Newton's Laws of Motion

The three laws, and what each one is for

Newton's first law says that a particle stays at rest, or carries on moving in a straight line at constant speed, unless a resultant force acts on it. In a question this is the licence to say 'the forces balance'. A car travelling at a steady speed on a level road has no resultant force, so its driving force equals the total resistance. A parcel hanging at rest on the end of a string has $T=mg$. Constant velocity does not mean no forces; it means the forces cancel.

Newton's second law says that the resultant force on a particle equals the rate of change of its momentum. For a particle of constant mass that is the equation $F=ma$: the resultant force in newtons equals the mass in kilograms times the acceleration in metres per second squared. This is the working equation of the whole topic. One newton is the force that gives a mass of 1 kg an acceleration of $1\mathrm{m}{\mathrm{s}}^{-2}$.

Newton's third law says that if body A exerts a force on body B, then B exerts a force of the same magnitude and opposite direction on A, and the two forces act on different bodies. Its use here is in connected particles: the tension in a tow-rope pulls the car backwards and the trailer forwards with the same magnitude, and that is what lets you write two equations and solve them together.

The syllabus sets every question in this topic inside one model: a particle of constant mass moving under constant forces. Constant forces give a constant acceleration, so once $F=ma$ has given you $a$, the constant-acceleration formulae from kinematics - $v=u+at$, $s=ut+\frac{1}{2}a{t}^2$, $v^2=u^2+2as$ - become available to finish the question.

• constant velocity: $a=0$, so the forces balance in every direction
• constant resultant force: constant acceleration, so $F=ma$ and the kinematics formulae work together
• third law: equal and opposite, but on two different bodies - never draw both arrows on the same diagram

$F=ma$ in one direction: the equation of motion

Choose the direction of the motion and state it as positive - 'taking the direction of motion as positive' is a line worth writing, because it fixes the sign of every force that follows. Then the equation of motion is: the sum of the components of the forces in that direction equals $ma$, where $a$ is the acceleration in that same direction. A negative value of $a$ means the particle is slowing down if it is moving in the positive direction, and you should say so in words.

The list of forces that can appear is short: the weight $mg$ acting vertically downwards; the normal reaction $R$ perpendicular to a surface and away from it; friction $F$ along a surface, opposing the tendency to slide; tension $T$ along a string, pulling; thrust along a rod, pushing; and the driving force and resistance of a vehicle. Identify them from the words of the question before you write anything down.

On a slope, work in two directions: parallel to the plane for the equation of motion and perpendicular to the plane for the reaction. The weight is the only force that needs resolving, and on a plane inclined at angle $\theta$ it splits into $mg\sin \theta$ down the plane and $mg\cos \theta$ perpendicular to the plane.

Never put the same force on the diagram twice, and never draw a force that the particle exerts on something else.

• state the positive direction before you resolve anything
• parallel to the plane: the equation of motion; perpendicular to the plane: the reaction
• a force at 90° to the motion contributes nothing to the equation of motion

Mass and weight

Mass is the amount of matter, measured in kilograms, and it does not change when the particle moves to another place. Weight is the gravitational force on the particle, measured in newtons, and it is given by $W=mg$. A parcel of mass 3 kg has weight $3\times 10=30N$.

The syllabus states that in this component questions are mainly numerical and that the approximate value 10 for $g$, in $\mathrm{m}{\mathrm{s}}^{-2}$, is expected. Take $g=10$ unless the question says otherwise, and write down the value you are using so the marker can follow your arithmetic.

A horizontal floor and a slope at 30° both have a weight of $mg$ on them; only the components differ.

A particle moving vertically

For motion in a vertical line the forces are the weight and whatever else the question supplies. The syllabus says that any force resisting the motion other than friction - air resistance, for instance - will be stated in the question, so if a resistance is given, it is acting opposite to the velocity.

Falling freely: taking downwards as positive, the only force is the weight, so $mg=ma$ and $a=g$. Every falling body has the same acceleration whatever its mass, which is why a feather and a hammer fall together in a vacuum.

Falling with air resistance: taking downwards as positive, $mg-F=ma$, so $a=g-\frac{F}{m}$, which is less than $g$ and depends on the mass. A heavy stone loses proportionally less of its acceleration to the same resistance than a light one does.

Thrown upwards: taking upwards as positive, the only force is the weight acting downwards, so $-mg=ma$ and $a=-g$. Without resistance the acceleration is the same on the way up and on the way down, so the particle passes any given height at the same speed in both directions. With air resistance the two differ: going up, the resistance acts downwards and adds to the deceleration; coming down, it acts upwards and reduces the acceleration.

A body in a lift is the same calculation seen from the floor. Taking upwards as positive, $R-mg=ma$ for a passenger of mass $m$, so a lift accelerating upwards gives $R>mg$ and one accelerating downwards gives $R<mg$.

• falling with no resistance: $a=g$
• falling with resistance $F$: $a=g-\frac{F}{m}$
• thrown upwards, no resistance: $a=-g$, so it slows at $g$
• lift accelerating upwards: $R=m(g+a)$; at constant speed $R=mg$

Motion on an inclined plane

Resolve parallel and perpendicular to the plane. Perpendicular to the plane the particle does not accelerate, so the components must balance: if nothing else has a component in that direction, $R=mg\cos \theta$. Parallel to the plane, $F=ma$ gives the equation of motion, and the component of the weight down the plane is $mg\sin \theta$.

