Week 5 In-Depth — Why Newton's Laws Solve These Problems

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

1 — Why a free body diagram is the whole game

A free body diagram is not artwork — it is the encoding step that turns a word problem into algebra. Once your FBD is correct, the equations write themselves: $\sum F_x=m{a}_x$ and $\sum F_y=m{a}_y$. Every mistake in this course can be traced to one of three FBD errors:

1. A force that doesn’t exist (centrifugal force on a rotating body, an “applied” force that’s actually a reaction to your own arrow).
2. A force in the wrong direction (friction on the wrong side, normal force not perpendicular to the surface).
3. A force missed entirely (tension on the other end of the rope, a normal from a surface you forgot about).

The lecturer’s explicit instruction is: spend most of your time on Visualise, not on Solve (slide 10). That’s because once you have a clean FBD, “Solve” is bookkeeping.

Why Newton’s 3rd law is so easy to forget

If body A pushes on body B with force $\vec{F}$, then B pushes back on A with $-\vec{F}$. The pair acts on different bodies, so it never cancels on a single FBD. Spring-coupled blocks are the standard place this matters: the spring pulls block 1 one way and block 2 the other way with equal magnitude $kx$. If you draw a single FBD and put both spring arrows on it, you’ve added the same force twice.

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2 — Why centripetal force is not a new force

This is the single most common error in the exam. The phrase “centripetal force” is shorthand for “the net inward force needed to keep the body going in a circle.” It is not an additional force you should draw on the FBD.

Here’s the test: if your FBD for a ball on a string in a horizontal circle has four arrows — weight, normal, tension, and a fourth labelled $F_c$ — you have over-counted. The correct FBD has weight and tension only. The centripetal direction is just the direction toward the centre, and the radial Newton equation is the projection of the real forces onto that direction.

The Taylor-style argument: why $a_c=v^2/r$

A particle on a circle of radius $r$ at speed $v$ has position

$$\vec{r}(t)=r(\cos (\omega t),\sin (\omega t)),\omega =v/r.$$

Differentiate once:

$$\vec{v}(t)=r\omega (-\sin (\omega t),\cos (\omega t)),|\vec{v}|=r\omega =v.$$

Speed is constant. Differentiate again:

$$\vec{a}(t)=-r{\omega }^2(\cos (\omega t),\sin (\omega t))=-{\omega }^2\vec{r}.$$

The acceleration points opposite to $\vec{r}$ — i.e. toward the centre — and has magnitude $r{\omega }^2=v^2/r$. That is why ${\vec{F}}_{net}=m\vec{a}$ becomes $F_c=m{v}^2/r$ pointing inward, whether the real force supplying it is tension, gravity, normal, friction, or a spring.

Worked example with every step — vertical circle (slide 25)

A 0.5 kg ball on a 1.5 m string is swung in a vertical circle at $v=12$ m/s. Find the tension at the top and at the bottom.

Setup. Inward (toward centre) is the positive radial direction. The only real forces on the ball are weight $mg$ (always down) and tension $T$ (always along the string toward the centre).

At the top of the loop. Both $T$ and weight point downward, which is toward the centre.

$$\sum F_{\text{rad}}=m{a}_c:T+mg=\frac{m{v}^2}{r}.$$

Solve:

$$T=\frac{m{v}^2}{r}-mg=\frac{0.5\times 12^2}{1.5}-0.5\times 9.8=48-4.9=43.1\text{N}.$$

At the bottom of the loop. $T$ points upward (toward centre); weight points downward (away from centre). Choosing inward positive flips the sign of weight:

$$\sum F_{\text{rad}}=m{a}_c:T-mg=\frac{m{v}^2}{r}.$$

Solve:

$$T=\frac{m{v}^2}{r}+mg=48+4.9=52.9\text{N}.$$

Sanity check. Tension is larger at the bottom. Intuition: at the bottom the rope has to do two jobs — hold the ball up against gravity and curve its path inward. At the top, gravity helps with the curving, so the rope can do less work.

