Newton's Laws of Motion

Newton’s First Law — Inertia (CED Unit 2)

An object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction, unless acted upon by a net external force.

Inertial Reference Frames

Newton’s laws are valid only in inertial reference frames — frames that are not accelerating. A Frame attached to an accelerating car is non-inertial: a pendulum hanging from the ceiling swings Backward even though no horizontal force acts on it (in the car’s frame). In an inertial frame (the Ground), the pendulum tends to stay at rest while the car accelerates forward.

Mass and Inertia

Mass is a measure of an object’s resistance to changes in its motion (inertia). It is an Intrinsic property of an object, independent of location. The SI unit is the kilogram (kg).

Mass is not weight. Mass is a scalar; weight is a vector. Mass does not change with location; weight Depends on the local gravitational field. An astronaut on the Moon has the same mass as on Earth but one-sixth the weight.

Newton’s Second Law — Force and Acceleration (CED Unit 2)

The net external force on an object equals the product of its mass and acceleration:

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

In component form:

$$\sum F_x = ma_x, \qquad \sum F_y = ma_y$$

The Kilogram, Newton, and Slug

• 1 Newton (N) = $1 \mathrm{ kg \cdot \mathrm{m/s^2}}$
• 1 slug = $14.59 \mathrm{ kg}$ (used in the US customary system)

Weight vs Mass

$$\vec\{W\} = m\vec\{g\}$$

• Weight is a force (measured in N) that depends on the local gravitational field.
• Mass is an intrinsic property (measured in kg) that does not change with location.

On Earth: $g \approx 9.8 \mathrm{ m/s^2}$. On the Moon: $g \approx 1.62 \mathrm{ m/s^2}$.

Proof That $F = ma$ Is Not a Definition

One might think $F = ma$ merely defines force. But force is defined independently through Interactions (contact forces, gravitational forces, spring forces). $F = ma$ is an empirical law that relates the net force to the resulting acceleration. The fact that the same proportionality Constant ($m$) works for all forces (gravitational, electromagnetic, contact) is a deep and non-trivial result. If $F = ma$ were merely a definition, it would not have predictive power, but it does: given the forces, we can predict the motion, and given the motion, we can infer the forces.

Newton’s Third Law — Action-Reaction Pairs (CED Unit 2)

For every action, there is an equal and opposite reaction:

$$\vec\{F\}_\{A \mathrm\{ on B\} = -\vec\{F\}_\{B \mathrm\{ on A\}$$

Key Points About Action-Reaction Pairs

1. The two forces act on different objects.
2. The forces are always equal in magnitude and opposite in direction.
3. The forces are the same type (both gravitational, both normal, both frictional, etc.).
4. Action-reaction pairs cannot cancel each other because they act on different objects.

Systematic Identification of Third Law Pairs

To correctly identify a third law pair, use this algorithm:

1. Identify the force on object $A$ and what object exerts it: “Object $B$ exerts force $\vec{F}$on object $A$.”
2. The third law pair is: “Object $A$ exerts force $-\vec{F}$on object $B$.”
3. Verify: same type, equal magnitude, opposite direction, different objects.

Newton’s Law of Universal Gravitation (CED Unit 3)

$$F = \frac\{Gm_1 m_2\}\{r^2\}$$

Where $G = 6.674 \times 10^{-11} \mathrm{ N \cdot \mathrm{m^2/\mathrm{kg^2}}}$.

Gravitational Field

The gravitational field strength at a distance $r$ from mass $M$ is:

$$G = \frac\{GM\}\{r^2\}$$

Gravitational Potential Energy

$$U = -\frac\{GMm\}\{r\}$$

The negative sign indicates that work must be done against gravity to move masses apart. The Reference point ($U = 0$) is at infinity.

