---
title: "Newton’s Laws"
book: "High School Physics"
subject: physics
language: en
chapter: 25
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/25-newtons-laws
---

# Chapter 25 — Newton’s Laws

Smuggle a bathroom scale into an elevator. As the cabin pulls upward the needle reads three [kilograms](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) heavy; near the top it reads light; between the two, the plain truth — nothing aboard changed but the motion. This chapter states the three laws behind that needle and forges them into the method that runs all of mechanics: choose a system, draw its [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force), project $\sum \vect F = m\vect a$.

## 25.1 Momentum, and the first law made exact

**Definition 25.1 (Momentum).**

The *momentum* of a body of mass $m$ ($\mathrm{kg}$) moving with velocity $\vect v$ is the vector

$$
\vect p = m \vect v \qquad (\mathrm{kg}\,\mathrm{m}/\mathrm{s}):
$$

the velocity, weighted by the mass; for a system, the vector sum over its parts.

**Definition 25.2 (Inertial reference frame).**

An *inertial reference frame* is one in which every isolated body — no [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force), or [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) that [compensate](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-compensate) — moves in a straight line at constant velocity ([Chapter 6](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#ch-g10-inertia)). The ground is one to excellent accuracy, and so is any cabin in uniform straight-line motion over it; a braking train or turning carousel is not: there, loose luggage accelerates unpushed.

**Theorem 25.3 (Newton’s first law).**

In an inertial frame, the [momentum](#def-g12-newtons-laws-momentum) of an isolated system does not change: $\vect p$ is constant, i.e. $d\vect p/dt = \vect 0$.

**Proof.** *Admitted at this level.* ∎

**Remark 25.4 (Inertia, sharpened twice).**

The principle of inertia of [Chapter 6](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#ch-g10-inertia), upgraded twice: it says *where* it holds — it defines the inertial frames as those where it does — and it guards a vector, $m\vect v$, whose bookkeeping pays off at the chapter’s end.

## 25.2 Newton’s second law

**Theorem 25.5 (Newton’s second law).**

In an inertial frame, the [net force](https://one-course.com/books/physics/2/en/chapter/16-forces-and-motion#def-g11-forces-and-motion-netforce) on a body equals the rate of change of its [momentum](#def-g12-newtons-laws-momentum); for constant mass, since $\vect p = m\vect v$,

$$
\sum \vect F = \frac{d\vect p}{dt} = m\,\frac{d\vect v}{dt} = m \vect a .
$$

**Proof.** *Admitted at this level.* ∎

**Remark 25.6 (A law of nature, and its unit).**

There is no proof: this is a law of nature, weighed against experiment for three centuries. Last year ([Chapter 16](https://one-course.com/books/physics/2/en/chapter/16-forces-and-motion#ch-g11-forces-and-motion)) gave its shadow, $\Delta v \propto F\,\Delta t/m$; the derivative ([Chapter 24](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#ch-g12-kinematics-2d)) makes it exact at every instant — its fine print waits in the Year 1 volume. [Force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) gets a [unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit), $\mathrm{N} = \mathrm{kg}\,\mathrm{m}/\mathrm{s}^{2}$; and a body under its [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) alone obeys $m\vect a = m\vect g$: every mass falls with $\vect a = \vect g$, and $g = 9.81\,\mathrm{N}/\mathrm{kg} = 9.81\,\mathrm{m}/\mathrm{s}^{2}$ — the two readings are one.

**Example 25.7 (Average force, no details needed).**

A $1300\,\mathrm{kg}$ car goes from rest to $100\,\mathrm{km}/\mathrm{h}$ ($27.8\,\mathrm{m}/\mathrm{s}$) in $8.0\,\mathrm{s}$: whatever the instant-by-instant story, the average [net force](https://one-course.com/books/physics/2/en/chapter/16-forces-and-motion#def-g11-forces-and-motion-netforce) is $\Delta p/\Delta t = 1300 \times 27.8/8.0 \approx 4.5 \times 10^{3}\,\mathrm{N}$ — [momentum](#def-g12-newtons-laws-momentum) turns a [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) question into before-and-after bookkeeping.

## 25.3 Newton’s third law

**Theorem 25.8 (Newton’s third law).**

If a body $A$ exerts a [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $\vect F_{A \to B}$ on a body $B$, then $B$ exerts on $A$ the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $\vect F_{B \to A} = -\vect F_{A \to B}$: same line of action, same magnitude, opposite directions — at every instant, whatever the motion, contact or action at a distance.

