Use Newton's laws to resolve forces and form the equation of motion: weight, tension, normal and resistive forces, equilibrium of concurrent forces, and the conical pendulum

What this dot point is asking

NESA wants you to take the forces acting on a body, weight, tension, the normal reaction from a surface and any resistive force, and use Newton's laws to build the equation of motion. The skill is resolving each force into components, adding them to get the resultant, and then writing $F = ma$ in each direction. Once the equation of motion is set up, the calculus of the rest of the mechanics module (the velocity and acceleration functions, resisted motion, simple harmonic motion) takes over. Throughout this page, take $g = 9.8$ m/s$^2$.

Force as a vector

A force has both a magnitude (in newtons, where $1$ N $= 1$ kg$\cdot$m/s$^2$) and a direction, so it is a vector. This single fact drives the whole topic. When several forces act on a body you cannot just add their magnitudes; you must add them as vectors, which in practice means adding their components. The vector sum of all the forces on a body is called the resultant force, and it is the resultant, not any individual force, that determines how the body accelerates.

Because forces are vectors, two questions become routine. First, given several forces, what is their resultant (magnitude and direction)? Second, given one force at an angle, what are its components along directions we care about? Both are answered with the same right-angled-triangle trigonometry used for vectors elsewhere in the course.

Newton's three laws

Newton's laws, stated in his Principia of 1687, are the foundation of the whole module.

• First law (inertia). A body stays at rest, or moves in a straight line at constant velocity, unless a resultant force acts on it. So constant velocity, including being at rest, means zero resultant force. This is exactly the condition for equilibrium.
• Second law. The resultant force is proportional to the rate of change of momentum, and acts in the direction of that change. With mass constant and SI units, the constant of proportionality is $1$, so $\vec{F} = m\vec{a}$. This is the law you compute with.
• Third law. Every action has an equal and opposite reaction: if body $A$ pushes on body $B$, then $B$ pushes back on $A$ with an equal force in the opposite direction. The normal reaction from a surface is a third-law pair with the push of the body on the surface.

The forces you will meet

Four kinds of force account for almost every Extension 2 problem.

• Weight. Gravity pulls every mass straight down with a force of magnitude $mg$, where $g = 9.8$ m/s$^2$. Weight always points vertically downward, regardless of any slope, so on an inclined plane it must be resolved into components along and across the slope.
• Tension. A taut string, rope or wire pulls along its own length, away from the body, with a force called the tension $T$. A light (massless) string has the same tension throughout, and an ideal pulley simply changes the direction of that tension.
• Normal reaction. A surface pushes on a body with a force perpendicular ("normal") to the surface, called the normal reaction $N$. It is whatever size is needed to stop the body sinking into the surface, so it is found from the equation of motion, not assumed equal to the weight. On a horizontal table with nothing else pushing vertically, $N = mg$; but a downward push raises $N$ and an upward pull lowers it, as the practice questions show.
• Resistive force. Friction and air resistance oppose the motion (or the tendency to move). Friction on a rough surface has magnitude up to $\mu N$, where $\mu$ is the coefficient of friction; a drag force from a fluid is often modelled as proportional to $v$ or to $v^2$. A resistive force is what links this page to the resisted-motion and projectile-with-resistance dot points: once you have written $ma = (\text{driving force}) - (\text{resistance})$, you switch to the calculus forms of acceleration to solve it.

Resolving a force into components

To resolve a force $F$ acting at an angle $\theta$ to a chosen axis, drop a right-angled triangle onto that axis. The component along the axis is $F\cos\theta$ and the component perpendicular to it is $F\sin\theta$. The cosine always goes with the direction the angle is measured from. Getting this right is the single most common place to lose marks: if the angle is measured from the vertical instead of the horizontal, the sine and cosine swap.

For a body on a slope inclined at angle $\alpha$ to the horizontal, the weight $mg$ (which points straight down) resolves into $mg\sin\alpha$ down the slope and $mg\cos\alpha$ into the slope. The $\sin$ goes with the down-slope direction because the angle between the weight and the into-slope normal is $\alpha$; a quick way to remember it is that a steeper slope (larger $\alpha$) gives a larger down-slope pull, and $\sin\alpha$ grows with $\alpha$ while $\cos\alpha$ shrinks.

The resultant of concurrent forces

Forces are concurrent when their lines of action all pass through one point, so they can be treated as acting at a single particle. To find their resultant, resolve every force into the same two perpendicular directions, add the components separately to get the totals $\sum F_x$ and $\sum F_y$, then recombine:

where $\phi$ is the angle of the resultant to the $x$-axis. The acceleration then follows from $\vec{a} = \vec{R}/m$, in the same direction as $\vec{R}$. This component method never fails, even with three or more forces at awkward angles, and it is far more reliable in an exam than trying to draw a scale diagram.

Equilibrium: when the resultant is zero

A body is in equilibrium when the resultant force on it is zero, so by the first law it is at rest or moving with constant velocity. Zero resultant means the components balance in every direction at once:

These two scalar equations let you find two unknowns, typically the tensions in two supporting cords, or one tension and the normal reaction. There is a neat geometric picture too: if only three forces act and they are in equilibrium, then drawn tip-to-tail they form a closed triangle (their vector sum returns to the start), so you can also solve a three-force equilibrium with the sine rule or cosine rule on the triangle of forces. The diagram below shows that closed triangle for a hanging sign.

Building the equation of motion from a force diagram

The procedure is always the same, and it pays to make it mechanical so you never freeze on a new context.

