Mechanics

Mechanics uses mathematics to describe motion and explain how forces change it. A successful solution has three layers:

1. Model: decide which features of the real situation to include.
2. Mathematics: translate the model into equations, graphs or vectors.
3. Interpretation: attach units, directions and physical meaning to the result.

The algebra is often short. The difficult part is deciding what each quantity means, which direction is positive, and which equation expresses the physics.

Prerequisites

You should be able to:

• rearrange formulae and solve linear and quadratic equations;
• use trigonometry in right angled triangles;
• interpret gradients and areas on graphs;
• work with vectors and resolve them into perpendicular components;
• differentiate and integrate polynomials for later kinematics work.

If any of these are insecure, revise them alongside the mechanics rather than waiting to begin.

The mechanics problem solving cycle

Use this cycle for almost every question.

1. Define the system and model

Identify the object or collection of objects being studied. Record assumptions such as:

• a particle has mass but negligible size;
• a light string has negligible mass;
• a smooth surface or pulley has no friction;
• a rigid body does not deform;
• uniform gravity gives constant acceleration $g$, usually taken as $9.8\ \mathrm{m\,s^{-2}}$ unless told otherwise.

These are mathematical idealisations, not claims that the real object literally has no size or that a real pulley has no friction. Learn how assumptions affect equations in modelling in mechanics.

2. Choose axes and define signs

State a positive direction before assigning signs. In one dimension, a negative velocity means motion opposite to the positive direction. A negative acceleration means acceleration opposite to the positive direction. It does not automatically mean slowing down.

An object slows down when velocity and acceleration have opposite signs:

Velocity	Acceleration	Effect on speed
same sign	same direction as motion	increasing
opposite signs	opposite direction to motion	decreasing

3. Draw and label

For motion, sketch the path or the relevant graph. For forces, isolate one body and draw every external force acting on it. This is a free body diagram. Never include velocity as a force.

4. Select a principle

Common choices are:

$$\text{constant acceleration equations},\qquad \mathbf F=m\mathbf a,$$ $$\sum \mathbf F=\mathbf 0 \quad \text{for equilibrium},$$

or

$$\text{moment}=\text{force}\times\text{perpendicular distance}.$$

5. Solve, then check

Keep exact values until the final line where practical. Check dimensions, signs and scale. A time should not normally be negative, and a force measured in metres signals that units have been mixed.

Core quantities and units

Quantity	Typical symbol	SI unit	Scalar or vector?
mass	$m$	$\mathrm{kg}$	scalar
time	$t$	$\mathrm s$	scalar
displacement	$s$ or $\mathbf r$	$\mathrm m$	vector
velocity	$v$ or $\mathbf v$	$\mathrm{m\,s^{-1}}$	vector
acceleration	$a$ or $\mathbf a$	$\mathrm{m\,s^{-2}}$	vector
force	$F$ or $\mathbf F$	$\mathrm N$	vector
moment	$M$	$\mathrm{N\,m}$	has a turning sense

One newton is defined by

$$1\ \mathrm N=1\ \mathrm{kg\,m\,s^{-2}}.$$

This identity is a useful equation check. See quantities and units for conversions and dimensional reasoning.

Kinematics: describing motion

Kinematics describes motion without asking what causes it. The fundamental distinctions are:

• distance is total path length, while displacement is change in position;
• speed is the magnitude of velocity;
• acceleration is the rate of change of velocity, not simply the rate of increase of speed.

For motion along a line,

$$v=\frac{\mathrm ds}{\mathrm dt}, \qquad a=\frac{\mathrm dv}{\mathrm dt} =\frac{\mathrm d^2s}{\mathrm dt^2}.$$

When acceleration is constant, these relationships lead to the SUVAT equations, including

$$v=u+at, \qquad s=ut+\frac12at^2, \qquad v^2=u^2+2as.$$

They apply only over an interval where $a$ is constant.

Worked example 1: signs in vertical motion

A ball is projected vertically upwards at $14\ \mathrm{m\,s^{-1}}$. Find the time taken to reach its highest point and its maximum displacement above the launch point. Ignore air resistance and use $g=9.8\ \mathrm{m\,s^{-2}}$.

Choose upwards as positive. Then

$$u=14,qquad a=-9.8.$$

At the highest point the instantaneous velocity is $v=0$. From $v=u+at$,

$$0=14-9.8t \quad\Longrightarrow\quad t=\frac{14}{9.8}=\frac{10}{7}\ \mathrm s.$$

Now use $v^2=u^2+2as$:

$$0^2=14^2+2(-9.8)s,$$

so

$$s=\frac{196}{19.6}=10\ \mathrm m.$$

The ball reaches its highest point after $\boxed{1.43\ \mathrm s}$ at a height of $\boxed{10\ \mathrm m}$ above the launch point.

At the highest point

The velocity is momentarily zero, but the acceleration is still $-9.8\ \mathrm{m\,s^{-2}}$. Zero velocity does not imply zero acceleration.

Develop this strand through kinematics language, constant acceleration, kinematics graphs and calculus in kinematics.

Dynamics: explaining motion

Dynamics relates force to acceleration. Newton’s second law is a vector equation:

$$\sum \mathbf F=m\mathbf a.$$

The left side is the resultant external force, not an individual force. Resolve the equation along convenient perpendicular axes. If acceleration is zero, the resultant force is zero, but the individual forces need not be zero.

Common forces include:

• weight $mg$, vertically downwards;
• normal reaction, perpendicular to a contact surface;
• tension, pulling along a taut string;
• friction, opposing actual or impending relative motion;
• thrust, resistance and drag.

