Week 4 In-Depth — Why Newton's Laws Explain Motion

Read the cheatsheet first. This note explains why each law works and walks through one substantial worked example per scenario, with every step on the page.

1 — Why force is a vector (and why the FBD comes first)

A force is whatever changes an object’s velocity. The change-in-velocity vector has both a magnitude (how much faster?) and a direction (which way?), so the cause — the force — must also have both. That’s why $\vec{F}$ wears the arrow.

The consequence: you cannot add forces like numbers. Two $10\text{N}$ forces at right angles do not give $20\text{N}$; they give $\sqrt{10^2+10^2}\approx 14.1\text{N}$ at $45^{\circ }$. To sum forces correctly, you decompose each into its $x$ and $y$ components, add the components separately, and reassemble.

The free body diagram is the visual setup for that decomposition. Drawing the body as a dot at the origin of your axes does two things:

1. It strips away the geometry of the body so you focus on the forces only.
2. It puts every force tail at the same point, which is exactly what you need to add them as vectors.

If your FBD is wrong, every formula you write afterwards is wrong. Spend the time on it.

The two-axis identity

Newton’s 2nd law in component form,

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

is just two ordinary equations. The “vector” part is bookkeeping. Once you commit to an axis, every force gets a sign.

A useful habit: one of the two axes is almost always equilibrium. A horizontal surface? $a_y=0$. A boat that’s floating? $a_y=0$. A car on a level road? $a_y=0$. That equation gives you $F_N$ or $F_B$ for free, and then the other axis is where the action is.

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2 — Newton’s 1st law: inertia, not “stuff stops”

The law says: with zero net force, an object continues exactly as it was — at rest, or moving at constant velocity.

The “moving at constant velocity” half is where intuition fails. Everyday experience says “things stop on their own.” But that’s because friction and drag are always quietly doing $-ma$ on them. In a frictionless world (or in deep space), the puck slides forever.

This is why the FBD has to be honest. If your block is “sliding to a stop,” the net force is not zero — kinetic friction is decelerating it, and $\sum F=ma$ with $a<0$.

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3 — Newton’s 2nd law: the master equation

$${\vec{F}}_{net}=m\vec{a}.$$

Three things to internalise:

1. Direction. $\vec{a}$ points the same way as ${\vec{F}}_{net}$. The mass $m$ is a scalar, so it can’t rotate the vector.
2. Linearity in $m$. Double the mass, same force, half the acceleration. This is why the unloaded train in Example 1(b) accelerates more.
3. Linearity in $F$. Double the force, same mass, double the acceleration.

That’s it. The whole apparatus of “find the force” or “find the acceleration” is the same equation rearranged.

Worked example — the loaded vs unloaded train

A train has mass $m_T=1.2\times 10^6\text{kg}$ and accelerates at $a_x=2.4{\text{m/s}}^2$.

FBD. Three forces: weight ${\vec{F}}_W=mg$ down, normal ${\vec{F}}_N$ up, applied (engine pull) ${\vec{F}}_{app}$ in $+x$. Vertical equilibrium: $F_N=mg$. Vertical isn’t where the answer lives — we just confirmed it’s balanced.

$x$-axis (where $a$ lives).

$$\sum F_x=m{a}_x⟹F_{app}=(1.2\times 10^6)(2.4)=2.88\times 10^6\text{N}.$$

Part (b). Same applied force, unloaded mass $m_T=9.8\times 10^5\text{kg}$:

$$a_x=\frac{F_{app}}{m_T}=\frac{2.88\times 10^6}{9.8\times 10^5}\approx 2.94{\text{m/s}}^2.$$

The takeaway. Same equation, two unknowns swapped. You will see this pattern in every Portfolio 3 question.

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4 — Two-axis problems and why you can’t skip the FBD

Some problems live on both axes. The boat from Example 2 is the cleanest case.

A $960\text{kg}$ motorboat accelerates from a dock at $a_x=2.1{\text{m/s}}^2$, driven by thrust $F_{thrust}=4100\text{N}$. Find the buoyancy $F_B$ holding it up and the drag $F_{drag}$ slowing it down.

FBD. Four forces on the boat:

• Weight ${\vec{F}}_W=mg$ down.
• Buoyancy ${\vec{F}}_B$ up.
• Thrust ${\vec{F}}_{thrust}$ forward ($+x$).
• Drag ${\vec{F}}_{drag}$ backward ($-x$).