**Proof.** *Admitted at this level.* ∎

**Example 25.9 (The book, the table and the Earth).**

A book rests on a table: two interactions, two pairs. Gravitational: the Earth pulls the book down ($\vect P$), the book pulls the Earth up. Contact: the table pushes the book up ($\vect N$), the book presses the table down. Walking is the same deal: push the ground backward; its reaction is the only [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) driving you forward.

![One book, two interactions, four arrows: the gravitational pair (the Earth pulls the book, the book pulls the Earth) and the contact pair (the table pushes the book, the book presses the table). P and N are not partners: they act on the same book — and cancel only because its acceleration is zero.](https://one-course.com/images/onecourse/chapters/physics-2/g12-newtons-laws/fig-3bd91f9f93b9.svg)

*One book, two interactions, four arrows: the gravitational pair (the Earth pulls the book, the book pulls the Earth) and the contact pair (the table pushes the book, the book presses the table). $\vect P$ and $\vect N$ are *not* partners: they act on the same book — and cancel only because its acceleration is zero.*

## 25.4 The method: choose, draw, project

**Method 25.10 (Solving a dynamics problem).**

1. *System* : name the body (or set of bodies) studied.
2. *Frame* : choose an inertial frame and axes fitted to the motion — slope axes, or the [Frenet frame](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-frenet) for a curve ( [Chapter 24](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#ch-g12-kinematics-2d) ).
3. *Inventory* : list every [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) — the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) , then one [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) per contact, nothing else — and draw the free-body diagram: one point, one arrow per [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) .
4. *Project* : write $\sum \vect F = m\vect a$ , project it on each axis.
5. *Solve and check* : [units](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) , signs, limiting cases ( $\mu \to 0$ , $\alpha \to 0$ , …).

**Definition 25.11 (Normal force and friction).**

A surface acts on a body through two components: the *normal force* $\vect N$, perpendicular to the surface, and [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) $\vect f$, tangential. Experiment gives the [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) laws: while the body does not slide, static [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) adjusts itself up to a ceiling, $f_s \leq \mu_s N$; once it slides, [kinetic friction](https://one-course.com/books/physics/2/en/chapter/16-forces-and-motion#def-g11-forces-and-motion-friction) is $f_k = \mu_k N$, against the sliding. The *coefficients of [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory)* $\mu_s$ and $\mu_k$ ($\mu_k \leq \mu_s$) depend only on the two materials.

**Example 25.12 (The incline, twice).**

A block slides down a slope of angle $\alpha$; axes: $x$ down the slope, $y$ perpendicular. On $y$: $N = mg\cos\alpha$. On $x$, frictionless: $ma = mg\sin\alpha$, so $a = g\sin\alpha$ — no mass; at $\alpha = 30^\circ$, $a = 4.9\,\mathrm{m}/\mathrm{s}^{2}$. With [kinetic friction](https://one-course.com/books/physics/2/en/chapter/16-forces-and-motion#def-g11-forces-and-motion-friction): $a = g(\sin\alpha - \mu_k\cos\alpha)$; for $\mu_k = 0.20$, $a = 9.81\,(0.500 - 0.173) = 3.2\,\mathrm{m}/\mathrm{s}^{2}$. And it stayed put in the first place only while the needed static [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) fit under its ceiling: $\tan\alpha \leq \mu_s$.

![Free-body diagram on the incline: weight P, normal force N, friction f up the slope for a block sliding down. Axes along and perpendicular to the slope split P into mg and mg (dashed).](https://one-course.com/images/onecourse/chapters/physics-2/g12-newtons-laws/fig-0b658db3e7ea.svg)

*Free-body diagram on the incline: [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) $\vect P$, [normal force](#def-g12-newtons-laws-friction) $\vect N$, [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) $\vect f$ up the slope for a block sliding down. Axes along and perpendicular to the slope split $\vect P$ into $mg\sin\alpha$ and $mg\cos\alpha$ (dashed).*

**Definition 25.13 (Apparent weight).**

The *apparent weight* of a body is the [normal force](#def-g12-newtons-laws-friction) between it and its support — what a scale under it reads.