This is also why a roller coaster feels heavier at the bottom of a loop and lighter at the top — the apparent weight is the normal force from the seat, and that follows exactly the same equation with $F_N$ instead of $T$.

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3 — Why an inclined plane needs two friction checks

The static-friction inequality $F_{fric}\le {\mu }_s{F}_N$ is not an equation. It tells you the maximum friction can supply; the actual value adjusts to whatever the rest of the FBD needs. That ambiguity is what trips students up.

The clean procedure has two stages.

Stage 1: does it slide? Pretend it doesn’t. If $a=0$, $\sum F_x=0$ along the slope gives

$$0=mg\sin \theta -F_{fric}⟹F_{fric}^{\text{required}}=mg\sin \theta .$$

For this to be possible, we need $F_{fric}^{\text{required}}\le {\mu }_s{F}_N={\mu }_{s}mg\cos \theta$, i.e.

$${\mu }_s\ge \tan \theta .$$

So the sliding criterion is just a comparison between ${\mu }_s$ and $\tan \theta$.

Stage 2: if it slides, switch to kinetic. Now there’s relative motion, so $F_{fric}={\mu }_k{F}_N$ as an equation, and

$$ma=mg\sin \theta -{\mu }_{k}mg\cos \theta ⟹a=g\sin \theta -{\mu }_{k}g\cos \theta .$$

Worked example with every step — 30° incline (slide 17, notes pp. 3–4)

A wooden block on a 30° wooden incline with ${\mu }_s=0.4$ and ${\mu }_k=0.2$. Does it slide? If so, find $a$.

Stage 1 — sliding test.

$$\tan 30°\approx \mathrm{0.577.}$$

Compare to ${\mu }_s=0.4$. Since $0.4<0.577$, static friction is insufficient — the block slides.

Stage 2 — acceleration.

$$a=g\sin 30°-{\mu }_{k}g\cos 30°=9.8(0.5)-0.2(9.8)(0.866)=4.9-1.697=3.203{\text{m/s}}^2.$$

Note the units cancel cleanly: $g$ is m/s${}^2$, the trig factors and ${\mu }_k$ are dimensionless.

Why mass cancelled. Both the driving force ($mg\sin \theta$) and the friction (${\mu }_{k}mg\cos \theta$) are proportional to $m$, so dividing by $m$ in $ma=\dots$ removes the mass entirely. This is why a heavy and a light block slide down the same ramp at the same rate — Galileo’s observation in modern dress.

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4 — Why coupled bodies have one acceleration

When two blocks are joined by an inextensible rope or a light spring with no time delay, the constraint is that they share a single acceleration $a$. That is the algebraic glue that lets you write two equations in two unknowns ($a$ and the internal force) and solve.

There are always two valid solution paths:

1. System method. Lump both bodies. Internal forces cancel by Newton’s 3rd law. $\sum F_{\text{external}}=(m_1+m_2)a$. Use this when you only need $a$.
2. Separate FBDs. Write each body’s equation, then add or subtract to eliminate the internal force. Use this when you need the internal force (rope tension, spring stretch, contact force between blocks).

Worked example with every step — spring-coupled blocks (Example 1, slide 16)

$m_1=2.0$ kg and $m_2=3.0$ kg on a horizontal frictionless surface, joined by a massless spring of $k=110$ N/m. A 15 N force is applied to $m_2$ pulling rightward. Find the spring extension $x$.

System method first (fast).

Total mass 5 kg. External force 15 N. Acceleration:

$$a=\frac{15}{5}=3{\text{m/s}}^2.$$

Now use the FBD of $m_1$ — the spring is the only horizontal force on it:

$$kx=m_{1}a=2\times 3=6\text{N},$$ $$x=\frac{6}{110}=0.0545\text{m}\approx 55\text{mm}.$$

Separate FBD method (slower but more general).

