while sliding

Once the block slides, kinetic friction acts up the slope:

```text
ΣF_x = m·g·sin θ − μ_k·m·g·cos θ = m·a
a = g · (sin θ − μ_k · cos θ)
```

Again the mass cancels. If the result is negative, friction is stronger than the down-slope pull: a block given a push down the ramp would slow down and stop.

## Worked example

A 10 kg crate sits on a ramp tilted at 30°. μ_s = 0.50 and μ_k = 0.40.

1. **Does it slide?** tan 30° = 0.577, which is greater than 0.50. Yes, static friction cannot hold it.
2. **Forces.**

```text
weight             = 10 × 9.81              = 98.1 N
along the ramp     = 98.1 × sin 30°         = 49.1 N
normal force N     = 98.1 × cos 30°         = 85.0 N
kinetic friction   = 0.40 × 85.0            = 34.0 N
net force          = 49.1 − 34.0            = 15.1 N
acceleration       = 15.1 / 10              = 1.51 m/s²
```

3. **Check with the formula.** a = 9.81 × (0.500 − 0.40 × 0.866) = 9.81 × 0.154 = 1.51 m/s². The numbers agree.

If the ramp were frictionless, a = 9.81 × sin 30° = 4.91 m/s².

## In code

Python's `math` module works in radians, so convert degrees first:

```python
import math

theta = math.radians(30)
print(round(math.tan(theta), 3))   # 0.577
```

## Common mistakes

- **Using degrees in `math.sin`.** `math.sin(30)` treats 30 as radians. Use `math.radians(30)`.
- **Setting N = m·g on a slope.** On an incline N = m·g·cos θ.
- **Swapping sin and cos.** Sanity-check with extremes: at θ = 0 the down-slope pull must be zero (sin 0 = 0); at θ = 90° the block is in free fall.
- **Always using μ_s · N.** Static friction is a maximum, not a fixed value.

## Safety note

Slopes, ramps and retaining structures in the real world involve soil, wet surfaces, dynamic loads and people. Their design follows codes and must be reviewed and sealed by a licensed professional engineer. Use these models to build intuition, not to approve a design.

## Quick recap

- Kinetic friction f_k = μ_k·N; static friction can be anything up to μ_s·N.
- On a ramp, resolve weight into m·g·sin θ (along) and m·g·cos θ (into); N = m·g·cos θ.
- It slides when tan θ > μ_s; then a = g(sin θ − μ_k cos θ).

## Try it

In the **Do** tab, write the sliding-acceleration formula and test it against the worked example.


Model static and kinetic friction and solve an inclined-plane problem by resolving forces along the slope.

Friction and inclines

Friction: Crash Course Physics #6

Introduction to Inclined Planes

Describe motion with the constant-acceleration equations, explain it with Newton's laws and free-body diagrams, and solve friction and inclined-plane problems.