Newton's laws of motionEdexcel International A Level Maths: Revision notes

A force is a vector, measured in newtons (N), that can change the motion of an object. The resultant force is the vector sum of all the forces acting. Newton's first law: a particle remains at rest or moves with constant velocity unless acted on by a non-zero resultant force. So constant velocity (including rest) means the resultant force is zero, and the particle is in equilibrium. Common forces: weight $W=mg$ (downwards), normal reaction $R$ (perpendicular to a surface), tension $T$ (along a string, away from the body), thrust and resistance.

Newton's second law: the resultant force on a particle of constant mass is $F=ma$, in the direction of the acceleration. $m$ is in kg, $a$ in m s⁻², $F$ in N. Method:

• draw a force diagram for each particle,
• choose a positive direction (the direction of acceleration if known),
• write resultant force (forces in that direction minus forces against) $=ma$. For vertical motion $W=mg$ with $g=9.8$ m s⁻². For a particle on a smooth horizontal plane the weight and the normal reaction balance, so $R=mg$ if there is no vertical acceleration or other vertical force. Example: a $15$ kg crate pulled by a rope at $30^\circ$ with tension $60$ N against $20$ N resistance: $60\cos30^\circ-20=15a$ gives $a=2.13$ m s⁻².

Newton's third law: when body A exerts a force on body B, B exerts an equal and opposite force on A. The two forces act on different bodies, so they never cancel in a single equation of motion. Example: a passenger of mass $70$ kg in a lift accelerating upwards at $0.6$ m s⁻². The lift floor exerts reaction $R$ upwards on the passenger: $R-70g=70(0.6)$, so $R=728$ N. The passenger pushes down on the floor with the same $728$ N. For the lift and the passenger together (mass $470$ kg), the cable tension satisfies $T-470g=470(0.6)$, so $T=4888$ N. A light string has negligible mass and the tension is the same all along it.

Forces and accelerations can be written as $a\mathbf{i}+b\mathbf{j}$, where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors. Newton's second law holds as a vector equation: $\mathbf{F}=m\mathbf{a}.$ Add forces component by component to find the resultant. Divide by the mass to find the acceleration. Example: forces $(5\mathbf{i}-2\mathbf{j})$ N and $(-\mathbf{i}+8\mathbf{j})$ N act on a $3$ kg particle. $\mathbf{F}=4\mathbf{i}+6\mathbf{j}$ N, so $\mathbf{a}=\frac43\mathbf{i}+2\mathbf{j}$ m s⁻². The magnitude of $a\mathbf{i}+b\mathbf{j}$ is $\sqrt{a^2+b^2}$; its direction is found with $\tan^{-1}\frac{b}{a}$ and a sketch to place the angle.

When the force is constant, the acceleration is constant, so the constant acceleration formulae hold as vector equations: $\mathbf{v}=\mathbf{u}+\mathbf{a}t,\qquad \mathbf{r}=\mathbf{u}t+\tfrac12\mathbf{a}t^2.$ Here $\mathbf{r}$ is the displacement from the starting position. Speed is the magnitude of the velocity vector. Example: $\mathbf{u}=2\mathbf{i}+3\mathbf{j}$, $\mathbf{a}=6\mathbf{i}-4\mathbf{j}$, $t=4$. Then $\mathbf{v}=26\mathbf{i}-13\mathbf{j}$ and speed $\sqrt{26^2+13^2}=29.1$ m s⁻¹. Also $\mathbf{r}=56\mathbf{i}-20\mathbf{j}$, a distance of $59.5$ m from the start. For motion in a straight line, work in scalars with a chosen positive direction instead.

State and use the assumptions that make a problem solvable:

• particle: size ignored and rotation ignored,
• smooth: no friction,
• light string: negligible mass, same tension throughout,
• inextensible string: both ends have the same acceleration,
• no air resistance. If the question mentions a resistance, include it as a force opposing motion. If a force changes (for example a rope breaks), start a new equation of motion for the new situation.