1. Draw the free-body diagram. Every force as a labelled arrow from the body, and nothing else (no velocities, no made-up forces).
2. Choose axes. Pick perpendicular directions that simplify the problem. Put one axis along the acceleration if you can.
3. Resolve every force onto those axes.
4. Write $\sum F = ma$ in each direction. Use $a = 0$ in any direction with no acceleration (this is where the normal reaction usually drops out).
5. Solve for the unknown, whether that is an acceleration, a tension, or a normal reaction.

The worked example "Acceleration of a block sliding down a rough slope" below runs this procedure end to end. The four panels build the same calculation one stage at a time: the picture, the free-body diagram, the resolution of the weight, and finally the equation of motion that gives the acceleration $a = 3.20$ m/s$^2$.

Stage 1, draw the situation. A $10$ kg block sits on a rough plane inclined at $30^\circ$ and is about to slide down. Nothing is drawn yet except the geometry and the mass; the angle of the slope is the key piece of information.

Stage 2, draw the free-body diagram. Three forces act on the block: its weight $mg$ straight down, the normal reaction $N$ perpendicular to and away from the slope, and the friction $F$ up the slope, opposing the downhill motion the block is about to make. No other force exists, so these three are all that go into the equation of motion.

Stage 3, choose axes and resolve. Take axes along and across the slope, because the block accelerates along the slope and the normal reaction lies across it. The weight is the only force not already on an axis, so resolve it: $mg\sin 30^\circ$ down the slope and $mg\cos 30^\circ$ into the slope. Across the slope there is no acceleration, so $N = mg\cos 30^\circ$, and the friction is $F = \mu N$.

Stage 4, write the equation of motion. Along the slope, taking down-slope as positive, Newton's second law gives $ma = mg\sin 30^\circ - \mu N = mg\sin 30^\circ - \mu\,mg\cos 30^\circ$. The mass cancels, leaving $a = g(\sin 30^\circ - \mu\cos 30^\circ)$. With $\mu = 0.2$ this is $a = 9.8(\sin 30^\circ - 0.2\cos 30^\circ) = 3.20$ m/s$^2$ down the slope.

The conical pendulum

A conical pendulum is a mass on a string, swung so that it travels in a horizontal circle at constant speed while the string sweeps out a cone. It is the classic test of resolving forces, because the tension does two different jobs at the same time, and it ties Newton's laws to circular motion.

Two forces act on the ball: the tension $T$ along the string toward the pivot, and the weight $mg$ straight down. There is no third force; in particular, the centripetal force is not a separate force but is provided by the horizontal part of the tension. Let the string make a constant angle $\theta$ with the vertical, and let $r$ be the radius of the circle. The ball does not move up or down, so the vertical forces balance; the ball does accelerate horizontally toward the centre (centripetal acceleration $\dfrac{v^2}{r}$), so the horizontal forces give that. Resolving:

Dividing the second equation by the first eliminates both $T$ and $m$ and gives the clean result

so $v = \sqrt{gr\tan\theta}$. The free-body diagram below shows the tension resolved into its vertical part $T\cos\theta$ (which holds the weight) and its horizontal part $T\sin\theta$ (which turns the ball).

From the same two equations you can read off everything else. With $r = L\sin\theta$ for a string of length $L$, the period is

a strikingly simple result: the period of a conical pendulum depends only on the height of the cone $L\cos\theta$, not on the mass or the speed. As the cone opens out ($\theta \to 90^\circ$) the tension $T = mg/\cos\theta$ grows without bound, which is why a string cannot be swung perfectly horizontally.

How exam questions ask about forces and Newton's laws

The wording tells you which part of the method to reach for.

• "A force $F$ acts on a mass $m$. Find the acceleration." Straight $F = ma$; if $F$ is a function of $t$, $x$ or $v$, this is the bridge to the velocity-and-acceleration-functions dot point.
• "Several forces act ... find the resultant / the acceleration." Resolve every force into components, add them, recombine with $R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$ and $\tan\phi = \sum F_y / \sum F_x$, then $a = R/m$.
• "Show that the body is in equilibrium" / "find the tensions." Set $\sum F_x = 0$ and $\sum F_y = 0$ and solve the two equations; for three forces you may instead use a triangle of forces.
• "A block on a plane inclined at $\alpha$ ..." Resolve along and across the slope: weight gives $mg\sin\alpha$ down-slope and $mg\cos\alpha$ into the slope; $N = mg\cos\alpha$ (plus any across-slope part of other forces); friction is $\mu N$.
• "Find the normal reaction" / "the apparent weight in a lift." Write the vertical equation of motion; the normal reaction is $m(g+a)$ when accelerating up and $m(g-a)$ when accelerating down, not simply $mg$.
• "A conical pendulum / mass in a horizontal circle ..." Resolve the tension: $T\cos\theta = mg$ vertically and $T\sin\theta = mv^2/r$ horizontally, then divide to get $\tan\theta = v^2/(gr)$.
• "Hence, find the velocity / position ..." Once the equation of motion is set up, switch to the calculus forms of acceleration (the velocity-and-acceleration-functions dot point) and integrate.

Why this matters for the rest of the module

Every later mechanics dot point begins where this one ends: with an equation of motion. Resisted motion is just $ma = -(\text{resistance})$, with the resistance proportional to $v$ or $v^2$; vertical resisted motion adds the weight, $ma = mg - kv$; projectile motion resolves the weight (and any resistance) into horizontal and vertical equations; and simple harmonic motion is the case where the resultant force is a restoring force proportional to displacement, $ma = -kx$. Mastering the step from a force diagram to $F = ma$ is therefore the foundation the whole module is built on, and it is why this material is worth getting fluent before the calculus takes over.