Smooth plane: there is no friction, so the only force along the plane is the component of the weight. Then $mg\sin \theta =ma$, giving $a=g\sin \theta$, which does not contain the mass at all - a heavy block and a light one slide down a smooth slope with the same acceleration. A particle released from rest slides down; a particle projected up the plane decelerates at $g\sin \theta$, stops, and returns with the same acceleration.

Rough plane: friction acts along the plane, always opposing the motion. Going up the plane, friction acts down the plane, so the deceleration is $g\sin \theta +\mu g\cos \theta$. Coming down, friction acts up the plane, so the acceleration is $g\sin \theta -\mu g\cos \theta$. These two are different, and the syllabus singles this case out: the acceleration while moving up the plane differs from the acceleration while moving down it.

The method is always the same: find $R$ from the perpendicular direction, find the friction from $F=\mu R$, then write the parallel equation. Notice that $m$ cancels out of both accelerations, so the answers do not depend on the mass - useful as a check.

A particle resting on a rough plane stays at rest while $\tan \theta \le \mu$. Once it is moving, friction is at its limiting value $F=\mu R$.

• perpendicular to the plane: $R=mg\cos \theta$
• smooth plane, either direction: $a=g\sin \theta$
• rough plane, moving up: $a=g(\sin \theta +\mu \cos \theta )$, a deceleration
• rough plane, moving down: $a=g(\sin \theta -\mu \cos \theta )$

Connected particles over a pulley

Two particles joined by a light inextensible string passing over a smooth pulley is the standard connected-particles question, and each word carries information. Light means the string has no mass, so no force is needed to accelerate it and the tension is the same at every point. Inextensible means it does not stretch, so the two particles have equal speeds and equal accelerations in magnitude. Smooth means no friction at the pulley, so the tension does not change from one side to the other.

Write an equation of motion for each particle separately, taking the direction in which each one moves as positive for that particle. The tension pulls each particle towards the pulley, so it appears with opposite signs in the two equations. Adding the equations eliminates $T$ and gives the acceleration directly; substituting back gives the tension.

For masses $m_1$ and $m_2$ hanging on either side, with $m_1>m_2$, the equations are $m_{1}g-T=m_{1}a$ and $T-m_{2}g=m_{2}a$. Adding gives $a=\frac{(m_1-m_2)g}{m_1+m_2}$ and then $T=m_1(g-a)$. Notice that the heavier particle accelerates downwards, and that $T$ lies between $m_{2}g$ and $m_{1}g$: it is always less than the weight of the heavier particle and greater than the weight of the lighter one while the system is moving.

If $m_1=m_2$ the system does not move at all: $a=0$ and $T=mg$, which is the same answer you would get from the first law.

One particle on a table joined by a string over a pulley at the edge to a hanging particle is the same problem. For the hanging mass $m_{2}g-T=m_{2}a$, and for the particle on a smooth table $T=m_{1}a$, so $a=\frac{m_{2}g}{m_1+m_2}$ and $T=\frac{m_1{m}_{2}g}{m_1+m_2}$. A rough table adds a friction term $\mu R$ to the second equation.

• light: the tension is the same throughout the string
• inextensible: the two particles have equal accelerations in magnitude
• smooth pulley: no friction, so no change in tension across the pulley
• add the two equations to find $a$, substitute back to find $T$

Cars, trailers, ropes and tow-bars

A car towing a trailer is a pair of connected particles moving with the same acceleration along a straight horizontal road. The method is two steps: treat car and trailer as one system to find the acceleration, then treat one of them alone to find the force in the coupling.

As one system, the coupling forces are internal and cancel, so the driving force minus every resistance equals the total mass times the acceleration: $D-F_{\text{car}}-F_{\text{trailer}}=(m_{\text{car}}+m_{\text{trailer}})a$. That single equation is the fastest route to $a$.

Then take the trailer alone: the only horizontal forces on it are the force from the coupling and its own resistance, so $T-F_{\text{trailer}}=m_{\text{trailer}}a$. This gives the force in the coupling, and it is where the marks for a second equation are earned.

A light rope can only pull, so the force it exerts is a tension: forwards on the trailer and backwards on the car. A light rigid tow-bar can push as well as pull, so the force in it is a thrust when it is pushing the trailer forwards - the usual case for a car accelerating a trailer on a level road - and a tension when the car is braking or the trailer is pushing the car along downhill. State which one it is; the direction is part of the answer.

A light rigid tow-bar has no mass of its own, so its weight never appears in an equation; all it does is transmit a force along its length.

• whole system: driving force minus all resistances equals total mass times acceleration
• trailer alone: coupling force minus the trailer's resistance equals the trailer's mass times acceleration
• rope: tension only, and it can go slack; tow-bar: tension or thrust
• put each resistance on its own body - never twice

Choosing between $F=ma$ and the constant-acceleration formulae

A 9709 mechanics question usually needs both. $F=ma$ is what converts the forces into an acceleration; the constant-acceleration formulae then convert that acceleration into a speed, a distance or a time. The order is always the same: forces first, then kinematics.

So a question that asks for the speed of a particle after it has moved 5 m up a rough plane is not a kinematics question with a missing $a$: find $a$ from the equation of motion, then use $v^2=u^2+2as$ with that $a$. Remember that on the way up $a$ and $s$ have opposite signs if you take up the plane as positive.

Either direction is fine as long as you keep one sign convention from beginning to end.

• forces to acceleration: $F=ma$
• acceleration to speed, distance or time: the constant-acceleration formulae
• keep one positive direction for the whole question
• a slack string exerts no force, so the two particles stop being connected