Mass is measured in kilograms. Weight is a force measured in newtons:

$$W=mg.$$

Worked example 2: force on an inclined plane

A $5\ \mathrm{kg}$ particle slides down a smooth plane inclined at $30^\circ$ to the horizontal. Find its acceleration and the normal reaction.

The forces are weight $5g$ vertically downwards and reaction $R$ perpendicular to the plane. Resolve weight parallel and perpendicular to the plane.

Down the plane:

$$5g\sin 30^\circ=5a.$$

Therefore

$$a=g\sin30^\circ=9.8\left(\frac12\right) =\boxed{4.9\ \mathrm{m\,s^{-2}}}.$$

Perpendicular to the plane there is no acceleration, so

$$R-5g\cos30^\circ=0.$$

Hence

$$R=49\left(\frac{\sqrt3}{2}\right) =\boxed{\frac{49\sqrt3}{2}\ \mathrm N} \approx\boxed{42.4\ \mathrm N}.$$

Notice that the reaction is not automatically equal to $mg$. It balances only the component of weight perpendicular to the plane.

Study forces and free body diagrams, resolving forces and equilibrium, Newton’s laws and friction before tackling connected particles and dynamics in a plane.

Moments and rigid bodies

A force can produce translation and rotation. The magnitude of the moment of a force about a point is

$$M=F d_{\perp},$$

where $d_{\perp}$ is the perpendicular distance from the point to the force’s line of action. For a rigid body in equilibrium,

$$\sum \mathbf F=\mathbf 0 \quad\text{and}\quad \sum M=0.$$

Worked example 3: an equilibrium beam

A uniform horizontal beam of length $4\ \mathrm m$ and weight $120\ \mathrm N$ is supported at both ends, $A$ and $B$. A load of $180\ \mathrm N$ is placed $1\ \mathrm m$ from $A$. Find the upward reactions $R_A$ and $R_B$.

The beam’s weight acts at its midpoint, $2\ \mathrm m$ from $A$. Take moments about $A$ so that $R_A$ has zero moment:

$$4R_B=120(2)+180(1)=420.$$

Thus

$$R_B=105\ \mathrm N.$$

Vertical equilibrium gives

$$R_A+R_B-120-180=0,$$

so

$$R_A=300-105=195\ \mathrm N.$$

Therefore $\boxed{R_A=195\ \mathrm N}$ and $\boxed{R_B=105\ \mathrm N}$. The larger reaction is sensibly nearer the added load. Continue with moments.

Two dimensional motion and projectiles

Perpendicular components can be analysed independently. For a projectile with initial speed $U$ at angle $\theta$ above the horizontal, ignoring air resistance,

$$u_x=U\cos\theta, \qquad u_y=U\sin\theta,$$

and

$$a_x=0, \qquad a_y=-g.$$

The horizontal and vertical motions share the same time $t$. This shared time links two otherwise independent SUVAT calculations. Build the required vector ideas in two dimensional motion before studying projectiles.

Common misconceptions

• Negative means slowing down. False. Compare the signs of velocity and acceleration.
• Constant speed means zero acceleration. Only in straight line motion. Changing direction changes velocity.
• The reaction always equals weight. Only when the perpendicular force balance makes it so.
• Friction is always $\mu R$. In equilibrium, friction adjusts up to a limiting value. The equation $F=\mu R$ applies at limiting equilibrium or in the model of sliding friction.
• Newton’s third law pairs cancel. The paired forces act on different bodies, so they do not cancel on one free body diagram.
• Every force has a moment $Fd$. The distance must be perpendicular to the force’s line of action.
• SUVAT always applies. It requires constant acceleration within the chosen stage.

Self-check

1. Direction and acceleration

A particle has velocity $v=-6\ \mathrm{m\,s^{-1}}$ and acceleration $a=-2\ \mathrm{m\,s^{-2}}$. Is it speeding up or slowing down?

Answer

It is speeding up. Velocity and acceleration have the same sign, so acceleration acts in the direction of motion. After one second, its velocity is $-8\ \mathrm{m\,s^{-1}}$, whose speed is $8\ \mathrm{m\,s^{-1}}$.

2. Equilibrium

A $3\ \mathrm{kg}$ lamp hangs at rest from a vertical cable. Find the tension, using $g=9.8\ \mathrm{m\,s^{-2}}$.

Answer

The acceleration is zero, so the upward tension balances the downward weight:

$$T-3g=0 \quad\Longrightarrow\quad T=29.4\ \mathrm N.$$

3. Modelling judgement

A question models a car as a particle. Which feature is ignored, and which properties may still be included?

Answer

Its dimensions and rotational effects are ignored. It may still have mass, position, velocity, acceleration and forces acting on it.

4. Choosing a method

A particle moves in a straight line with $v(t)=3t^2-12t+9$. What method finds its displacement between $t=1$ and $t=4$? What extra step is needed to find total distance?

Answer

Displacement is the signed integral

$$\int_1^4 v(t)\,\mathrm dt.$$

For total distance, first solve $v(t)=0$ and split the integral at every change of direction in $1\leq t\leq4$. Add the magnitudes of the resulting displacements, equivalently integrate $|v(t)|$ piecewise.

Recommended study order

Start with language and modelling, then separate the subject into motion and forces:

1. Quantities and units
2. Modelling in mechanics
3. Kinematics language
4. Constant acceleration and kinematics graphs
5. Calculus in kinematics
6. Forces and free body diagrams
7. Resolving forces and equilibrium, Newton’s laws and friction
8. Moments
9. Two dimensional motion and projectiles
10. Connected particles and dynamics in a plane

At every stage, practise translating words into a diagram before reaching for a formula. That habit unifies the entire mechanics course.