Chapter 03 • Newton's Laws of Motion

Newton's Laws of Motion

Explore accelerating reference frames, pseudo force, gravitation, orbital motion, centripetal effects, air resistance, variation of acceleration due to gravity with altitude and depth, and the turning effect of force.

01

Accelerating Frames & Pseudo Force

Understand inertial and accelerating frames and why pseudo force is introduced in a non-inertial frame.

02

Gravitation & Orbital Motion

Understand the role of gravity in orbital motion and the relationship between tangential motion and centripetal force.

03

Air Resistance & Gravity

Compare falling objects in air and vacuum and study the variation of g above and below Earth's surface.

04

Turning Forces & Torque

Learn moment of force, lever arm, angle of application and why a long spanner is effective.

Chapter Learning Path

  1. Start with the limitation of Newton's laws when the observer is in an accelerating frame.
  2. Use the pseudo-force idea to describe motion from inside such a frame.
  3. Connect gravitation with the orbital motion of Earth and the Moon.
  4. Study air resistance and the effect of altitude and depth on acceleration due to gravity.
  5. Finish with turning effects of forces and torque.

Core Formulae

$$F_{\text{pseudo}}=-ma_{\text{frame}}$$
$$F=mg$$
$$g_h=g\left(\frac{R}{R+h}\right)^2$$
$$g_d=g\left(1-\frac{d}{R}\right)$$
$$\tau=Fd\sin\theta$$

Newton's Laws of Motion: Learn the Concepts

3.1 Accelerating Frames and Pseudo Force

01

First, understand the idea

Imagine you are standing inside a bus. The bus is initially at rest. Suddenly, the driver accelerates the bus forward. You feel as if you are pushed backward. Nothing is actually pulling you backward. Your body simply tends to maintain its previous state of rest.

What is a Reference Frame?

Reference frame: A chosen point of view or coordinate system from which the position and motion of an object are described.

The same event can look different to different observers. A passenger inside a moving bus and a person standing on the road do not describe the passenger's motion in exactly the same way.

Inertial and Non-Inertial Frames

Inertial Frame

A frame that is at rest or moving with constant velocity. Newton's laws can be applied directly in this frame.

$$a_{\text{frame}}=0$$

Non-Inertial Frame

A frame that is accelerating. To apply Newton's laws from this frame, an additional apparent force called pseudo force is introduced.

$$a_{\text{frame}}\neq0$$

What is Pseudo Force?

Pseudo force is an apparent force introduced when motion is observed from an accelerating reference frame. It acts opposite to the acceleration of that frame.

$$\boxed{F_{\text{pseudo}}=-m a_{\text{frame}}}$$

Worked Example: Accelerating Lift

A 60 kg person is inside a lift accelerating upward at $4.5\,m/s^2$. Find the magnitude of the pseudo force.

Step 1: Write the formula

$$F_{\text{pseudo}}=ma_{\text{frame}}$$

Step 2: Substitute the values

$$F=60\times4.5$$

Step 3: Calculate

$$\boxed{F=270\,N}$$

Since the lift accelerates upward, the pseudo force is directed downward.

3D Interactive Simulation: Accelerating Frame & Pseudo Force
Three.js 3D Engine
Drag to orbit 3D elevator
Lift Acceleration: a = +4.5 m/s² (Upward)
Pseudo Force: F_pseudo = −270 N (Downward) | Scale Reading: N = m(g + a) = 858 N (Feels Heavier)
Elevator Motion:
Quick Check — Pseudo Force
  1. What is a reference frame?
  2. What is the difference between an inertial and a non-inertial frame?
  3. Write the formula for pseudo force.
  4. Why does a passenger appear to move backward when a bus suddenly accelerates forward?

3.2 Gravitation and Orbital Motion

Why Does Earth Orbit the Sun?

Earth has a tendency to continue moving in a straight line because of inertia. The Sun continuously attracts Earth toward itself through gravity. These two ideas together explain the curved orbital path.

Remember the two directions

  • Tangential direction: the direction in which Earth would move if gravity did not bend its path.
  • Inward direction: the direction of the Sun's gravitational pull, which provides the centripetal effect.

Think Like a Student

If the Sun suddenly disappeared, there would be no gravitational pull to bend Earth's path. Earth would continue approximately along the tangent to its orbit.

If Earth lost its tangential motion while gravity remained, it would move toward the Sun. In the real situation, both effects are present continuously.