**Example 25.14 (The elevator).**

A $70\,\mathrm{kg}$ passenger stands on a scale in an elevator of vertical acceleration $a_z$ (upward positive): $N - mg = ma_z$, so $N = m(g + a_z)$. Pulling up at $a_z = 1.5\,\mathrm{m}/\mathrm{s}^{2}$: $792\,\mathrm{N}$, heavy; braking near the top: $582\,\mathrm{N}$, light; steady cruise: $mg = 687\,\mathrm{N}$ — the first law; free fall, $a_z = -g$: $N = 0$, weightlessness on a scale.

**Example 25.15 (Two blocks and a pulley).**

A block $m_1 = 4.0\,\mathrm{kg}$ on a frictionless table is tied by a light rope, over an ideal pulley, to a hanging block $m_2 = 1.0\,\mathrm{kg}$. The rope transmits the same [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) $T$ at both ends; the blocks share one magnitude of acceleration $a$. One second law each, along each motion: $m_1 a = T$ and $m_2 a = m_2 g - T$, so $a = m_2\,g/(m_1 + m_2) = 1.96\,\mathrm{m}/\mathrm{s}^{2}$ and $T = m_1 a = 7.8\,\mathrm{N}$ — smaller than $m_2 g = 9.8\,\mathrm{N}$, as it must be for $m_2$ to accelerate downward.

![Two blocks, one light rope, one ideal pulley: the same tension T pulls both ends, the blocks share one magnitude of acceleration — one second law each, two equations, two unknowns.](https://one-course.com/images/onecourse/chapters/physics-2/g12-newtons-laws/fig-2e421ec922ba.svg)

*Two blocks, one light rope, one ideal pulley: the same [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) $T$ pulls both ends, the blocks share one magnitude of acceleration — one second law each, two equations, two unknowns.*

**Example 25.16 (The flat curve).**

A car of mass $m$ rounds a flat curve of radius $R$ at constant speed $v$. In the [Frenet frame](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-frenet) ([Chapter 24](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#ch-g12-kinematics-2d)) the acceleration is purely normal, $a_N = v^2/R$, aimed at the center. [Vertically](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight), $N = mg$; horizontally, the only centripetal [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on offer is the static [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) of road on tyres: $f = mv^2/R \leq \mu_s N = \mu_s mg$, so $v \leq \sqrt{\mu_s\, g R}$, whatever the mass. For $R = 90\,\mathrm{m}$, dry road ($\mu_s = 0.70$): $v_{\max} \approx 25\,\mathrm{m}/\mathrm{s}$ ($89\,\mathrm{km}/\mathrm{h}$); on ice the ceiling collapses — [Exercise 25.13](#exo-g12-newtons-laws-13) banks the road; [Exercise 25.10](#exo-g12-newtons-laws-10) swings the same projection on a wire.

![A car in a flat curve at constant speed: from above (left) the velocity is tangent and the net force aims at the center; from behind (right) N balances P, and the sideways static friction of the road is the entire centripetal force.](https://one-course.com/images/onecourse/chapters/physics-2/g12-newtons-laws/fig-b7ec3633779a.svg)

![A car in a flat curve at constant speed: from above (left) the velocity is tangent and the net force aims at the center; from behind (right) N balances P, and the sideways static friction of the road is the entire centripetal force.](https://one-course.com/images/onecourse/chapters/physics-2/g12-newtons-laws/fig-57a039f45b63.svg)

*A car in a flat curve at constant speed: from above (left) the velocity is tangent and the [net force](https://one-course.com/books/physics/2/en/chapter/16-forces-and-motion#def-g11-forces-and-motion-netforce) aims at the center; from behind (right) $\vect N$ balances $\vect P$, and the sideways static [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) of the road is the entire centripetal [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force).*

## 25.5 Isolated systems: momentum is conserved

**Proposition 25.17 (Conservation of momentum).**

For two bodies whose external [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) vanish or [compensate](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-compensate), the total [momentum](#def-g12-newtons-laws-momentum) $\vect p_1 + \vect p_2$ is constant, whatever the [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) they exert on each other.

**Proof.** Add the two second laws: $\frac{d}{dt}(\vect p_1 + \vect p_2) = \vect F_{2 \to 1} +
\vect F_{1 \to 2} + \vect F_{\text{ext}}$: the internal pair cancels by the third law, and $\vect F_{\text{ext}} = \vect 0$ by hypothesis. ∎

**Example 25.18 (Recoil).**

A $60\,\mathrm{kg}$ skater at rest on smooth ice throws a $4.0\,\mathrm{kg}$ medicine ball horizontally at $6.0\,\mathrm{m}/\mathrm{s}$. Before: $\vect p = \vect 0$; after: $0 = 60\,V + 4.0 \times 6.0$, so $V = -0.40\,\mathrm{m}/\mathrm{s}$ — she drifts backward. Nothing external pushed her: the ball did (third law), and the books stay at zero.

