Newton Laws of Motion
Apply Newton first, second, and third laws to explain inertia, acceleration, interactions, and action-reaction force pairs.
A lesson of Physics, in Forces and Newton Laws, about 10 minutes.
Newton's Laws of Motion
By the end of this lesson, you will be able to state and apply all three of Newton's laws, apply Newton's Third Law to identify action-reaction pairs, and solve multi-object system problems.
Newton's First Law (Law of Inertia)
"An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted upon by a net external force."
Inertia: the tendency of an object to resist changes in its state of motion. Measured by mass.
Implication: uniform circular motion is NOT natural — it requires a centripetal force. A satellite in orbit is constantly accelerating toward Earth.
Newton's Second Law
$$\vec{F}_{\text{net}} = m\vec{a}$$
- $\vec{F}_{\text{net}}$: vector sum of all forces on the object (N)
- $m$: mass (kg)
- $\vec{a}$: acceleration (m/s²)
Units: $1\text{ N} = 1\text{ kg}\cdot\text{m/s}^2$
Newton's Third Law
"For every action, there is an equal and opposite reaction."
More precisely: when object A exerts a force on object B, object B exerts an equal and opposite force on object A.
Key features of action-reaction pairs:
- They act on DIFFERENT objects
- They are always equal in magnitude and opposite in direction
- They are the same type of force
Example 1 — Third Law Pairs
A book rests on a table:
Action: Earth pulls book downward (gravity) — this is a force on the book.
Reaction: Book pulls Earth upward (gravity) — same type, different object.
NOT a Third Law pair: book's weight (Earth on book) and table's normal force (table on book) — these are both on the same object (the book). They're equal only because the book is in equilibrium.
Example 2 — Two-Object System (Atwood Machine)
Two masses connected by a string over a frictionless pulley: $m_1 = 3$ kg, $m_2 = 5$ kg. Find acceleration and tension.
The system has one acceleration ($a$ = same for both):
For $m_2$ (heavier, going down): $m_2g - T = m_2a$
For $m_1$ (lighter, going up): $T - m_1g = m_1a$
Add equations: $(m_2 - m_1)g = (m_1 + m_2)a$
$a = \frac{(5-3)(9.8)}{8} = \frac{19.6}{8} = 2.45$ m/s²
$T = m_1(g+a) = 3(9.8+2.45) = 36.75$ N
Inertial Reference Frames
Newton's laws are valid in inertial reference frames — frames not themselves accelerating.