Newton's laws of motionAQA A-Level Physics: Revision notes

Newton's first law: an object remains at rest or moves with constant velocity unless acted on by a resultant force. Constant velocity means both speed and direction are unchanged.

If the resultant force is zero the object is in equilibrium: it is stationary or moving at constant velocity. A car at a steady speed has a driving force equal to the resistive forces. The first law also defines inertia, the reluctance of an object to change its velocity; the greater the mass, the greater the inertia.

Newton's second law: the resultant force on an object is proportional to the rate of change of its momentum and acts in the direction of that change. For constant mass this becomes

$F = ma$

where $F$ is the resultant force in newtons, $m$ the mass in kg and $a$ the acceleration in m s⁻². One newton is the force that gives a mass of 1 kg an acceleration of 1 m s⁻². Force and acceleration are both vectors with the same direction.

Always find the resultant first: for the car with driving force 3000 N and resistive force 800 N, $F = 2200$ N, so $a = 2200 \div 1200 = 1.8$ m s⁻².

Newton's third law: when body A exerts a force on body B, body B exerts a force on body A that is equal in magnitude, opposite in direction and of the same type.

The two forces of a third-law pair always act on different bodies, so they never cancel each other when you consider the motion of one body. For a person standing on the ground, the weight (Earth pulling the person) and the contact force (ground pushing the person) are not a third-law pair: they act on the same body and are different types of force. The pair of the weight is the gravitational pull of the person on the Earth.

A free-body diagram shows a single object isolated from its surroundings with every force acting on it drawn as an arrow from the object, labelled and pointing in the correct direction. Typical forces are weight (down), normal contact force (perpendicular to the surface), friction or drag (opposing motion), tension and driving force.

On a slope, resolve the weight into a component down the slope, $mg\sin\theta$, and a component perpendicular to the slope, $mg\cos\theta$. The perpendicular component balances the normal contact force, so the resultant force down the slope is $mg\sin\theta - F_{friction}$.

Slope. A crate of mass 8.0 kg slides down a 25° slope with friction 12 N. Down-slope weight component: $8.0\times9.81\times\sin25^\circ = 33.2$ N. Resultant: $33.2-12=21.2$ N. Acceleration: $21.2\div8.0=2.6$ m s⁻².

Lift. A person of mass 70 kg in a lift accelerating upwards at 1.5 m s⁻²: $R - mg = ma$, so $R = 70(9.81+1.5) = 792$ N. At constant velocity $R = mg = 687$ N.

Connected bodies. For a truck and trailer, find the acceleration of the whole system using the external forces, then apply $F = ma$ to one body to find the tension, which is internal to the system.