Week 4 Cheatsheet — Forces and Newton's Laws

How this week breaks down

Three workflows, all flowing from one master technique — isolate body, sum forces, equate to $m\vec{a}$. Skim this once, then revise from the in-depth note.

Topic	What you do
Identify forces	List every push and pull on the body. Tag each as contact / long-range.
Draw the FBD	Body as a dot at origin. Draw each force as a labelled vector. Pick axes.
Apply Newton’s laws	Sum forces per axis. Set equal to $ma$ (or $0$ in equilibrium). Solve.

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1 — The catalogue of forces

Force	Symbol	Contact?	Direction	Magnitude / formula
Gravity / weight	${\vec{F}}_G$, ${\vec{F}}_W$	Long-range	Vertically down	$F_G=mg$ ($g=9.81{\text{m/s}}^2$)
Spring	${\vec{F}}_S$	Contact	Along spring, toward natural length	$F_S=kx$
Normal	${\vec{F}}_N$	Contact	Perpendicular to surface	Whatever it takes for vertical equilibrium
Tension	${\vec{F}}_T$	Contact	Along the rope/string	Whatever the rope is pulling with
Kinetic friction	${\vec{F}}_{fric}$	Contact	Parallel to surface, opposite to velocity	$F_{fric}={\mu }_k{F}_N$
Static friction	${\vec{F}}_{fric}$	Contact	Parallel to surface, opposite to impending motion	$F_{fric}\le {\mu }_s{F}_N$
Thrust	${\vec{F}}_{thrust}$	Contact	Opposite to exhaust expulsion	Given in problem
Drag	${\vec{F}}_{drag}$	Contact	Opposite to motion through fluid	Given (ignore unless stated)
Buoyancy	${\vec{F}}_B$	Contact	Up (in fluid)	$F_B=$ weight of fluid displaced
Universal gravitation	${\vec{F}}_{A\text{ on }B}$	Long-range	Along line joining centres	$F=\frac{G{m}_A{m}_B}{r^2}$

Universal constant: $G=6.67\times 10^{-11}\text{N}{\text{m}}^2/{\text{kg}}^2$.

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2 — Newton’s three laws

#	Statement	Working form
1st	Inertia — no net force means no change in velocity.	${\vec{F}}_{net}=0\iff \vec{a}=0$ (at rest or constant velocity).
2nd	Net force causes acceleration in the same direction.	${\vec{F}}_{net}=m\vec{a}$, or equivalently $\sum \vec{F}=m\vec{a}$.
3rd	Forces come in equal-and-opposite pairs.	${\vec{F}}_{A\text{ on }B}=-{\vec{F}}_{B\text{ on }A}$.

Newton’s 2nd law in component form

Always split into axes:

$$\sum F_x=m{a}_x,\sum F_y=m{a}_y.$$

This is two scalar equations, one per axis. One of them is usually a trivial equilibrium (e.g. vertical $\sum F_y=0$) and gives you $F_N$ or $F_B$ for free.

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3 — The FBD recipe

1. Pick the body. One object. Treat it as a particle (dot).
2. Pick axes. Usually $x$ horizontal, $y$ vertical. On an incline, rotate so $x$ is along the slope.
3. Draw every force as a labelled arrow from the dot. Magnitude (if known) and direction.
4. Sum forces per axis. Decompose tilted forces into components.
5. Equate to $ma$ on the axis with acceleration, $0$ on the axis without.
6. Solve. Assess. Units? Sign? Believable?

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4 — Worked snippets (mirror the lecture examples)

Problem	Setup	Answer
Train, $m=1.2\times 10^6\text{kg}$, $a=2.4{\text{m/s}}^2$	$\sum F_x=ma$	$F_{app}=2.88\times 10^6\text{N}$
Same train unloaded ($m=9.8\times 10^5\text{kg}$), same $F_{app}$	$a=F/m$	$a=2.94{\text{m/s}}^2$
Boat, $m=960\text{kg}$, $F_{thrust}=4100\text{N}$, $a=2.1{\text{m/s}}^2$	$y$: $F_B=mg$; $x$: $F_{thrust}-F_{drag}=ma$	$F_B=9408\text{N}$, $F_{drag}=2084\text{N}$
Spring, $k=270\text{N/m}$, stretched $0.48\text{m}$	$F_S=kx$	$F_S=129.6\text{N}$
Same spring, vertical, $5\text{kg}$ on top	$kx=mg$	$x=0.181\text{m}$
$1\text{kg}$ on Earth’s surface	$F=GMm/r^2$	$F=9.8\text{N}$ — recovers $g$

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Common mistakes

1. Treating $F$ as a scalar. Every force is a vector. On the FBD, draw an arrow. In equations, decompose into components.
2. Drawing forces the body exerts. The FBD shows forces on the body, never by it. The reaction force lives on the other body.
3. Missing weight or normal. They’re almost always there. If the body is on a surface, draw ${\vec{F}}_N$.
4. Confusing 3rd-law pairs with balanced forces. Gravity and normal on a resting book act on the same body — they’re balanced by Newton’s 2nd law, not a 3rd-law pair. The 3rd-law partner of gravity (Earth on book) is the book’s pull on the Earth.
5. Kinetic friction direction. Opposes the velocity, not the applied force. A sliding block decelerating still has friction opposite its current motion.
6. Static friction set equal to ${\mu }_s{F}_N$ always. Only at the verge of slipping is static friction maximal. Otherwise $F_{fric}\le {\mu }_s{F}_N$ and adjusts to whatever is needed.
7. Spring sign. Compressed spring pushes back toward natural length; stretched spring pulls back. In both cases use $|F_S|=k|x|$.
8. Units. Convert cm→m before $F_S=kx$. Final force should reduce to $\text{kg}\cdot {\text{m/s}}^2=\text{N}$.
9. Ignoring drag/air. Unless the question says so, neglect it.

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Key formulas

$${\vec{F}}_{net}=m\vec{a}\sum F_x=m{a}_x\sum F_y=m{a}_y$$ $$F_G=mg{F}_S=kx{F}_{fric}=\mu F_N$$ $$F_{A\text{ on }B}=\frac{G{m}_A{m}_B}{r^2},G=6.67\times 10^{-11}\text{N}{\text{m}}^2/{\text{kg}}^2$$ $${\vec{F}}_{A\text{ on }B}=-{\vec{F}}_{B\text{ on }A}$$

For the why and many more worked examples, see the in-depth note.