$y$-axis: no vertical acceleration, $a_y=0$.

$$\sum F_y=0:F_B-mg=0⟹F_B=(960)(9.8)=9408\text{N}.$$

$x$-axis: horizontal acceleration, $a_x=2.1{\text{m/s}}^2$.

$$\sum F_x=m{a}_x:F_{thrust}-F_{drag}=m{a}_x.$$

Solve for the unknown:

$$F_{drag}=F_{thrust}-m{a}_x=4100-960(2.1)=4100-2016=2084\text{N}.$$

Why this is the canonical 2nd-law problem. Two unknowns ($F_B$, $F_{drag}$), two axes, two equations. You don’t need to be clever — you just need to be tidy. The FBD makes the tidiness automatic.

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5 — Why Hooke’s law $F_S=kx$ has a sign convention

A spring resists being deformed. Compress it, it pushes back. Stretch it, it pulls back. Either way, the spring force points toward the natural length.

The magnitude is proportional to how far you’ve deformed it — and the constant of proportionality is $k$, the spring constant, in $\text{N/m}$. Stiff spring, large $k$; floppy spring, small $k$.

Worked example — vertical spring with a mass

A spring of stiffness $k=270\text{N/m}$ is held vertically. A $5\text{kg}$ block is placed gently on top. How far does the spring compress?

FBD of the block at equilibrium. Two forces:

• Weight ${\vec{F}}_W=mg$ down.
• Spring ${\vec{F}}_S$ up (compressed spring pushes back toward its natural length).

The block isn’t accelerating (it’s resting), so $\sum F_y=0$:

$$F_S-mg=0⟹kx=mg⟹x=\frac{mg}{k}=\frac{(5)(9.8)}{270}\approx 0.181\text{m}.$$

That’s $18.1\text{cm}$ of compression. The intuition: the spring force adjusts itself until it exactly balances gravity. At equilibrium, $kx=mg$, so $x$ is whatever satisfies that.

The same trick works for “weighing a fish on a spring scale” — the spring stretches until $kx=mg$, and you read $x$ off the scale (calibrated to mass).

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6 — Friction: two regimes, one formula

Friction comes in two flavours, with different physics even though the formula looks similar.

Static friction — the adjustable one

Block sitting still on a surface. You push gently — it doesn’t move. You push harder — still doesn’t. Eventually, you push hard enough that it breaks loose.

Static friction adjusts itself to whatever value cancels your push, up to a maximum:

$$F_{fric}^{static}\le {\mu }_s{F}_N.$$

So if you push with $20\text{N}$ and ${\mu }_s{F}_N=39.2\text{N}$, static friction is $20\text{N}$, not $39.2\text{N}$. The block is in equilibrium.

Kinetic friction — the constant one

Once the block slides, kinetic friction takes over with a fixed magnitude:

$$F_{fric}^{kinetic}={\mu }_k{F}_N,$$

pointing opposite to the velocity. Typically ${\mu }_k<{\mu }_s$ (it’s easier to keep something sliding than to start it).

Worked example — does the block move?

A $10\text{kg}$ block sits on a horizontal surface with ${\mu }_s=0.4$, ${\mu }_k=0.2$. Two scenarios:

(a) Applied force $F=20\text{N}$.

Max static friction: ${\mu }_s{F}_N={\mu }_{s}mg=0.4\times 10\times 9.8=39.2\text{N}$. Applied force is $20\text{N}<39.2\text{N}$. Block stays put. Friction is exactly $20\text{N}$ (matching the push), and $a=0$.

(b) Applied force $F=100\text{N}$.

$100\text{N}>39.2\text{N}$, so the block breaks loose. Once sliding, kinetic friction is ${\mu }_{k}mg=0.2\times 10\times 9.8=19.6\text{N}$.

$$\sum F_x=m{a}_x:100-19.6=10a_x⟹a_x=8.04{\text{m/s}}^2.$$

The key insight. You always check the static threshold first. Skip that check and you can get nonsense — like a “negative acceleration” from a force too small to start motion.

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7 — Newton’s 3rd law: pairs live on different bodies

The law: every force has an equal-and-opposite partner acting on the other body.

The common trap: thinking that gravity and the normal force on a book sitting on a table are a 3rd-law pair. They’re not. They both act on the book, and they happen to be balanced by the 2nd law (because $a=0$ vertically).

The true 3rd-law pairs are:

Force	Reaction partner
Earth pulls book down (gravity)	Book pulls Earth up (gravity)
Table pushes book up (normal)	Book pushes table down (normal)

The pair always swaps the “by” and “on” labels.