Activity 3.2

Thread-and-Bob Model

  1. Tie a small object to a thread.
  2. Swing it in a horizontal circle.
  3. The thread pulls the object inward.
  4. The object's inertia makes it tend to continue tangentially.

3D Interactive Simulation: Gravitation & Orbital Motion
Three.js 3D Engine
Drag to orbit 3D solar system
Gravitational Force: F_g = GMm/r² (Centripetal pull) | Tangential Velocity: v = √(GM/r)
Stable Orbit: Continual free fall curved toward Sun without crashing due to forward inertia.
Quick Check — Orbital Motion
  1. What force keeps Earth in its orbit?
  2. What would happen to Earth's path if the Sun suddenly disappeared?
  3. Why does Earth not simply fall straight into the Sun?
  4. What is meant by tangential motion?

3.3 Air Resistance: Why Shape Matters

What happens when objects fall through air?

Gravity pulls a falling object downward, while air resistance acts opposite to its motion. The amount of air resistance depends on factors such as the object's cross-sectional area, shape and speed.

In Air

  • Air resistance is present.
  • A larger cross-sectional area generally produces greater drag.
  • For equal-mass objects of different shapes, the smaller-area object can fall faster in air.

In Vacuum

  • There is no air.
  • Therefore there is no air resistance.
  • Objects fall under gravity without drag differences due to shape.

Everyday Connection: Paper and a Crumpled Paper Ball

A flat sheet presents a larger area to the air than the same sheet crumpled into a ball. The difference in fall behaviour in air helps us see the effect of air resistance.

3D Interactive Simulation: Air Resistance vs. Vacuum Fall
Three.js 3D Engine
Drag to orbit 3D chambers
Chamber 1 (In Air): Flat Sheet Lags (High Air Drag) vs Compact Sphere.
Chamber 2 (In Vacuum): Both Fall Together with identical acceleration $g = 9.8\text{ m/s}^2$.
Quick Check — Air Resistance
  1. In which direction does air resistance act for a falling object?
  2. Name two factors that affect air resistance.
  3. Why does a larger cross-sectional area generally experience greater drag?
  4. Why do objects of different shapes behave more similarly in a vacuum?

3.4 Acceleration Due to Gravity at Height

The acceleration due to gravity depends on the distance from Earth's centre. At the surface, that distance is $R$. At height $h$, the distance becomes $R+h$.

Derivation: Value of $g$ at Height $h$

Step 1 — At Earth's surface

$$g=\frac{GM}{R^2}$$

Step 2 — At height $h$

$$g_h=\frac{GM}{(R+h)^2}$$

Step 3 — Divide the two equations

$$\frac{g_h}{g}=\frac{R^2}{(R+h)^2}$$

Step 4 — Final result

$$\boxed{g_h=g\left(\frac{R}{R+h}\right)^2}$$

Worked Example

Find $g$ at a height of 800 km if $R=6400$ km and $g=9.8\,m/s^2$.

Substitute:

$$g_h=9.8\left(\frac{6400}{6400+800}\right)^2$$

Therefore:

$$\boxed{g_h\approx7.74\,m/s^2}$$
Quick Check — Gravity at Height
  1. Why does $g$ decrease as height increases?
  2. Write the formula for $g_h$.
  3. Calculate $g$ at a height of 400 km if $R=6400$ km and $g=9.8\,m/s^2$.

3.5 Acceleration Due to Gravity Below Earth's Surface

As we move from the surface toward the centre of Earth, the acceleration due to gravity decreases. In the uniform-density model used here, the decrease is linear with depth.

Derivation: Value of $g$ at Depth $d$

Step 1 — Mass of Earth

$$M=\frac{4}{3}\pi R^3\rho$$

Step 2 — Mass inside radius $(R-d)$

$$M_d=\frac{4}{3}\pi(R-d)^3\rho$$

Step 3 — Gravity at depth $d$

$$g_d=\frac{GM_d}{(R-d)^2}$$

Step 4 — Simplify

$$g_d=\frac{4}{3}\pi G\rho(R-d)$$

Step 5 — Compare with surface value

$$g=\frac{4}{3}\pi G\rho R$$

Step 6 — Final result

$$\boxed{g_d=g\left(1-\frac{d}{R}\right)}$$

Worked Example: When does $g$ become one-tenth?