**Remark 25.19 (How a rocket pushes on nothing).**

A rocket is a machine for throwing mass. Each second its engines hurl a batch of gas backward; the gas carries away backward [momentum](#def-g12-newtons-laws-momentum), so the rocket gains the same amount forward — a steady [thrust](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory), roughly the mass expelled per second times the ejection speed. The rocket pushes on its own exhaust, not on air: it [works](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) best in empty space.

## 25.6 Exercises

**Exercise 25.1 ★.**

Compute the [momentum](#def-g12-newtons-laws-momentum) of (a) a $70\,\mathrm{kg}$ sprinter at $10\,\mathrm{m}/\mathrm{s}$; (b) a $1300\,\mathrm{kg}$ car at $50\,\mathrm{km}/\mathrm{h}$; (c) an $8.0\,\mathrm{g}$ bullet at $800\,\mathrm{m}/\mathrm{s}$. Rank them. Why does the fastest rank last?

**Solution of Exercise 25.1.**

(a) $p = 70 \times 10 = 7.0 \times 10^{2}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}$. (b) $v = 13.9\,\mathrm{m}/\mathrm{s}$: $p \approx 1.8 \times 10^{4}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}$. (c) $p = 0.0080 \times 800 = 6.4\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}$. Car $>$ sprinter $>$ bullet: five orders of magnitude of mass are more than speed can buy.

**Exercise 25.2 ★.**

A $70\,\mathrm{kg}$ passenger stands on a scale in an elevator. What does it read when the cabin (a) accelerates upward at $1.2\,\mathrm{m}/\mathrm{s}^{2}$; (b) climbs at a constant $2.0\,\mathrm{m}/\mathrm{s}$; (c) accelerates downward at $1.2\,\mathrm{m}/\mathrm{s}^{2}$? Which law settles (b)?

**Solution of Exercise 25.2.**

$N = m(g + a_z)$: (a) $70 \times 11.01 = 771\,\mathrm{N}$; (b) $a_z = 0$, so $N = mg = 687\,\mathrm{N}$ — the first law: constant velocity, [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) [compensate](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-compensate); (c) $70 \times 8.61 = 603\,\mathrm{N}$.

**Exercise 25.3 ★.**

A block is released on a frictionless slope of angle $30^\circ$. Compute its acceleration, then its speed after sliding $3.0\,\mathrm{m}$. Where did the mass go?

**Solution of Exercise 25.3.**

$a = g\sin 30^\circ = 4.9\,\mathrm{m}/\mathrm{s}^{2}$; $v = \sqrt{2 \times 4.905 \times 3.0} = 5.4\,\mathrm{m}/\mathrm{s}$. The mass cancels between $ma$ and $mg\sin\alpha$.

**Exercise 25.4 ★.**

A $58\,\mathrm{g}$ tennis ball arrives at $15\,\mathrm{m}/\mathrm{s}$ and is returned along the same line at $25\,\mathrm{m}/\mathrm{s}$; contact lasts $5.0\,\mathrm{ms}$. Compute the change of [momentum](#def-g12-newtons-laws-momentum) and the average [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) of the strings; compare that [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) with the ball’s [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight).

**Solution of Exercise 25.4.**

Reversal: $\Delta p = 0.058 \times (25 + 15) = 2.3\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}$; $F = 2.32/0.0050 = 4.6 \times 10^{2}\,\mathrm{N}$ — about $800$ times the $0.57\,\mathrm{N}$ [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight), which is why gravity is ignored during contact.

**Exercise 25.5 ★.**

A book rests on a table. Name the two [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) acting on the book and, for each, its action–reaction partner (which body? which direction?). Why are the book’s [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) and the table’s push not partners of each other?

**Solution of Exercise 25.5.**

[Weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) (Earth on book, down): partner = book pulls the Earth, up. [Normal force](#def-g12-newtons-laws-friction) (table on book, up): partner = book presses the table, down. [Weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) and normal act on the *same* body and belong to two different interactions — equal here only because $\vect a = \vect 0$.

