Fundamentals of Mechanical Engineering
This chapter establishes the working vocabulary: units, forces, vectors, and free body diagrams. Master these and the rest of the course is variations on a theme.
Engineering Units and Systems
The SI System
Mechanical engineers primarily use SI units (the metric system). Many engineering failures trace back to unit errors, so taking units seriously pays off immediately.
Base SI Units for Mechanical Engineering
| Quantity | Unit | Symbol |
|---|---|---|
| Length | meter | m |
| Mass | kilogram | kg |
| Time | second | s |
| Force | Newton | N |
| Pressure | Pascal | Pa |
| Energy | Joule | J |
| Power | Watt | W |
| Temperature | Kelvin or Celsius | K or °C |
Derived Units
| Quantity | Formula | Unit |
|---|---|---|
| Force | F = m × a | 1 N = 1 kg⋅m/s² |
| Pressure | P = F / A | 1 Pa = 1 N/m² |
| Energy/Work | W = F × d | 1 J = 1 N⋅m |
| Power | P = W / t | 1 W = 1 J/s |
Common Prefixes
| Prefix | Symbol | Factor | Example |
|---|---|---|---|
| kilo | k | 10³ | 1 km = 1,000 m |
| Mega | M | 10⁶ | 1 MPa = 1,000,000 Pa |
| Giga | G | 10⁹ | 1 GW = 1,000,000,000 W |
| milli | m | 10⁻³ | 1 mm = 0.001 m |
| micro | µ | 10⁻⁶ | 1 µm = 0.000001 m |
| nano | n | 10⁻⁹ | 1 nm = 0.000000001 m |
US Customary Units (Still Common)
Many US industries still use Imperial or US units:
| Quantity | Common Units |
|---|---|
| Length | inch (in), foot (ft), mile |
| Force | pound-force (lbf) |
| Mass | pound-mass (lbm), slug |
| Pressure | psi (lbf/in²) |
| Energy | British Thermal Unit (BTU) |
| Power | horsepower (hp) |
Key Conversions
- 1 inch = 25.4 mm
- 1 foot = 0.3048 m
- 1 lbf = 4.448 N
- 1 psi = 6,895 Pa
- 1 hp = 746 W
Unit Conversion Tips
Always use dimensional analysis.
Example: Convert 60 mph to m/s
60 miles/hour × (1 hour/3600 s) × (5280 ft/1 mile) × (0.3048 m/1 ft)
= 60 × (5280 × 0.3048) / 3600
= 26.8 m/s
Write units in every step of the calculation. Errors show up as soon as the units stop matching.
Forces
What is a Force?
A force is a push or pull that can accelerate, deform, or change the state of an object. Measured in Newtons (N) or pounds-force (lbf).
Types of Forces
Contact Forces
- Normal force (perpendicular to surface)
- Friction force (opposes motion)
- Tension (pulling through rope or cable)
- Compression (pushing)
Body Forces
- Gravity / weight (W = m × g)
- Magnetic force
- Electrostatic force
Applied Forces
- External loads
- Pressures
- Distributed loads
Newton's Laws of Motion
First Law (Inertia)
An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted on by a net external force.
Second Law (F = ma)
The acceleration of an object is directly proportional to the net force and inversely proportional to its mass.
F = m × a
Where:
- F = net force (N)
- m = mass (kg)
- a = acceleration (m/s²)
Third Law (Action-Reaction)
For every action, there is an equal and opposite reaction.
When you push a wall with 100 N, the wall pushes back on you with 100 N.
Weight vs Mass
Mass (m): amount of matter, measured in kg. Weight (W): force due to gravity, W = m × g.
On Earth: g = 9.81 m/s² (or approximately 10 m/s²)
Example:
- Mass: 70 kg (same everywhere)
- Weight on Earth: W = 70 kg × 9.81 m/s² = 686.7 N
- Weight on Moon: W = 70 kg × 1.62 m/s² = 113.4 N
Vectors
Forces have both magnitude (size) and direction, which makes them vectors.
Vector Representation
^ y
|
| F (force vector)
| /
| /
| /θ
|/________> x
Component Form
- Fx = F × cos(θ), horizontal component
- Fy = F × sin(θ), vertical component
Magnitude
- |F| = sqrt(Fx² + Fy²)
Direction
- θ = atan(Fy / Fx)
Vector Addition
To add vectors, add their components.
