Laws of Motion • Topic 1 of 3

Newton's Laws & Inertia

Why do objects move the way they do? For centuries people believed a force was needed to keep something moving. Newton overturned that idea with three laws that together form the foundation of mechanics. They tell us not just that things move, but why their motion changes.

Newton's First Law (law of inertia). A body continues in its state of rest or of uniform motion in a straight line unless an external unbalanced force acts on it. The natural tendency of an object to resist any change in its state of motion is called inertia. A heavier body has greater inertia, so mass is a measure of inertia. There are three kinds: inertia of rest (a passenger lurches backward when a bus starts), inertia of motion (you lurch forward when it stops), and inertia of direction (you are thrown outward on a sharp turn).

Newton's Second Law. The rate of change of linear momentum of a body is directly proportional to the applied force and takes place in the direction of the force. Writing momentum as $\vec{p}=m\vec{v}$, the law is $\vec{F}=\frac{d\vec{p}}{dt}$. For a body of constant mass this reduces to the familiar $\vec{F}=m\vec{a}$. The SI unit of force is the newton (N), where $1\ \text{N}=1\ \text{kg}\cdot\text{m/s}^2$ — the force that gives a 1 kg mass an acceleration of $1\ \text{m/s}^2$.

Newton's Third Law. To every action there is an equal and opposite reaction. If body A exerts a force on body B, then B exerts an equal and opposite force on A: $\vec{F}_{AB}=-\vec{F}_{BA}$. The crucial point is that action and reaction act on different bodies, so they never cancel each other.

  • Free-body diagram (FBD): a sketch showing all the forces acting on a single chosen body — weight $mg$ downward, normal reaction $N$ perpendicular to the surface, tension $T$ along strings, applied force and friction. Solving problems means setting $\sum \vec{F}=m\vec{a}$ along chosen axes.
  • Apparent weight in a lift: the normal reaction (what a weighing scale reads) is $N=m(g+a)$ when the lift accelerates upward, $N=m(g-a)$ when it accelerates downward, and $N=0$ (weightlessness) during free fall ($a=g$).
Free-body diagram: weight, normal reaction, applied force and frictionmN (normal)mg (weight)F (applied)f (friction)Sum of forces = m a (Newton's 2nd law)
Solved Examples
1
Worked Example
A force of 20 N acts on a body of mass 4 kg. Calculate the acceleration produced.
Solution
  1. Step 1: Use Newton's second law $\vec{F}=m\vec{a}$, so $a=\frac{F}{m}$.
  2. Step 2: Substitute: $a=\frac{20}{4}$.
  3. Step 3: Compute: $a=5\ \text{m/s}^2$.

Answer: $a=5\ \text{m/s}^2$

2
Worked Example
A car of mass 1000 kg accelerates from rest to 20 m/s in 10 s. Find the net force acting on it.
Solution
  1. Step 1: Find acceleration: $a=\frac{v-u}{t}=\frac{20-0}{10}=2\ \text{m/s}^2$.
  2. Step 2: Apply $F=ma$.
  3. Step 3: Substitute: $F=1000 \times 2$.
  4. Step 4: Compute: $F=2000\ \text{N}$.

Answer: $F=2000\ \text{N}$

3
Worked Example
A person of mass 60 kg stands on a weighing machine in a lift. Find the reading (apparent weight) when the lift accelerates upward at $2\ \text{m/s}^2$. (Take $g=10\ \text{m/s}^2$.)
Solution
  1. Step 1: For upward acceleration, $N=m(g+a)$.
  2. Step 2: Substitute: $N=60(10+2)$.
  3. Step 3: Compute: $N=60 \times 12=720\ \text{N}$.

Answer: $N=720\ \text{N}$ (apparent weight is greater than the true weight $600\ \text{N}$).

4
Worked Example
The same 60 kg person stands in a lift that accelerates downward at $3\ \text{m/s}^2$. Find the apparent weight. (Take $g=10\ \text{m/s}^2$.)
Solution
  1. Step 1: For downward acceleration, $N=m(g-a)$.
  2. Step 2: Substitute: $N=60(10-3)$.
  3. Step 3: Compute: $N=60 \times 7=420\ \text{N}$.

Answer: $N=420\ \text{N}$ (apparent weight is less than the true weight).

5
Worked Example
Two blocks of masses 3 kg and 2 kg are placed in contact on a frictionless surface. A force of 25 N is applied to the 3 kg block. Find the common acceleration and the contact force between them.
Solution
  1. Step 1: Treat both blocks as one system: $a=\frac{F}{m_1+m_2}=\frac{25}{3+2}=5\ \text{m/s}^2$.
  2. Step 2: The contact force $C$ alone accelerates the 2 kg block: $C=m_2 a$.
  3. Step 3: Substitute: $C=2 \times 5=10\ \text{N}$.

Answer: $a=5\ \text{m/s}^2$ and contact force $C=10\ \text{N}$.

6
Worked Example
A 5 kg block hangs from a string attached to the ceiling of a lift. Find the tension in the string when the lift moves up with acceleration $2\ \text{m/s}^2$. (Take $g=10\ \text{m/s}^2$.)
Solution
  1. Step 1: Forces on the block: tension $T$ up, weight $mg$ down; net upward force $=ma$.
  2. Step 2: Apply Newton's second law: $T-mg=ma$, so $T=m(g+a)$.
  3. Step 3: Substitute: $T=5(10+2)=5 \times 12$.
  4. Step 4: Compute: $T=60\ \text{N}$.

Answer: $T=60\ \text{N}$

Key Points

  • First law (inertia): a body keeps its state of rest or uniform motion unless an unbalanced external force acts; mass measures inertia.
  • Second law: $\vec{F}=\frac{d\vec{p}}{dt}$, which gives $\vec{F}=m\vec{a}$ for constant mass; $1\ \text{N}=1\ \text{kg}\cdot\text{m/s}^2$.
  • Third law: action and reaction are equal and opposite ($\vec{F}_{AB}=-\vec{F}_{BA}$) but act on different bodies, so they never cancel.
  • A free-body diagram shows all forces ($mg$, $N$, $T$, applied, friction) on one chosen body before applying $\sum\vec{F}=m\vec{a}$.
  • Apparent weight in a lift: $N=m(g+a)$ going up, $N=m(g-a)$ going down, $N=0$ in free fall (weightlessness).
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Mass of a body is a measure of its:
Explanation: A larger mass resists changes in motion more, so mass measures inertia.
Q2.The SI unit of force, the newton, equals:
Explanation: From $F=ma$, $1\ \text{N}=1\ \text{kg}\cdot\text{m/s}^2$.
Q3.Action and reaction forces:
Explanation: They are equal and opposite but act on different bodies, so they do not cancel.
Q4.A person feels heavier in a lift when the lift:
Explanation: Upward acceleration gives $N=m(g+a)>mg$, so apparent weight increases.
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