Orbital Mechanics

For a circular orbit of radius $r$ around mass $M$:

$$\frac\{GMm\}\{r^2\} = \frac\{mv^2\}\{r\} \implies v = \sqrt\{\frac\{GM\}\{r\}\}$$ $$T = \frac\{2\pi r\}\{v\} = 2\pi\sqrt\{\frac\{r^3\}\{GM\}\}$$

These are Kepler’s Third Law: $T^2 \propto r^3$.

Why All Orbits Are Conic Sections

Newton showed that any orbit under an inverse-square gravitational force is a conic section (ellipse, parabola, or hyperbola). Bound orbits (total energy negative) are ellipses. Escape Trajectories (total energy zero) are parabolas. Unbound trajectories (total energy positive) are Hyperbolas. Circular orbits are a special case of elliptical orbits with eccentricity zero.

Intuition

Newton’s laws answer the question why do objects move the way they do? The key insight is that forces cause changes in motion, not motion itself. An object moving at constant velocity needs no net force, it’s the acceleration that requires a force.

First Law intuition: This is really about reference frames. In an inertial frame (not accelerating), objects maintain their state of motion unless acted upon by a net force. The “law of inertia” is the baseline, it tells you what happens when nothing pushes.

Second Law intuition: $\vec{F} = m\vec{a}$is not a definition, it’s an empirical law. Force is defined independently through interactions (gravity, springs, contact). The deep fact is that the same proportionality constant ($m$) works for all forces. This is why we can predict motion from forces and infer forces from motion.

Third Law intuition: Forces always come in pairs acting on different objects. When you push a wall, the wall pushes back on you with equal force. But these forces don’t cancel because they act on different bodies. This is why a book on a table doesn’t accelerate, not because $\vec{N}$and $\vec{W}$are an action-reaction pair (they aren’t), but because they act on the same object and happen to balance.

Free-body diagrams: The single most important skill in mechanics. Isolate the object, draw every force acting on it, and never include forces the object exerts on other things. Get this right, and the rest is algebra.

Common Pitfalls

1. Including forces in the wrong direction on FBDs. Always draw forces in the direction they actually act, not in the direction of motion.
2. Confusing action-reaction pairs. $\vec{N}$and $\vec{W}$on the same object are not an action-reaction pair.
3. Using $\mu_s N$ as the static friction force. $f_s \le \mu_s N$; static friction adjusts to the applied force. Only use $\mu_s N$ for the maximum.
4. Forgetting that tension in a massless string is the same on both sides of a pulley (assuming a frictionless, massless pulley).
5. Assuming $\mu_k = \mu_s$. Kinetic friction is generally less than static friction.
6. Using $g = 9.8$ when a different value is specified or when the problem takes place far from Earth’s surface.
7. Ignoring the direction of the normal force. The normal force is always perpendicular to the contact surface, not necessarily vertical.
8. Adding mass to a pulley system without accounting for the pulley’s inertia. The standard Atwood machine assumes a massless pulley. A massive pulley has its own moment of inertia.
9. Assuming the acceleration of a connected system is the same for all parts. This is true only if the connections are rigid or inextensible.

Practice Questions

1. A $10 \mathrm{ kg}$ box is pushed across a floor with $\mu_k = 0.25$ by a horizontal force of $50 \mathrm{ N}$. Find the acceleration of the box.

2. Two blocks of masses $4.0 \mathrm{ kg}$ and $6.0 \mathrm{ kg}$ are connected by a string over a frictionless pulley (Atwood machine). Find the acceleration and the tension.

3. A $60 \mathrm{ kg}$ person stands on a scale in an elevator that accelerates upward at $2.0 \mathrm{ m/s^2}$. What does the scale read?

4. A $2.0 \mathrm{ kg}$ block on a $35^\circ$ incline has $\mu_s = 0.5$ and $\mu_k = 0.3$. Find the acceleration if the block is given an initial push up the incline.

5. A satellite orbits Earth at an altitude of $400 \mathrm{ km}$. Find the orbital velocity and period. (Earth’s mass = $5.97 \times 10^{24} \mathrm{ kg}$ Earth’s radius = $6371 \mathrm{ km}$.)