**Exercise 25.6 ★★.**

A $50\,\mathrm{kg}$ crate sits on a horizontal floor, $\mu_s = 0.45$, $\mu_k = 0.35$. (a) What horizontal push just starts it? (b) If that push is maintained, what is its acceleration? (c) What push keeps it moving at constant velocity?

**Solution of Exercise 25.6.**

$N = mg = 490\,\mathrm{N}$. (a) $F > \mu_s N = 221\,\mathrm{N}$. (b) $a = (220.7 - 0.35 \times 490.5)/50 = 0.98\,\mathrm{m}/\mathrm{s}^{2}$. (c) $F = \mu_k N = 172\,\mathrm{N}$.

**Exercise 25.7 ★★.**

The same crate is set on an adjustable ramp ($\mu_s = 0.45$, $\mu_k = 0.35$). (a) At what angle does it start to slide? (b) At $30^\circ$, compute its acceleration. (c) Once started at the angle of (a), does it keep accelerating? Why?

**Solution of Exercise 25.7.**

(a) Slides when $\tan\alpha > \mu_s$: $\alpha = \arctan 0.45 \approx
24^\circ$. (b) $a = g(\sin 30^\circ - 0.35\cos 30^\circ) =
9.81 \times 0.197 = 1.9\,\mathrm{m}/\mathrm{s}^{2}$. (c) Yes: once sliding, [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) drops to $\mu_k < \mu_s$, so $a = g\cos\alpha\,(\tan\alpha - \mu_k) \approx 0.9\,\mathrm{m}/\mathrm{s}^{2} > 0$.

**Exercise 25.8 ★★.**

A $3.0\,\mathrm{kg}$ block on a frictionless table is tied over an ideal pulley to a hanging $2.0\,\mathrm{kg}$ block. Compute the acceleration and the [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory). Why did the [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) have to come out smaller than the hanging [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight)?

**Solution of Exercise 25.8.**

$a = m_2 g/(m_1 + m_2) = 19.62/5.0 = 3.9\,\mathrm{m}/\mathrm{s}^{2}$; $T = m_1 a = 11.8\,\mathrm{N}$. The hanging block accelerates downward, so the [net force](https://one-course.com/books/physics/2/en/chapter/16-forces-and-motion#def-g11-forces-and-motion-netforce) on it points down: $T$ must undercut $m_2 g$.

**Exercise 25.9 ★★.**

A curve of radius $50\,\mathrm{m}$ is flat. Compute a car’s maximum speed for $\mu_s = 0.70$ (dry) and $0.40$ (wet), in $\mathrm{km}/\mathrm{h}$. Why does the answer hold for a scooter and a truck alike?

**Solution of Exercise 25.9.**

$v_{\max} = \sqrt{\mu_s g R}$: dry $\sqrt{0.70 \times 9.81 \times 50} = 18.5\,\mathrm{m}/\mathrm{s} \approx
67\,\mathrm{km}/\mathrm{h}$; wet $14.0\,\mathrm{m}/\mathrm{s} = 50\,\mathrm{km}/\mathrm{h}$. Both $mv^2/R$ and $\mu_s mg$ carry $m$: it cancels.

**Exercise 25.10 ★★.**

A $0.30\,\mathrm{kg}$ bob on a $1.2\,\mathrm{m}$ wire traces a horizontal circle, the wire holding a constant $25^\circ$ to the vertical (a conical pendulum). Project the second law [vertically](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight), then along the radius: compute the [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) and the speed.

**Solution of Exercise 25.10.**

Vertical: $T\cos\theta = mg$, so $T = 2.943/\cos 25^\circ = 3.2\,\mathrm{N}$. Radius $R = L\sin\theta =
0.507\,\mathrm{m}$; radial: $T\sin\theta = mv^2/R$, so $v = \sqrt{gR\tan\theta} = \sqrt{9.81 \times 0.507 \times 0.466}
= 1.5\,\mathrm{m}/\mathrm{s}$.

**Exercise 25.11 ★★.**

A $5.0\,\mathrm{kg}$ parcel rides a scale in a freight elevator; the readings:

| phase | $0\text{ to }2\,\mathrm{s}$ | $2\text{ to }8\,\mathrm{s}$ | $8\text{ to }10\,\mathrm{s}$ |
| --- | --- | --- | --- |
| scale ($\mathrm{N}$) | 54.0 | 49.0 | 44.1 |

Compute the acceleration in each phase and describe a possible ride. Could the elevator have been moving *downward* the whole time?

