ME3120 · Dynamic Modelling
Theme 2 · Newton's method
Newton's laws & free-body diagrams
The oldest modelling tool: isolate a body, draw its forces, and write force = mass × acceleration. The recipe we'll run on every machine.
Source: course notes, Week 1 foundations.
Before you start
What you need first
- Generalized coordinates \(q\) — one per degree of freedom.
- State \([q,\dot q]\) — and that the laws give acceleration \(\ddot q\).
What you'll be able to do
- Use the 5-step recipe for any machine.
- Write Newton's law for straight-line and turning motion.
- Draw a correct free-body diagram (FBD).
The 5-step recipe — used on every machine
1Draw it & set the boundary. A clear picture. What is inside the system, what is outside?
2Choose coordinates \(q\). One per DOF. Mark the positive direction.
3Account for forces. Draw a free-body diagram (today) — or write the energies (next theme).
4Write the balance. Newton: \(\sum F=m\ddot x,\ \ \sum\tau=I\ddot\theta\).
5Check. Units match? Sensible at the limits? Where does it rest?
We run these same five steps on every machine — from a spring next, to a
full robot arm later.
Newton's law — straight-line motion
$$\sum F = m\,\ddot x$$
| Symbol | Meaning | SI unit |
|---|---|---|
| \(\sum F\) | all forces added together (with sign) | N |
| m | mass | kg |
| \(\ddot x\) | acceleration | m/s² |
In words: (total push) = (mass) × (acceleration).
Newton's law — turning motion
$$\sum \tau = I\,\ddot\theta$$
| Symbol | Meaning | SI unit |
|---|---|---|
| \(\sum \tau\) | all torques added together (with sign) | N·m |
| I | moment of inertia — the "rotational mass" | kg·m² |
| \(\ddot\theta\) | angular acceleration | rad/s² |
In words: (total twist) = (rotational inertia) × (angular acceleration).
The same idea, twice
| Straight line | Turning |
|---|---|
| force \(F\) | torque \(\tau\) |
| mass \(m\) | inertia \(I\) |
| acceleration \(\ddot x\) | angular acc. \(\ddot\theta\) |
| \(\sum F = m\ddot x\) | \(\sum\tau = I\ddot\theta\) |
🔭 Looking ahead: robots turn at their joints, so the
torque version is the one we use most once we reach the arm.
How to draw a free-body diagram
- Isolate one body — imagine cutting it free from everything else.
- Draw every outside force as an arrow: gravity, normal, friction, applied push.
- Pick a positive direction and keep it for the whole problem.
- Add the forces along each axis → that sum is \(\sum F\).
Isolate the block; draw only the outside forces acting on it.
Golden rule: only forces from outside the
body go on the diagram. There is no "\(m\ddot x\) arrow" — that lives on the
other side of the equation.
Find the acceleration from an FBD
A crate of mass \(m = 5\ \text{kg}\) is pushed along the floor by a horizontal force \(F = 20\ \text{N}\). Friction opposes the motion with a force \(f = 5\ \text{N}\). Find the acceleration.
Free-body diagram of the crate, with the numbers from the problem.
1Pick \(+x\) in the push direction and sum the horizontal forces:
$$\sum F = F - f = 20 - 5 = 15\ \text{N}$$
2Apply \(\sum F = m\ddot x\) and solve for \(\ddot x\):
$$\ddot x = \frac{\sum F}{m} = \frac{15}{5} = 3\ \text{m/s}^2$$
The vertical forces (weight \(mg\) down, normal \(N\) up) cancel, so they
don't affect the horizontal motion — exactly why the FBD keeps the axes separate.
✏️ Try it yourself
A crate of mass \(m = 4\ \text{kg}\) is pushed with \(F = 22\ \text{N}\); friction opposes it with \(f = 6\ \text{N}\). Find the acceleration.
Step 1.
\(\sum F = 22 - 6 = 16\ \text{N}\)
Step 2.
\(\ddot x = \dfrac{16}{4} = 4\ \text{m/s}^2\)
Recap — the whole topic on one screen
| Idea | What you own now |
|---|---|
| The 5-step recipe | Draw → coordinates → forces → balance → check |
| Newton (line) | \(\sum F = m\ddot x\) |
| Newton (turning) | \(\sum\tau = I\ddot\theta\) |
| Free-body diagram | Isolate the body; only outside forces; pick a + direction |