FBD of $m_1$ (force on it from spring is $kx$ rightward):

$$m_{1}a=kx⟹2a=110x⟹a=55x.\text{\hspace{1em}(1)}$$

FBD of $m_2$ (applied force 15 N rightward, spring pulls leftward by Newton’s 3rd law):

$$m_{2}a=15-kx⟹3a=15-110x.\text{\hspace{1em}(2)}$$

Substitute (1) into (2):

$$3(55x)=15-110x⟹165x+110x=15⟹275x=15,$$ $$x=\frac{15}{275}=0.0545\text{m}.$$

Same answer. The 3rd law sign on $kx$ is the only subtle step — block 1 sees the spring pull right, block 2 sees the spring pull left, with the same magnitude.

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5 — How these three pieces connect

The cheat is that all of this is one technique:

Setup	What’s "${\vec{F}}_{net}=m\vec{a}$" doing?
Coupled blocks	Two FBDs, one shared $a$, eliminate internal forces
Incline + friction	Tilt the axes, project weight, decide if friction is static or kinetic
Uniform circular motion	The radial component of ${\vec{F}}_{net}$ equals $m{v}^2/r$
Banked curve	Same as above; the surface tilt rotates which forces have radial components
Conical pendulum	Same again; vertical balance fixes $T$, horizontal balance gives $v$

If you understand FBDs deeply, every set-up in this course is just the same three lines — coordinate axes, $\sum F_x$, $\sum F_y$ — with a different drawing.

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6 — Exam-style sample, end-to-end

A 0.9 kg rock is attached to a 1.3 m string and whirled in a horizontal circle, so the string makes some angle $\theta$ with the horizontal. The string breaks at $F_T=120$ N tension. Find the minimum angle (below which the string can’t support the rock) and the corresponding speed.

FBD. Two forces only: tension $F_T$ along the string (inward and upward) and weight $mg$ down.

Coordinate choice. $y$ vertical, $x$ horizontal (toward the centre of the circle).

Geometry. With $\theta$ measured from the horizontal, the string’s horizontal projection is $\cos \theta$, so the orbit radius is

$$r=L\cos \theta =1.3\cos \theta .$$

Vertical equilibrium (no vertical motion):

$$\sum F_y=0:F_T\sin \theta -mg=0⟹F_T=\frac{mg}{\sin \theta }.\text{\hspace{1em}(1)}$$

For (1) to be valid with $F_T=120$ N (the breaking limit):

$$\sin \theta =\frac{mg}{F_T}=\frac{0.9\times 9.8}{120}=0.0735,$$ $$\theta ={\sin }^{-1}(0.0735)\approx 4.215°.$$

That’s the minimum angle — any shallower and the vertical component of tension can’t support gravity.

Radial equation.

$$\sum F_x=m{a}_c:F_T\cos \theta =\frac{m{v}^2}{r}=\frac{m{v}^2}{L\cos \theta },$$ $$v^2=\frac{F_{T}L{\cos }^2\theta }{m}.$$

Plug in $F_T=120$ N, $L=1.3$ m, $\cos \theta \approx 0.9973$, $m=0.9$ kg:

$$v^2=\frac{120\times 1.3\times 0.9946}{0.9}\approx 172.4,v\approx 13.13\text{m/s}.$$

Sanity check. As $\theta$ gets small (cord nearly horizontal), $\cos \theta \to 1$ and the radius approaches $L$, so the rock has to move fast to keep $m{v}^2/r$ matched to the horizontal tension. That tracks the answer.

Template that works for every circular-motion question on this exam.

1. Two-axis FBD with one axis pointing toward the centre.
2. $\sum F_{\text{non-radial}}=0$ if there’s no acceleration in that direction.
3. $\sum F_{\text{radial}}=m{v}^2/r$.
4. Solve simultaneously.

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Where to look next

• The cheatsheet for fast revision plus the quiz.
• The lecture summary for the original lecture order and tutorial outlines.
• The workshop PDFs cited in the frontmatter for more practice problems.