Let $g_d=g/10$.

$$\frac{g}{10}=g\left(1-\frac{d}{R}\right)$$ $$\frac{1}{10}=1-\frac{d}{R}$$ $$\frac{d}{R}=\frac{9}{10}$$ $$\boxed{d=\frac{9R}{10}}$$
Quick Check — Gravity with Depth
  1. Write the formula for $g_d$.
  2. What happens to $g$ as we move toward the centre of Earth?
  3. At what depth will $g$ become half its surface value?
  4. What is the value of $g$ at the centre according to this model?

3.6 Turning Effect of Force — Torque

Start with a Door

A door rotates about its hinges. A force can therefore do more than simply move an object; it can also produce rotation.

Moment of force / Torque: The turning effect produced by a force about a pivot.

Torque Formula

$$\boxed{\color{#0284c7}{\tau} = \color{#e11d48}{F} \color{#059669}{d} \color{#7c3aed}{\sin\theta}}$$
τ Torque / Turning Moment (N·m)
F Applied Force (N)
d Lever Arm Distance from Pivot (m)
θ Angle between Force and Arm
Hinge / Pivot (O) Lever Arm Distance d Force F (θ = 90°) Torque τ = F × d
Figure 3.1: Turning moment produced about a pivot. Longer lever arm d produces greater rotational torque τ.

Here:

  • $\tau$ = torque or moment of force
  • $F$ = applied force
  • $d$ = distance from pivot to point of application
  • $\theta$ = angle between the force and the line joining the pivot to the point of application

When is Torque Maximum?

Since $\sin90^\circ=1$, torque is maximum when the force is perpendicular to the lever arm.

$$\boxed{\tau_{\max}=Fd\quad\text{when }\theta=90^\circ}$$

When is Torque Zero?

$$\tau=0\quad\text{when }\theta=0^\circ\text{ or }180^\circ$$

Worked Example

A 20 N force acts perpendicular to a wrench at a distance of 0.5 m from the pivot. Find the torque.

$$\tau=Fd\sin90^\circ$$ $$=20\times0.5\times1$$ $$\boxed{\tau=10\,Nm}$$
3D Interactive Simulation: Torque & Lever Arm ($\tau = F \cdot d \cdot \sin\theta$)
Three.js 3D Engine
Drag to orbit 3D wrench
Lever Arm: d = 0.58 m | Force: F = 20 N | Angle: θ = 90°
Torque Generated: τ = F·d·sinθ = 11.60 N·m (Maximum Torque)
Wrench Length (d):
Force Angle (θ):
Quick Check — Torque
  1. What is the turning effect of a force called?
  2. Write the formula for torque.
  3. At what angle is torque maximum?
  4. Why is a long spanner easier to use than a short spanner?
  5. A 20 N force acts perpendicular to a wrench at 0.5 m from the pivot. Calculate the torque.

Exercise 3.1 – Newton's Laws of Motion

Question 1. In which type of reference frame are Newton's laws valid?
View Answer

According to the supplied chapter, Newton's laws are strictly valid in a non-accelerating (inertial) frame.

Question 2. Define pseudo force and write its formula.
View Answer

A pseudo force is an apparent force observed only in an accelerating frame of reference.

$$F_{\text{pseudo}}=-ma_{\text{frame}}$$
Question 3. A lift accelerates upward at $4.5\,m\,s^{-2}$. Calculate the pseudo force experienced by a 60 kg person inside the lift.
View Answer

Using the pseudo-force magnitude $F=ma$:

$$F=60\times4.5=270\,N$$

The pseudo force acts opposite to the acceleration of the frame, i.e. downward for the upward-accelerating lift.

Question 4. Why does pseudo force disappear in an inertial frame?
View Answer

Pseudo force is introduced only because the reference frame itself is accelerating. In an inertial frame the frame acceleration is zero, so the pseudo-force term is zero.

Question 5. Why is the Sun's gravitational pull important for Earth's orbit?
View Answer

The Sun's gravitational pull provides the centripetal force needed to continuously change Earth's direction of motion and keep it on its curved orbit.

Question 6. Why does an object with smaller cross-sectional area fall faster than an equal-mass object with larger cross-sectional area in air?
View Answer

The larger cross-sectional area experiences greater air resistance. The smaller object faces less opposition, giving it a greater net downward force and allowing it to fall faster.

Question 7. Why do two equal-mass objects reach the ground together in a vacuum?
View Answer

In a vacuum there is no air resistance. Objects then fall under gravity without the drag difference caused by their shape or cross-sectional area.