6. A $1500 \mathrm{ kg}$ car rounds a banked curve of radius 100 m at $20 \mathrm{ m/s}$. The banking angle is $15^\circ$. Find the minimum coefficient of static friction.

7. Three blocks of masses $m_1$, $m_2$ And $m_3$ are connected by strings on a frictionless table, with $m_3$ hanging off the edge over a pulley. Derive the acceleration in terms of the masses.

8. A $5.0 \mathrm{ kg}$ block is on a frictionless table connected to a $3.0 \mathrm{ kg}$ block hanging over the edge. Find the acceleration of the system and the tension in the string.

9. A block of mass $m$ is placed on a wedge of mass $M$ with angle $\theta$. All surfaces are frictionless. Find the acceleration of the wedge relative to the ground.

10. A $2000 \mathrm{ kg}$ car towing a $1000 \mathrm{ kg}$ trailer accelerates at $1.5 \mathrm{ m/s^2}$. If the tension in the tow bar is $1800 \mathrm{ N}$ Find the friction force on the car.

10. Friction: A Deeper Look

Microscopic Origin of Friction

At the microscopic level, even apparently smooth surfaces are rough. When two surfaces are pressed Together, the microscopic peaks (asperities) interlock. To slide the surfaces, these interlocking Features must be broken or deformed. The force needed is friction.

Static friction involves cold welding: the atoms at the contact points form temporary bonds. These bonds must be broken simultaneously to initiate motion, which requires a larger force. Once Sliding begins, the surfaces do not have time to form as many bonds, and the kinetic friction force is lower.

The Angle of Repose

The steepest angle at which an object on an inclined plane remains stationary is the angle of Repose, given by:

$$\theta_r = \arctan(\mu_s)$$

This is because the block is on the point of sliding when $mg\sin\theta = \mu_s mg\cos\theta$ I.e., $\tan\theta = \mu_s$. This relationship is used in geology to predict landslides and in industry to Design hoppers and chutes.

Worked Example: Friction on a Banked Curve Without Friction

A circular road of radius $80 \mathrm{ m}$ is banked at $15^{\circ}$. Find the speed at which a car can round the curve without needing any friction.

At the ideal speed, the horizontal component of the normal force provides the centripetal force, and the vertical component balances gravity:

$$N\sin\theta = \frac{mv^2}{r}, \qquad N\cos\theta = mg$$

Dividing: $\tan\theta = \frac{v^2}{rg}$

$$v = \sqrt{rg\tan\theta} = \sqrt{80 \times 9.8 \times \tan 15^{\circ}} = \sqrt{80 \times 9.8 \times 0.2679} = \sqrt{209.9} = 14.5 \mathrm{ m/s}$$

At this speed, the car can negotiate the curve even on ice. Below this speed, friction acts up the Slope; above it, friction acts down the slope.

11. Connected Bodies: Systematic Approach

Worked Example: Three-Body System with a Pulley

Three blocks are connected as follows: $m_1 = 4 \mathrm{ kg}$ and $m_2 = 6 \mathrm{ kg}$ on a Frictionless table, connected by a string over a pulley at the edge, with $m_3 = 5 \mathrm{ kg}$ Hanging vertically.

System approach: The net force on the system is $m_3 g$ (the weight of the hanging mass). The Total mass being accelerated is $m_1 + m_2 + m_3 = 15 \mathrm{ kg}$.

$$a = \frac{m_3 g}{m_1 + m_2 + m_3} = \frac{5 \times 9.8}{15} = \frac{49}{15} = 3.27 \mathrm{ m/s^2}$$

Individual approach for tension: For $m_3$:

$$m_3 g - T = m_3 a \implies T = m_3(g - a) = 5(9.8 - 3.27) = 5 \times 6.53 = 32.7 \mathrm{ N}$$

For the string between $m_1$ and $m_2$ on the table: $T' = m_2 a = 6 \times 3.27 = 19.6 \mathrm{ N}$.

