Newton’s laws of motion

$F = ma$. Resultant force = mass × acceleration.

Method.

1. Identify all forces on body.
2. Find resultant (vector sum) in each direction.
3. Apply $F = ma$ in that direction.

Example. 4 kg on smooth surface, 20 N horizontal push. $F_{\text{net}} = 20$ → $20 = 4a$ → $a = 5$ m/s².

With friction. Subtract friction from applied force.

On incline. Resolve weight along/perp to slope.

Cambridge tip. Always sketch with forces labelled. Mark scheme rewards diagram.

Setup. Block mass $m$ on slope angle $\theta$, friction coefficient $\mu$.

Forces.

• Weight $W = mg$ vertically down.
• Normal reaction $R$ perpendicular to slope.
• Friction $F = \mu R$ along slope, opposing motion.
• Applied force (if any).

Components.

• Weight along slope (down): $mg\sin\theta$.
• Weight perpendicular: $mg\cos\theta = R$.

Sliding down (no other force). Net force along slope (taking down positive): $ma = mg\sin\theta - \mu R = mg\sin\theta - \mu mg\cos\theta$. $a = g(\sin\theta - \mu\cos\theta)$.

Cambridge tip. Take 'down slope' as positive when block slides down.

Setup. Two bodies connected by inextensible string. String passes over smooth pulley.

Key facts.

• Tension $T$ same throughout string.
• Both bodies have SAME magnitude of acceleration $a$.

Method.

1. Apply $F = ma$ to EACH body in its direction of motion.
2. Two equations, two unknowns ($T$ and $a$).

Example. 2 kg on smooth table, 3 kg hanging.

• Table mass: $T = 2a$.
• Hanging: $3g - T = 3a$.
• Add: $3g = 5a \Rightarrow a = 5.88$ m/s².
• $T = 11.76$ N.

Cambridge tip. State 'inextensible string → same $a$' and 'smooth pulley → same $T$'.