Why this matters in problems

When you draw the FBD of the book, you draw only the forces on the book. You do not draw the book’s pull on the Earth or the book’s push on the table — those belong on the Earth’s FBD and the table’s FBD.

If you confuse the partners, you double-count or under-count, and your $\sum F$ will be wrong.

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8 — Universal gravitation: where $g$ comes from

Newton’s law of universal gravitation says every two masses attract with

$$F=\frac{G{m}_1{m}_2}{r^2},G=6.67\times 10^{-11}\text{N}{\text{m}}^2/{\text{kg}}^2.$$

Why we ignore it between everyday objects

Two $1\text{kg}$ masses $1\text{m}$ apart attract each other with

$$F=\frac{(6.67\times 10^{-11})(1)(1)}{1^2}=6.67\times 10^{-11}\text{N}.$$

That’s around a hundred-billionth of a newton. Friction from a passing puff of air dominates this. So for ordinary problems we treat $F_G$ between everyday objects as zero.

Why we cannot ignore it between an object and the Earth

A $1\text{kg}$ mass at Earth’s surface, with $M_{\text{Earth}}=5.98\times 10^{24}\text{kg}$ and $r=6.37\times 10^6\text{m}$:

$$F=\frac{(6.67\times 10^{-11})(5.98\times 10^{24})(1)}{(6.37\times 10^6{)}^2}\approx 9.8\text{N}.$$

That’s a serious force, and it’s where $g\approx 9.8{\text{m/s}}^2$ comes from. The whole $F_G=mg$ shortcut is Newton’s law of universal gravitation, evaluated at Earth’s surface. Use $mg$ when you’re near the surface; use the full $G{m}_1{m}_2/r^2$ when you’re not.

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9 — How these all connect

Technique	Where Newton’s 2nd law shows up
Identifying forces	Building the input — every term in $\sum \vec{F}$.
Drawing an FBD	The visual algebra of the sum. Wrong picture, wrong equation.
Newton’s 1st law	The special case $\sum \vec{F}=0$ (which is just $a=0$).
Newton’s 3rd law	Tells you what does not belong on this body’s FBD.
Universal gravitation	Provides the ${\vec{F}}_G$ term that goes into $\sum \vec{F}$.

So really, this whole week is one equation (${\vec{F}}_{net}=m\vec{a}$) applied with care. Once the FBD is right, the algebra is mechanical.

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10 — Exam-style sample, end-to-end

A $52\text{kg}$ passenger stands on a bathroom scale in an elevator. The elevator accelerates upward at $1.8{\text{m/s}}^2$. What does the scale read?

A bathroom scale reads the normal force it exerts on the person.

Setup.

Item	Detail
Body	The passenger (treated as a particle)
Axes	$y$ up, $x$ horizontal (no $x$-motion)
Acceleration	$a_y=+1.8{\text{m/s}}^2$ (upward)
Forces on passenger	Gravity ${\vec{F}}_W=mg$ down; normal ${\vec{F}}_N$ up from the scale

FBD. Dot at origin. Arrow ${\vec{F}}_N$ upward (long). Arrow ${\vec{F}}_W$ downward (also long).

Newton’s 2nd law along $y$.

$$\sum F_y=m{a}_y⟹F_N-mg=m{a}_y.$$

Solve.

$$F_N=m(g+a_y)=52\times (9.8+1.8)=52\times 11.6=603.2\text{N}.$$

Newton’s 3rd law. The passenger pushes down on the scale with $603.2\text{N}$ (the reaction partner of ${\vec{F}}_N$).

Assess. Heavier-than-resting reading (since at rest it would be $52\times 9.8=509.6\text{N}$). The accelerating elevator means the passenger feels heavier — this matches everyday experience. Units: $\text{kg}\cdot {\text{m/s}}^2=\text{N}$. Looks right.

Answer. Scale reads $603.2\text{N}$ (or about $61.5\text{kg}$ if it’s calibrated in mass).

That layout — Body, Axes, Acceleration, Forces, FBD, $\sum F=ma$ per axis, Solve, Assess — is the template for every Portfolio 3 question and every exam question on this material.

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Where to look next

• The cheatsheet for fast revision + the quiz.
• The lecture summary for the original synthesis.
• Tutorial 4 PDF for more practice problems (especially the inclined-skier — a teaser for next week).
• Whatever you add to this folder — Lecture Atlas picks up any new Markdown note automatically.