Newton's third law, pulleys and connected particlesAQA A-Level Maths: Revision notes

Newton's third law: when body $A$ exerts a force on body $B$, body $B$ exerts an equal and opposite force on body $A$. The two forces act on different bodies, so they never cancel in the equation of motion of a single body. Examples: a tow bar pulling a trailer with $600$ N is pulled back by the trailer with $600$ N; a string with tension $T$ pulls each particle it joins with $T$ towards the string; a particle pressing on a surface with force $R$ is pushed by the surface with a normal reaction $R$.

For a particle in equilibrium, resolve forces in two perpendicular directions and set each resultant to zero (Newton's first law). For motion in a straight line, apply $F=ma$ along the line, with the resultant force on that particle only. Forces are restricted to two perpendicular directions, or simple 2D vector forces. For vectors, $\mathbf{F}_1+\mathbf{F}_2=m\mathbf{a}$, or $=\mathbf{0}$ in equilibrium.

Two particles joined by a light inextensible string move with the same speed and acceleration along the string, so long as it stays taut. A light string has negligible mass, so the tension is the same at both ends, and a light rod or tow bar behaves in the same way. Car ($1200$ kg) and trailer ($400$ kg), driving force $2400$ N, no resistance: for the whole system $a=\frac{2400}{1600}=1.5$ m s⁻². For the trailer alone, $T=400\times1.5=600$ N. You may treat the connected particles as one body to find $a$, but you must apply $F=ma$ to a single particle to find a tension.

A smooth pulley exerts no friction, so the tension in the string is the same on both sides, and it just changes the direction of the string. Hanging masses (Atwood): $A=3$ kg, $B=5$ kg. $B$: $5g-T=5a$; $A$: $T-3g=3a$; so $a=\frac{2g}{8}=2.45$ m s⁻² and $T=36.75$ N. Table and pulley: $P=4$ kg on a smooth table, $Q=6$ kg hanging: $T=4a$ and $6g-T=6a$, so $a=5.88$ m s⁻² and $T=23.5$ N. The force of the string on the pulley is the vector sum of the two tensions: $2T$ vertically for two vertical strings, or $T\sqrt2$ for perpendicular strings of equal tension.

On a smooth plane at $\alpha$ to the horizontal, the weight component along the plane is $mg\sin\alpha$. For $A=2$ kg on a $30^\circ$ plane joined to $B=3$ kg hanging: $T-2g\sin30^\circ=2a$ and $3g-T=3a$, so $a=\frac{2g}{5}=3.92$ m s⁻² and $T=17.6$ N. If $B$ lands after falling $1.2$ m, $v^2=2(3.92)(1.2)=9.408$. The string goes slack, and $A$ decelerates at $g\sin30^\circ=4.9$ m s⁻², travelling a further $\frac{9.408}{2(4.9)}=0.96$ m before stopping. Always re-form the equations of motion when the situation changes.