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Newton's Laws of Motion

Source lecture(s): SC133 Lec 6

Intuition

Before Newton, "natural" motion was rest: things stop unless pushed. Newton flipped it: motion continues unless pushed — what needs explaining is change of motion, and force is the thing that changes it. Three laws capture the whole story: what happens with no force (1st), how force produces acceleration (2nd), and where forces come from (3rd: always in pairs).

The three laws

First law (inertia). If the net force on a body is zero, its velocity does not change. Constant velocity — including rest — is the force-free state. (This law also defines inertial reference frames: the frames where it holds.)

Second law. The net force equals mass times acceleration:

\[\boxed{\,\sum \vec F = m\vec a\,}\]

— a vector equation: one equation per axis. More generally \(\sum\vec F = d\vec p/dt\) with momentum \(\vec p = m\vec v\), the form that survives into collisions and rocket motion. Details on the equation page.

Third law. If A pushes B with force \(\vec F\), then B pushes A with \(-\vec F\) — equal magnitude, opposite direction, acting on different bodies. Forces are interactions; nothing pushes without being pushed back.

The everyday forces

Force Nature Magnitude
Weight gravity on mass \(mg\), downward
Normal force \(N\) surface pushes back whatever geometry demands (⊥ surface)
Tension \(T\) rope pulls along itself same throughout an ideal (massless) rope
Friction resists sliding \(\leq \mu_s N\) static; \(\mu_k N\) kinetic
Spring restoring \(F = -kx\) (Hooke's law)

The method: free-body diagrams

Every Newton's-law problem is the same five steps:

  1. Isolate one body; draw it alone.
  2. Draw every force acting on it (not forces it exerts on others).
  3. Choose axes — align one with the acceleration if you know its direction.
  4. Write \(\sum F = ma\) per axis.
  5. Solve; sanity-check limits (does \(\theta \to 0\) make sense?).

Worked example: two blocks and a pulley

Mass \(m_1 = 3\) kg on a frictionless table, string over an ideal pulley to hanging \(m_2 = 2\) kg. Find the acceleration and tension.

Two bodies, two diagrams: \(m_1\): \(T = m_1 a\). \(m_2\): \(m_2 g - T = m_2 a\). Add: \(m_2 g = (m_1 + m_2)a \Rightarrow a = \frac{2(9.8)}{5} = 3.92\,\text{m/s}^2\), \(T = 3(3.92) = 11.8\,\text{N}\) — less than \(m_2g = 19.6\) N, as it must be (the hanging block accelerates down).

Common mistakes

  • Third-law pairs on one diagram. Weight and normal force are not a third-law pair (both act on the same body); the pair of your weight is your pull on Earth.
  • "Force of motion". A puck sliding at constant velocity needs zero net force — velocity is not evidence of force; acceleration is.
  • Assuming \(N = mg\) always. On inclines, in elevators, or with extra vertical forces, \(N\) adjusts — solve for it, don't assume it.
  • Applying \(\sum F = ma\) to the wrong body or mixing forces from different bodies in one equation.

Knowledge graph position

Prerequisites: Kinematics, Vectors. Leads to: essentially everything — work & energy, momentum, rotation, gravitation, oscillations.

Quiz

Q1 (conceptual). A horse pulls a cart; by the third law the cart pulls the horse back equally. How does anything ever move?

Answer

The pair acts on different bodies. The cart accelerates because the horse's pull exceeds friction on the cart; the horse accelerates because the ground's friction on its hooves exceeds the cart's backward pull on the horse. Sum forces per body, never across bodies.

Q2 (computational). A 70 kg person stands on a scale in an elevator accelerating upward at \(2\,\text{m/s}^2\). What does the scale read?

Answer

\(N - mg = ma \Rightarrow N = m(g + a) = 70(11.8) = 826\,\text{N}\) — about 84 kg "apparent weight". Descending at the same rate it would read 546 N.

Q3 (multiple choice). An object moves at constant velocity. The net force on it is: (a) constant and nonzero (b) in the direction of motion (c) zero

Answer

(c) — first law. Any nonzero net force would change the velocity.

Q4 (conceptual). Why does a small car and a huge truck experience the same force magnitude in a collision, yet the car fares worse?

Answer

Third law guarantees equal forces. But \(a = F/m\): the same force gives the lighter car a much larger acceleration (and its occupants larger accelerations too).