Question 8. Calculate acceleration due to gravity at a height of 400 km if $R=6400$ km and $g=9.8\,m/s^2$.
View Answer
$$g_h=9.8\left(\frac{6400}{6400+400}\right)^2$$ $$g_h=9.8\left(\frac{6400}{6800}\right)^2$$ $$g_h\approx8.68\,m/s^2$$
Question 9. At what depth will $g$ become half of its surface value?
View Answer
$$g_d=g\left(1-\frac dR\right)$$ $$\frac g2=g\left(1-\frac dR\right)$$ $$\frac12=1-\frac dR$$ $$\frac dR=\frac12$$ $$d=\frac R2$$
Question 10. Why is it easier to open a door when you push at the handle rather than near the hinges?
View Answer

Torque depends on the distance from the pivot. A larger lever arm gives a greater turning effect for the same force.

Question 11. A force is applied to a wrench at different angles. At which angle will the rotating force be maximum? What happens when the force is applied parallel to the wrench?
View Answer

Torque is maximum when $\theta=90^\circ$ because $\sin90^\circ=1$. If the force is parallel to the wrench, $\theta=0^\circ$ and the torque is zero.

Question 12. Two students apply the same force to open a gate. One pushes perpendicular to the gate at 20 cm from the hinge. The other pushes perpendicular at 80 cm. Who produces greater torque? Justify.
View Answer

The student pushing at 80 cm produces greater torque because $\tau=Fd$ for perpendicular force. The lever arm is four times as large, so the torque is four times as large.

Question 13. Is it possible for a force to act on a body and still produce zero turning about a given fixed point? Give a real-life example.
View Answer

Yes. If the force acts along the line through the pivot, its perpendicular lever arm is zero, so torque is zero. For example, pushing directly toward a door hinge does not produce the same turning effect as a perpendicular push at the handle.

Question 14. Two forces act on a rod pivoted at its centre: (I) 10 N downward at 0.5 m on the left and (II) 10 N downward at 0.5 m on the right. Will the rod rotate?
View Answer

The two torques have equal magnitudes but opposite turning directions, so their net torque is zero. The rod will not rotate due to these two forces.

Question 15. How can a mechanic loosen a tight bolt using a long spanner instead of applying a very large force?
View Answer

For a perpendicular force, $\tau=Fd$. Increasing the spanner length increases the lever arm $d$, so the mechanic can produce the required torque with a smaller force.

Question 16. A force of 20 N is applied to a door at 0.8 m from the hinge. Calculate the torque when the force is applied at (a) 90°, (b) 60°, (c) 30° to the door surface.
View Answer

Use $\tau=Fd\sin\theta$, with $F=20\,N$ and $d=0.8\,m$.

$$\tau_{90}=20(0.8)\sin90^\circ=16\,Nm$$
$$\tau_{60}=20(0.8)\sin60^\circ\approx13.86\,Nm$$
$$\tau_{30}=20(0.8)\sin30^\circ=8\,Nm$$

Newton's Laws of Motion Worksheets

Worksheets

Worksheet content was not included in the supplied Chapter 3 material. This tab is reserved for printable practice on pseudo force, gravitation, orbital motion, gravity variation and torque.

Coming Soon

Newton's Laws of Motion – Quick Revision

Concept Quick Revision
Inertial frame A non-accelerating frame in which Newton's laws are used without the pseudo-force correction.
Non-inertial frame An accelerating frame in which pseudo force is introduced to describe motion consistently.
Pseudo force Apparent force observed only from an accelerating frame; opposite to frame acceleration.
Orbital motion Gravity provides the inward/centripetal pull while inertia gives the tangential tendency.
Air resistance Drag opposes motion and depends on factors including cross-sectional area, shape, speed and air density.
Gravity at height According to the chapter, $g_h=g(R/(R+h))^2$.
Gravity at depth For the uniform-density model, $g_d=g(1-d/R)$.
Torque Turning effect of force: $\tau=Fd\sin\theta$.
Maximum torque Occurs at $\theta=90^\circ$.
Zero torque Occurs at $\theta=0^\circ$ or $180^\circ$.

Must-Remember Formulae

$$F_{\text{pseudo}}=-ma_{\text{frame}}$$
$$F=mg$$
$$g_h=g\left(\frac{R}{R+h}\right)^2$$
$$g_d=g\left(1-\frac{d}{R}\right)$$
$$\tau=Fd\sin\theta$$

Newton's Laws of Motion – Topic Tests

Interactive Tests

Test content was not included in the supplied Chapter 3 material. This tab is reserved for graded tests covering pseudo force, gravitation, air resistance, variation of $g$, orbital motion and torque.

Coming Soon