Newton's three laws of motion, weight, resolving forces, equilibrium of a particle, friction and the coefficient of friction, motion on an inclined plane, and connected particles.

What this dot point is asking

OCR wants you to apply Newton's three laws, use weight $W = mg$, resolve forces into perpendicular components, set up the equilibrium conditions for a particle, model friction with $F \le \mu R$ (and $F = \mu R$ when moving or on the point of moving), analyse motion on an inclined plane, and solve connected-particle problems such as masses over a pulley.

The answer

Newton's laws

Weight and resolving

Weight is the downward force $W = mg$. When a force acts at an angle, resolve it into perpendicular components, usually horizontal and vertical, or along and perpendicular to a slope.

Equilibrium of a particle

A particle is in equilibrium when the resultant force is zero, so the components balance in every direction. Resolving in two perpendicular directions and setting each sum to zero gives the equations.

Friction

Friction opposes motion (or attempted motion) along a surface, up to a maximum.

Examples in context

Newton's second law with friction

The standard method is: resolve perpendicular to the motion to find $R$, compute friction $\mu R$, then apply $F = ma$ along the motion.

Motion on an inclined plane

On a slope, resolve along and perpendicular to the surface. The weight contributes $mg\sin\theta$ down the slope and $mg\cos\theta$ into it (so $R = mg\cos\theta$ for a particle on the slope with no other perpendicular force).

Connected particles

For two masses connected by a light inextensible string over a smooth pulley, the tension is the same throughout and both share one acceleration. Write Newton's second law for each mass and add the equations to eliminate the tension.

Try this

Q1. A $2$ kg mass sits on a rough surface with $\mu = 0.3$. Find the maximum friction force ($g = 9.8$). [2 marks]

• Cue. $R = 2 \times 9.8 = 19.6$ N, so $F_{\max} = 0.3 \times 19.6 = 5.88$ N.

Q2. A $5$ kg block on a smooth slope inclined at $20^\circ$ is released. Find its acceleration ($g = 9.8$). [2 marks]

• Cue. $a = g\sin 20^\circ = 9.8 \times 0.342 \approx 3.35$ m s$^{-2}$.