F1 = (F1x, F1y)
F2 = (F2x, F2y)
Fresultant = (F1x + F2x, F1y + F2y)
Example
F1 = 50 N at 0° (horizontal) -> (50, 0)
F2 = 30 N at 90° (vertical) -> (0, 30)
Fresultant = (50, 30)
|F| = sqrt(50² + 30²) = sqrt(2500 + 900) = sqrt(3400) = 58.3 N
θ = atan(30/50) = 31°
3D Vectors
In 3D space, vectors have three components.
F = (Fx, Fy, Fz)
Where:
Fx = F × cos(α) x-component
Fy = F × cos(β) y-component
Fz = F × cos(γ) z-component
|F| = sqrt(Fx² + Fy² + Fz²)
Free Body Diagrams (FBDs)
The most important skill in mechanics. A free body diagram shows all forces acting on an isolated object.
How to Draw an FBD
- Isolate the object of interest.
- Draw all forces as vectors.
- Label magnitude and direction.
- Include reactions at supports.
Example: Block on a Table
^ N (Normal force from table)
|
[========] block (mass m)
|
v W = mg (Weight)
Forces on the block:
- Weight W = mg (downward)
- Normal force N (upward from table)
If at rest: N = W (equilibrium).
Example: Block on Inclined Plane
N ^ (perpendicular to surface)
|
|
[====] block
/ v W = mg
/ θ
/________
Components:
- Weight parallel to plane: W_parallel = mg sin(θ)
- Weight perpendicular to plane: W_perp = mg cos(θ)
- Normal force: N = mg cos(θ)
Common Support Types
| Support | Reactions |
|---|---|
| Pin / Hinge | Fx, Fy (no moment) |
| Roller | Single perpendicular force |
| Fixed | Fx, Fy, M (moment) |
| Rope / Cable | Tension along rope only |
Equilibrium
An object is in equilibrium when:
- ΣF = 0 (sum of all forces equals zero)
- ΣM = 0 (sum of all moments equals zero)
In 2D (Planar Problems)
ΣFx = 0 sum of horizontal forces = 0
ΣFy = 0 sum of vertical forces = 0
ΣM = 0 sum of moments = 0
Example Problem: Hanging Sign
/|\
/ | \
/ | \
T1 | T2 tension cables at 45° and 30°
|
[SIGN] W = 200 N
Given: sign weighs 200 N, T1 at 45°, T2 at 30° from horizontal. Find: T1 and T2.
Solution
ΣFx = 0: T1 cos(45°) - T2 cos(30°) = 0
ΣFy = 0: T1 sin(45°) + T2 sin(30°) - 200 = 0
From equation 1:
T1 × 0.707 = T2 × 0.866
T1 = 1.225 T2
Substitute into equation 2:
1.225 T2 × 0.707 + T2 × 0.5 = 200
0.866 T2 + 0.5 T2 = 200
1.366 T2 = 200
T2 = 146.4 N
T1 = 1.225 × 146.4 = 179.3 N
Practice Problems
Problem 1: Unit Conversion
Convert 100 km/h to m/s.
<details> <summary>Solution</summary>
100 km/h × (1000 m / 1 km) × (1 h / 3600 s)
= 100 × 1000 / 3600
= 27.8 m/s
</details>
Problem 2: Vector Addition
Two forces act on a bolt: F1 = 100 N east, F2 = 150 N north. Find the resultant force.
<details> <summary>Solution</summary>
F1 = (100, 0) N
F2 = (0, 150) N
Fresultant = (100, 150) N
Magnitude: |F| = sqrt(100² + 150²) = sqrt(10000 + 22500) = 180.3 N
Direction: θ = atan(150/100) = 56.3° from east
</details>
Problem 3: Weight Calculation
What is the weight of a 500 kg machine on Earth?
<details> <summary>Solution</summary>
W = m × g
W = 500 kg × 9.81 m/s²
W = 4,905 N ≈ 4.9 kN
</details>
Key Takeaways
- Units are critical: include them at every step and convert carefully.
- Forces are vectors: magnitude and direction both matter.
- FBDs are essential: master them for every problem.
- Equilibrium means no acceleration (ΣF = 0, ΣM = 0).
- Break vectors into components: every other calculation gets easier.
Next Steps
Continue to 02-statics.md to apply equilibrium to structures and systems at rest.