**Solution of Exercise 25.11.**

$a_z = (N - mg)/m$ with $mg = 49.05\,\mathrm{N}$: $+1.0\,\mathrm{m}/\mathrm{s}^{2}$, $0$, $-1.0\,\mathrm{m}/\mathrm{s}^{2}$. E.g. an upward trip: speed up, cruise, brake. Yes: entering already descending, the same readings mean brake, cruise, speed up again downward — the scale reads acceleration, not velocity.

**Exercise 25.12 ★★★.**

A $60\,\mathrm{kg}$ skater at rest on smooth ice throws a $4.0\,\mathrm{kg}$ ball horizontally. (a) The ball leaves at $8.0\,\mathrm{m}/\mathrm{s}$ *over the ground*: her recoil speed? (b) The $8.0\,\mathrm{m}/\mathrm{s}$ is now measured *relative to her*: recompute both speeds. (c) Which [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) pushed her backward?

**Solution of Exercise 25.12.**

(a) $0 = 60V + 4.0 \times 8.0$: $V = -0.53\,\mathrm{m}/\mathrm{s}$. (b) Ball at $8.0 + V$ over the ground: $0 = 60V + 4.0(8.0 + V)$, so $V = -32/64 = -0.50\,\mathrm{m}/\mathrm{s}$ and the ball flies at $7.5\,\mathrm{m}/\mathrm{s}$. (c) The ball’s push on her hands — the third-law partner of her throw.

**Exercise 25.13 ★★★.**

A highway curve of radius $80\,\mathrm{m}$ is designed for $20\,\mathrm{m}/\mathrm{s}$ with no help from [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory). Show that the roadway must be banked at the angle $\theta$ given by $\tan\theta = v^2/(gR)$ and compute it. What must [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) do for a car taking the curve slower — or faster?

**Solution of Exercise 25.13.**

Frictionless: $N\cos\theta = mg$ (vertical) and $N\sin\theta =
mv^2/R$ (radial); dividing, $\tan\theta = v^2/(gR) = 400/784.8$: $\theta \approx 27^\circ$. Slower: the car tends to slip down the banking, [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) must point up-slope; faster: the reverse.

**Exercise 25.14 ★★★.**

A rocket’s engines eject $250\,\mathrm{kg}$ of gas per second at $2500\,\mathrm{m}/\mathrm{s}$. (a) How much [momentum](#def-g12-newtons-laws-momentum) leaves per second — what [thrust](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) do the gases return? (b) The rocket masses $5.0 \times 10^{4}\,\mathrm{kg}$ at ignition: its [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight)? its lift-off acceleration? (c) Why does the engine [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) just as well in empty space?

**Solution of Exercise 25.14.**

(a) $250 \times 2500 = 6.3 \times 10^{5}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}$ each second: [thrust](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) $6.3 \times 10^{5}\,\mathrm{N}$. (b) $P = 4.9 \times 10^{5}\,\mathrm{N}$; $a = (6.25 - 4.905) \times 10^5 / 5.0 \times 10^{4} = 2.7\,\mathrm{m}/\mathrm{s}^{2}$. (c) It pushes on its own exhaust (third law), not on the air.

**Exercise 25.15 ★★★.**

A $4.0\,\mathrm{kg}$ block on a table ($\mu_s = 0.35$, $\mu_k = 0.25$) is tied over an ideal pulley to a hanging mass $m_2$. (a) What minimum $m_2$ starts the motion? (b) For $m_2 = 2.0\,\mathrm{kg}$, compute acceleration and [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory). (c) Check that $\mu_k = 0$ recovers the frictionless formula of [Exercise 25.8](#exo-g12-newtons-laws-8).

**Solution of Exercise 25.15.**

(a) $m_2 g > \mu_s m_1 g$: $m_2 > 0.35 \times 4.0 = 1.4\,\mathrm{kg}$. (b) $a = (m_2 - \mu_k m_1)\,g/(m_1 + m_2) = 1.0 \times 9.81/6.0 =
1.6\,\mathrm{m}/\mathrm{s}^{2}$; $T = m_2(g - a) = 16.4\,\mathrm{N}$. (c) $\mu_k = 0$ gives $a = m_2 g/(m_1 + m_2)$: the formula of [Exercise 25.8](#exo-g12-newtons-laws-8).

