Newton's third law and connected particlesEdexcel A-Level Maths: Revision notes

Newton's third law: when body $A$ exerts a force on body $B$, $B$ exerts a force on $A$ that is equal in magnitude, opposite in direction and of the same type. The two forces act on different bodies, so they never cancel in the equation of motion of one body. A book on a table: the table pushes up on the book with $R$, and the book pushes down on the table with $R$. These are a third-law pair. The weight of the book (pulled down by Earth) is paired with the book's gravitational pull up on Earth, not with $R$. At a taut string, the tension pulls inwards on each end with the same magnitude. In a lift, the force the person exerts on the floor equals the reaction of the floor on the person.

For a particle, the resultant force $F=ma$ in each direction. Choose a positive direction (the direction of motion or acceleration) and write one equation for each particle. Equilibrium: the acceleration is zero, so the resultant force is zero in every direction. This includes a particle moving at constant velocity, not just one at rest. Forces in two perpendicular directions are treated separately. Use $g=9.8\ \text{m s}^{-2}$ unless told otherwise, and give answers to 2 or 3 significant figures.

A person of mass $m$ stands on the floor of a lift. The forces on the person are the weight $mg$ down and the normal reaction $R$ up. Accelerating upwards at $a$: $R-mg=ma$, so $R=m(g+a)$ and the person feels heavier. Accelerating downwards at $a$: $mg-R=ma$, so $R=m(g-a)$ and the person feels lighter. Constant speed: $R=mg$. For the cable tension $T$, treat lift and person together: $T-(M+m)g=(M+m)a$. Example: lift 250 kg, person 70 kg, $a=1.5$ upwards: $R=70(11.3)=791$ N and $T=320(11.3)=3616$ N.

Particles joined by a light inextensible string have the same acceleration (inextensible) and the string has the same tension throughout, even over a smooth pulley (light string, smooth pulley). Write $F=ma$ for each particle, then add or substitute to eliminate $T$. Hanging particles $m_1<m_2$ over a smooth pulley: $m_2g-T=m_2a$ and $T-m_1g=m_1a$, so $a=\frac{(m_2-m_1)g}{m_1+m_2}$. For 3 kg and 5 kg: $a=\frac{2(9.8)}{8}=2.45\ \text{m s}^{-2}$ and $T=3(9.8+2.45)=36.75$ N. Particle on a smooth table joined to a hanging particle: the table particle has $T=m_Aa$ and the hanging one $m_Bg-T=m_Ba$.

Two bodies in contact (for example blocks pushed along a surface) have the same acceleration. The contact force is an action-reaction pair: equal and opposite on the two blocks. For the whole system treat them as one body of total mass, then apply $F=ma$ to one block to find the contact force. Example: 30 N on $P$ (4 kg) pushing $Q$ (6 kg), smooth surface: $a=\frac{30}{10}=3$ and the contact force is $6\times3=18$ N. With resistances 5 N on $P$ and 7 N on $Q$: $a=\frac{18}{10}=1.8$ and $R-7=6(1.8)$, $R=17.8$ N. When a hanging particle lands, its tension disappears. The string becomes slack, so the other particle moves at constant velocity (smooth table) or decelerates under the other forces (rough table, slope). Use the suvat equations in each stage, with the final velocity of one stage as the start of the next.