Note that the tension in the string connecting to the hanging mass ($32.7 \mathrm{ N}$) is greater than the tension in the string between $m_1$ and $m_2$ ($19.6 \mathrm{ N}$), because the latter string Only needs to accelerate $m_2$ While the former must also provide the net force to accelerate $m_3$.

12. Gravitational Field Strength: Derivation of Key Results

Variation of $g$ with Altitude

At the Earth’s surface, $g \approx 9.8 \mathrm{ m/s^2}$. At height $h$ above the surface:

$$g(h) = \frac{GM}{(R_E + h)^2}$$

At $h = R_E$ (one Earth radius above the surface): $g = \frac{GM}{4R_E^2} = \frac{g_0}{4} \approx 2.45 \mathrm{ m/s^2}$.

This rapid decrease is why astronauts in low Earth orbit (altitude about 400 km) are not truly Weightless because of distance from Earth — they are weightless because they are in free fall. The Gravitational field strength at 400 km altitude is $g \approx 8.7 \mathrm{ m/s^2}$ About 89% of the Surface value.

Gravitational Field Strength Inside a Uniform Sphere

For a point at distance $r$ from the centre of a uniform sphere of radius $R$ and mass $M$:

$$g(r) = \frac{GMr}{R^3} \quad (r \le R)$$ $$g(r) = \frac{GM}{r^2} \quad (r \gt R)$$

Inside the sphere, $g$ increases linearly with $r$ and reaches its maximum at the surface. This is because only the mass within radius $r$ contributes to the gravitational force at that point (by Newton’s shell theorem).

13. Orbital Mechanics: Extended Worked Examples

Worked Example: Satellite Speed and Period

A satellite orbits at an altitude of $500 \mathrm{ km}$ above Earth’s surface.

$$r = R_E + h = 6.371 \times 10^6 + 5 \times 10^5 = 6.871 \times 10^6 \mathrm{ m}$$ $$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{6.871 \times 10^6}} = \sqrt{5.80 \times 10^7} = 7615 \mathrm{ m/s}$$ $$T = \frac{2\pi r}{v} = \frac{2\pi \times 6.871 \times 10^6}{7615} = 5671 \mathrm{ s \approx 94.5 \mathrm{ min}}$$

Worked Example: Escape Velocity from a Planet

The escape velocity from a planet of mass $M$ and radius $R$ is:

$$v_e = \sqrt{\frac{2GM}{R}}$$

For Earth: $v_e = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{6.371 \times 10^6}} = 11186 \mathrm{ m/s \approx 11.2 \mathrm{ km/s}}$.

A useful approximation: $v_e \approx \sqrt{2} \times v_{\mathrm{orbit}}$ for a surface-skimming orbit. The factor of $\sqrt{2}$ arises because escape requires twice the kinetic energy of a circular Orbit: $\frac{1}{2}mv_e^2 = 2 \times \frac{1}{2}mv_{\mathrm{orbit}^2}$.

14. Summary Table: Forces and Their Characteristics

15. Common Pitfalls: Extended

1. Assuming the normal force equals $mg$ on an inclined plane. The normal force is $mg\cos\theta$ Not $mg$. Always resolve forces perpendicular to the surface.

2. Forgetting that tension has a maximum value. A string can only pull, not push. If the calculated tension is negative, the string has gone slack and the tension is zero.

3. Applying $F = ma$ to individual objects when using the system approach. The system approach gives the acceleration of the entire system but does not give tensions or internal forces.

4. Ignoring air resistance in orbital calculations. While this is a valid approximation for satellites above the atmosphere, it fails for objects in the lower atmosphere.

5. Using $g = 9.8 \mathrm{ m/s^2}$ for objects far from Earth’s surface. Use $g = GM/r^2$ with the actual distance from Earth’s centre.

