Forces and equilibrium

Forces and force diagrams Paper 4

In Paper 4 almost every body is modelled as a particle: its size is ignored, so every force acts at one point and the body cannot turn. Before any calculation, decide which forces act. These are the ones you will meet.

• Weight, $W=mg$, acting vertically downwards. Paper 4 uses $g=10{\text{m s}}^{-2}$, so a $3\text{kg}$ particle has weight $30\text{N}$.
• Normal reaction, $R$, from a surface, at right angles to the surface and pushing away from it.
• Friction, $F$, from a rough surface, along the surface and opposing the motion, or the motion that would happen without it.
• Tension, $T$, in a string, pulling the particle along the string, away from the particle.
• Thrust in a rod, which can push as well as pull.
• Applied forces: a pull, a push, a driving force, a resistance. The question tells you about these.

Draw a force diagram every time: a dot for the particle and an arrow for each force, labelled with its size or a letter. A diagram is how you find the forces; it is not a way to measure them. The syllabus is clear that calculations are always required, not approximate solutions by scale drawing, so a force read off a ruler earns nothing.

A force diagram shows forces that act on the particle and nothing else. Velocity and acceleration are not forces, so they never appear as force arrows.

Force as a vector: components and resultants Paper 4

A force has a size (its magnitude, in newtons) and a direction, so it is a vector. Two vectors at an angle cannot be added like numbers; instead, split each force into components in two perpendicular directions and add those.

A force of $F$ newtons at an angle $\theta$ to a direction has

• a component $F\cos \theta$ along that direction, and
• a component $F\sin \theta$ at right angles to it.

This comes straight from the right-angled triangle with the force as its hypotenuse: the side next to $\theta$ is $F\cos \theta$ and the side opposite is $F\sin \theta$. A component pointing the negative way takes a minus sign.

To find the resultant of several forces, add their components: $X$ is the sum of the components in the $x$-direction and $Y$ the sum in the $y$-direction. Then

$$\text{magnitude}=\sqrt{X^2+Y^2},\text{angle with the}x\text{-direction}={\tan }^{-1}\frac{\left|Y\right|}{\left|X\right|}$$

Always sketch $X$ and $Y$ to see which way the resultant points. The calculator gives an acute angle; the sketch tells you which side of the axis it is on.

Equilibrium: resolving in two directions Paper 4

A particle is in equilibrium when the vector sum of the forces on it is zero. Splitting into components, this is the same as saying that the sum of the components in any direction is zero. In two dimensions, two perpendicular directions are enough: if the components balance in both, they balance in every direction.

So an equilibrium problem gives you two equations, and you can find two unknowns. The method:

1. Draw the force diagram.
2. Choose two perpendicular directions. Horizontal and vertical usually suit a hanging particle; along and perpendicular to the plane usually suit an inclined plane, because the weight is then the only force you need to split.
3. Resolve: write (sum of components one way) $=$ (sum of components the other way) for each direction.
4. Solve the pair of equations.

Solutions by resolving are what examiners expect. Other methods, such as a triangle of forces or Lami's theorem, are acceptable where they suit, but they are not required knowledge and questions will not refer to them. Resolving always works, so make it your method.

Contact forces: normal reaction and friction Paper 4

When two surfaces touch, each pushes on the other with a single contact force. It is usually easier to treat this force as two components:

• the normal component $R$, at right angles to the surface;
• the frictional component $F$, along the surface.

The total contact force has magnitude $\sqrt{R^2+F^2}$, but in almost every question you work with $R$ and $F$ separately.

A smooth contact is a model in which the frictional component is zero, so the contact force is entirely normal. It is a good model when friction is small compared with the other forces: ice, a polished floor, a well-oiled track, a pulley that turns freely. Its limitations: no real surface is perfectly smooth, so a smooth model predicts that a particle will slide when in fact friction might hold it, overestimates speeds and accelerations, and underestimates the force needed to pull something along. If friction matters to the answer, the question will say the surface is rough.

Friction, limiting equilibrium and the coefficient of friction Paper 4

Friction is not a fixed force. On a rough surface it takes whatever value is needed to stop sliding, up to a maximum. That maximum is $\mu R$, where $\mu$ is the coefficient of friction between the two surfaces and $R$ is the normal reaction. So

$$F⩽\mu R$$

• If the other forces need less than $\mu R$ to balance them, friction is exactly that smaller amount and the particle stays at rest.
• If they would need more, friction cannot supply it and the particle moves; while it slides, $F=\mu R$.
• When the particle is on the point of moving, friction has reached its maximum: $F=\mu R$. This is limiting friction, and the particle is in limiting equilibrium. Questions also say "about to slip" or "on the point of sliding" for this.

The coefficient of friction is defined by this relationship: in limiting friction, $\mu =\frac{F}{R}$. It has no units, and it depends only on the two surfaces.

Two habits avoid most errors. First, find $R$ by resolving perpendicular to the surface; $R=mg$ only on a horizontal surface with no other vertical forces. A pull upwards at an angle reduces $R$; a push downwards increases it. Second, decide the direction of friction from the way the particle would move if the surface were smooth: friction acts the opposite way.

Newton's third law Paper 4

When body A exerts a force on body B, body B exerts a force on body A that is equal in size and opposite in direction. For example, the force exerted by a particle on the ground is equal and opposite to the force exerted by the ground on the particle: a crate pressing down on the floor with $400\text{N}$ is pushed up by the floor with $400\text{N}$.

The two forces of a pair act on different bodies, so they never appear on the same force diagram and never cancel each other. This matters as soon as two bodies touch:

• A book on a box on the floor: the book pushes down on the box with a force equal to the reaction of the box on the book. On the box's diagram, that downward push from the book appears alongside the box's own weight.
• Friction obeys the law too: if the ground exerts a frictional force on a sliding block in one direction, the block exerts an equal frictional force on the ground in the other.
• In a string or rod joining two bodies, the force on one body is equal and opposite to the force on the other.

Draw a separate diagram for each body, and use the third law to link them.