## 25.7 Problem: The Cable Car

**Problem 25.1.**

Weekend problem — certifying the cable car: the tension the slope demands at rest, the surcharge of the start-up, the relief over the tower, the price of an emergency stop — and the one number the placard in the cabin may promise

A mountain cable car — cabin plus carriage, empty mass $M_0 = 2.0 \times 10^{3}\,\mathrm{kg}$ — rolls on a track cable inclined at $\alpha = 30^\circ$, pulled by a haul cable parallel to the slope with [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) $\vect T$. Rolling [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) is negligible; the cabin floor stays horizontal throughout. The cabin seats $25$ passengers of average mass $80\,\mathrm{kg}$; you are the certifying engineer. Unless stated otherwise, the cabin is full: $M = 4.0 \times 10^{3}\,\mathrm{kg}$.

**Part I — At rest on the slope.**

1. Choose the system; say why the mountain frame will do as an inertial frame; inventory the three [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) , with directions.
2. Which law governs the wait? Project it along and perpendicular to the slope to express $T$ and the track cable’s [normal force](#def-g12-newtons-laws-friction) $N$ .
3. Compute $T$ and $N$ .
4. Recompute $T$ for the empty cabin. Why is $T$ proportional to the total mass?
5. The haul cable snaps, brakes off: show that the runaway acceleration is $g \sin\alpha$ , loaded or empty, and compute it.

**Part II — Start-up.** The winch accelerates the full cabin at $a = 0.60\,\mathrm{m}/\mathrm{s}^{2}$ up the slope until it cruises at $v = 6.0\,\mathrm{m}/\mathrm{s}$.

6. Express and compute the [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) $T'$ during this phase.
7. By what fraction does starting up raise the [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) above its value at rest?
8. How long does the phase last, and over what distance?
9. An $80\,\mathrm{kg}$ passenger stands on the horizontal floor. Project the second law [vertically](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) to express and compute her [apparent weight](#def-g12-newtons-laws-apparent) ; compare with $mg$ .
10. Which [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) accelerates her horizontally, and how big? Why do passengers lean?

**Part III — Over the tower.** At a constant $6.0\,\mathrm{m}/\mathrm{s}$, the track cable bends over a support tower along a vertical circular arc of radius $R = 40\,\mathrm{m}$; at the crest the velocity is horizontal.

11. In the [Frenet frame](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-frenet) at the crest ( [Chapter 24](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#ch-g12-kinematics-2d) ), give both components of the acceleration, with their values.
12. Project the second law along the normal to express and compute the [normal force](#def-g12-newtons-laws-friction) $N'$ of the cable on the carriage.
13. Same question for the passenger’s [apparent weight](#def-g12-newtons-laws-apparent) at the crest: by what fraction is she lighter?
14. At what speed would $N'$ vanish? Compare with the cruising speed and comment.
15. What [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) does the carriage exert on the cable over the tower — which law says so, with what magnitude and direction?

**Part IV — Braking, a jump, and the placard.**

16. On the horizontal arrival stretch, an emergency stop takes the full cabin from $6.0\,\mathrm{m}/\mathrm{s}$ to rest in $1.5\,\mathrm{s}$ : read $\sum \vect F = d\vect p/dt$ as an average over the stop to compute the braking [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) .
17. What horizontal [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) must stop an $80\,\mathrm{kg}$ passenger, and can shoe–floor static [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) ( $\mu_s = 0.40$ ) supply it? Conclude.
18. At the platform, brakes off, the empty cabin hangs free and still; the last passenger ( $80\,\mathrm{kg}$ ) jumps off horizontally at $2.5\,\mathrm{m}/\mathrm{s}$ *relative to the cabin* . Compute the cabin’s recoil speed and the passenger’s speed over the ground.
19. In one sentence: which propulsion device lives on this recoil principle, and what fixes its [thrust](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) ?
20. *Certification.* The haul cable breaks at $100\,\mathrm{kN}$ ; the rules cap the working [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) at one fifth of that. Show the worst case met above is the start-up, deduce the maximum total mass, and print the placard: how many $80\,\mathrm{kg}$ passengers may ride? Check the safety factor.

**Solution of Problem 25.1.**

**1.** System: cabin $+$ carriage ($+$ passengers). The mountain is fixed to the ground, inertial to excellent accuracy. [Forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force): [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) $M\vect g$ (down), [normal force](#def-g12-newtons-laws-friction) $\vect N$ of the track cable (perpendicular to it), haul [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) $\vect T$ (up the slope).