6. Confusing the mass of an orbiting body with the mass of the central body. In $v = \sqrt{GM/r}$, $M$ is the mass of the body being orbited (e.g., Earth), not the satellite.

Practice Questions (Additional)

1. A block of mass $m$ is placed on a wedge of mass $M$ with angle $\theta$. All surfaces are frictionless. Find the acceleration of the wedge. (Hint: use conservation of momentum in the horizontal direction.)

2. A $3 \mathrm{ kg}$ block on a frictionless table is connected by a string over a pulley to a $2 \mathrm{ kg}$ block hanging off the edge. The pulley has mass $0.5 \mathrm{ kg}$ and radius $0.1 \mathrm{ m}$ (moment of inertia $I = \frac{1}{2}mr^2$). Find the acceleration of the system and the tension on each side of the pulley.

3. A rocket of mass $1000 \mathrm{ kg}$ is launched vertically. The engine produces a thrust of $15000 \mathrm{ N}$ for $30 \mathrm{ s}$. Find the maximum height reached. (Assume $g$ is constant at $9.8 \mathrm{ m/s^2}$ and neglect air resistance.)

4. Two masses $m_1 = 2 \mathrm{ kg}$ and $m_2 = 3 \mathrm{ kg}$ are connected by a light string over a smooth pulley. $m_1$ rests on a rough table with $\mu_k = 0.3$. Find the acceleration and the tension in the string.

5. Calculate the gravitational field strength at a point halfway between the Earth and the Moon. (Earth-Moon distance $= 3.84 \times 10^8 \mathrm{ m}$, $M_E = 5.97 \times 10^{24} \mathrm{ kg}$ $M_M = 7.35 \times 10^{22} \mathrm{ kg}$.)

Extended Worked Examples

Example 16: Atwood Machine with Massive Pulley

Two masses $m_1 = 4 \mathrm{ kg}$ and $m_2 = 6 \mathrm{ kg}$ are connected by a light string over a Pulley of mass $M = 2 \mathrm{ kg}$ and radius $R = 0.1 \mathrm{ m}$ (solid disk, $I = \frac{1}{2}MR^2$). Find the acceleration of the system and the tensions on each side of the Pulley.

Step 1: Write equations for each mass and the pulley

For $m_1$ (lighter, accelerates up):

$$T_1 - m_1 g = m_1 a$$

For $m_2$ (heavier, accelerates down):

$$m_2 g - T_2 = m_2 a$$

For the pulley (rotates clockwise):

$$T_2 R - T_1 R = I\alpha = \frac{1}{2}MR^2 \times \frac{a}{R} = \frac{1}{2}MaR$$ $$T_2 - T_1 = \frac{1}{2}Ma$$

Step 2: Solve the system

From the first two equations:

$$T_1 = m_1(g + a), \quad T_2 = m_2(g - a)$$

Substituting into the pulley equation:

$$m_2(g - a) - m_1(g + a) = \frac{1}{2}Ma$$ $$(m_2 - m_1)g - (m_1 + m_2)a = \frac{1}{2}Ma$$ $$a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2}M} = \frac{(6 - 4) \times 9.8}{4 + 6 + 1} = \frac{19.6}{11} = 1.78 \mathrm{ m/s^2}$$

Step 3: Find the tensions

$$T_1 = 4(9.8 + 1.78) = 4 \times 11.58 = 46.3 \mathrm{ N}$$ $$T_2 = 6(9.8 - 1.78) = 6 \times 8.02 = 48.1 \mathrm{ N}$$

Cross-References

• Kinematics — Newton’s laws explain the forces that cause the motion described by kinematic equations.
• Work, Energy, and Power — The work-energy theorem connects force and displacement to energy changes.
• Momentum and Impulse — Newton’s second law in the form F = dp/dt directly links force to the rate of change of momentum.
• AP Calculus, Differential Equations: Newton’s second law F = ma is a second-order differential equation; solving it requires integration to find velocity and position from acceleration.