Laws of Motion & Friction

Newton's three laws form the foundation of classical mechanics. The First Law (Law of Inertia) states that a body remains at rest or in uniform motion in a straight line unless acted upon by a net external force. Inertia is the resistance to change in motion and is directly proportional to mass. The Second Law provides the quantitative link: F = ma = dp/dt, where F is the net force (N), m is the mass (kg), a is the acceleration (m/$s^{2}$), and p = mv is the momentum (kg m/s). The dimensional formula of force is [$M^{1} \; L^{1} \; T^{-2}$]. The Third Law states that for every action, there is an equal and opposite reaction — crucially, these forces act on DIFFERENT bodies. The normal force N and weight mg on a body resting on a surface are NOT an action-reaction pair; they act on the same body. The true action-reaction pair is the body's weight pulling the Earth up and the Earth's gravity pulling the body down.

Momentum p = mv [$M^{1} \; L^{1} \; T^{-1}$] (kg m/s) is conserved when the net external force on a system is zero: $m_{1} \; u_{1}$ + $m_{2} \; u_{2}$ = $m_{1} \; v_{1}$ + $m_{2} \; v_{2}$.
Impulse J = F $\Delta$-t = $\Delta$-p [$M^{1} \; L^{1} \; T^{-1}$] (N s = kg m/s) represents the change in momentum and equals the area under a force-time graph.

The apparent weight of a person of mass m in a lift depends on the lift's acceleration. When the lift accelerates upward, W' = m(g + a) — the person feels heavier. When it accelerates downward, W' = m(g - a) — the person feels lighter. In free fall (a = g), W' = 0 — weightlessness. Sign convention: take upward as positive throughout lift problems.

For the Atwood machine (two masses $m_{1}$ > $m_{2}$ connected by a string over a frictionless pulley): acceleration a = ($m_{1}$ - $m_{2}$)g / ($m_{1}$ + $m_{2}$) and tension T = $2m_{1} \; m_{2}$ g / ($m_{1}$ + $m_{2}$). Note that T is always between the two weights: $m_{2}$ g < T < $m_{1}$ g.

Friction opposes relative motion between surfaces. Static friction is self-adjusting: $f_{s}$ ranges from 0 up to the maximum value $\mu_{s}$ N (limiting friction), where $\mu_{s}$ is the coefficient of static friction (dimensionless) and N is the normal force (N).
Kinetic friction $f_{k}$ = $\mu_{k}$ N acts once motion begins, with $\mu_{k}$ < $\mu_{s}$ always.
Rolling friction $f_{r}$ = $\mu_{r}$ N, with $\mu_{r}$ much less than $\mu_{k}$.
The angle of repose on an inclined plane satisfies tan($\theta$) = $\mu_{s}$.

For circular motion dynamics, the centripetal force F = $mv^{2}$/r [$M^{1} \; L^{1} \; T^{-2}$] (N) must be provided by real forces (friction, tension, gravity, or normal force).
On a level road, friction provides centripetal force: $\mu$ mg = $mv^{2}$/r, giving $v_{max}$ = $\sqrt{\mu r g}$.
On a banked road with no friction: tan($\theta$) = $v^{2}$/(rg), where $\theta$ is the banking angle.

Solved Numerical 1: A 60 kg person in a lift accelerating upward at a = 2 m/$s^{2}$ (g = 10 m/$s^{2}$): Apparent weight W' = m(g + a) = 60(10 + 2) = 60 x 12 = 720 N. In free fall (a = g = 10 m/$s^{2}$): W' = m(g - g) = 60(0) = 0 N (weightlessness).

Solved Numerical 2: Atwood machine with $m_{1}$ = 5 kg, $m_{2}$ = 3 kg (g = 10 m/$s^{2}$).
Drawing FBD: For $m_{1}$: $m_{1}$ g - T = $m_{1}$ a, so 50 - T = 5a.
For $m_{2}$: T - $m_{2}$ g = $m_{2}$ a, so T - 30 = 3a.
Adding: 50 - 30 = (5 + 3)a, giving 20 = 8a, so a = 2.5 m/$s^{2}$.
Tension: T = $m_{2}$(g + a) = 3(10 + 2.5) = 37.5 N.
Check: $m_{2}$ g = 30 N < T = 37.5 N < $m_{1}$ g = 50 N.

Solved Numerical 3: Banked road, radius r = 100 m, banking angle $\theta$ = 30 degrees, no friction (g = 10 m/$s^{2}$).
From tan($\theta$) = $v^{2}$/(rg): $v^{2}$ = rg tan($\theta$) = 100 x 10 x tan 30 = 1000 x (1/$\sqrt{3}$) = 577.35 $\frac{m^{2}}{s^{2}}$.
v = $\sqrt{577.35}$ = 24.03 m/s = 86.5 km/h.

The key testable concept is free body diagram analysis with correct identification of action-reaction pairs and the self-adjusting nature of static friction ($f_{s}$ is NOT always equal to $\mu_{s}$ N).