**2.** At rest, the first law: $\sum \vect F = \vect 0$. Along the slope: $T = Mg\sin\alpha$; perpendicular: $N = Mg\cos\alpha$.

**3.** $T = 4000 \times 9.81 \times 0.500 = 19.6\,\mathrm{kN}$; $N = 4000 \times 9.81 \times 0.866 = 34.0\,\mathrm{kN}$.

**4.** Empty: $T = 9.8\,\mathrm{kN}$. Every [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) in the balance is proportional to $M$, so $T$ is too.

**5.** Along the slope, $Ma = Mg\sin\alpha$: $M$ cancels, $a = g\sin\alpha = 4.9\,\mathrm{m}/\mathrm{s}^{2}$ down the slope.

**6.** $T' = M(g\sin\alpha + a) = 4000 \times 5.505 =
22.0\,\mathrm{kN}$.

**7.** $a/(g\sin\alpha) = 0.60/4.905 \approx 12\%$.

**8.** $t = v/a = 10\,\mathrm{s}$; $d = v^2/(2a) = 30\,\mathrm{m}$.

**9.** Vertical component of $\vect a$: $a\sin\alpha =
0.30\,\mathrm{m}/\mathrm{s}^{2}$, so $N = m(g + a\sin\alpha) = 80 \times 10.11 =
809\,\mathrm{N}$ against $mg = 785\,\mathrm{N}$: $3\%$ heavier.

**10.** The floor’s static [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory), horizontal: $ma\cos\alpha = 80 \times 0.52 \approx 42\,\mathrm{N}$. Leaning puts the combined floor push through her [center of mass](https://one-course.com/books/physics/2/en/chapter/16-forces-and-motion#def-g11-forces-and-motion-com), so it does not tip her.

**11.** $a_T = dv/dt = 0$ (constant speed); $a_N = v^2/R = 36/40 = 0.90\,\mathrm{m}/\mathrm{s}^{2}$, aimed at the center — straight down at the crest.

**12.** Normal axis (down positive): $Mg - N' = Mv^2/R$, so $N' = M(g - v^2/R) = 4000 \times 8.91 = 35.6\,\mathrm{kN}$.

**13.** $N = m(g - v^2/R) = 80 \times 8.91 = 713\,\mathrm{N}$: lighter by $v^2/(gR) = 0.90/9.81 \approx 9\%$.

**14.** $N' = 0$ at $v = \sqrt{gR} = 19.8\,\mathrm{m}/\mathrm{s} \approx
71\,\mathrm{km}/\mathrm{h}$ — more than triple the cruise: the carriage rides the cable with a wide margin.

**15.** [Newton’s third law](#thm-g12-newtons-laws-third): $35.6\,\mathrm{kN}$, pressing down on the cable (hence on the tower) — slightly less than the resting [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight), relieved by the curved flight.

**16.** $F = \Delta p/\Delta t = 4000 \times 6.0/1.5 =
16\,\mathrm{kN}$, opposite the motion ($a = 4.0\,\mathrm{m}/\mathrm{s}^{2}$).

**17.** $f = ma = 80 \times 4.0 = 320\,\mathrm{N}$; the ceiling is $\mu_s mg = 0.40 \times 785 = 314\,\mathrm{N} < 320\,\mathrm{N}$: feet alone slip — hold the handrail.

**18.** $0 = M_0 V + m(v_{\text{rel}} + V)$: $V = -80 \times 2.5/2080 = -0.096\,\mathrm{m}/\mathrm{s}$; the passenger moves at $2.5 - 0.096 \approx 2.4\,\mathrm{m}/\mathrm{s}$ over the ground.

**19.** The rocket: its [thrust](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) is fixed by the mass ejected per second times the ejection speed relative to it.

**20.** The largest [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) met is the start-up $T' = M \times 5.505\,\mathrm{N}/\mathrm{kg}$ (rest gives $Mg\sin\alpha$, the crest even less). Cap: $T' \leq 20\,\mathrm{kN}$, so $M \leq 20000/5.505 = 3.63 \times 10^{3}\,\mathrm{kg}$: payload $1.63 \times 10^{3}\,\mathrm{kg}$, i.e. $20$ passengers of $80\,\mathrm{kg}$ (not $25$). Placard: *20 persons*; then $M = 3.6 \times 10^{3}\,\mathrm{kg}$, $T' = 19.8\,\mathrm{kN}$, safety factor $100/19.8 \approx 5.1 \geq 5$.