Mechanics (alternative to Unit 20)

Mechanics is the part of mathematics that describes how bodies push, pull and move. It is the oldest application of the algebra and calculus you have already learned, and it is the reason those subjects were invented. In this unit a force is a vector, so every rule from Unit 7 (Vectors) applies unchanged: forces add by the parallelogram law, they resolve into components, and a body stays still only when the vector sum of the forces on it is zero. Motion is described by three quantities — displacement, velocity, acceleration — and velocity is the derivative of displacement, acceleration the derivative of velocity, exactly as in Unit 18.

Mechanics splits into two halves, and this unit follows that split.

• Statics studies bodies at rest. The question is always: what combination of forces leaves the body in equilibrium?
• Dynamics studies bodies in motion. The question is always: given the forces (or the acceleration they produce), where is the body and how fast is it going?

Throughout, forces are measured in newtons (N), lengths in metres (m), time in seconds (s), and the acceleration due to gravity is taken as g = 9.8\ \text{m/s}^2 unless a question says otherwise. Some board questions use g = 10\ \text{m/s}^2 to keep the arithmetic short; always use the value printed in the question.

21.1 Statics: forces and resultant forces; parallelogram law of forces

What a force is

A force is a push or a pull. To describe it completely you need three things: its magnitude (how hard, in newtons), its direction, and its point of application. Magnitude and direction together make a force a vector, so we write it as \vec{F} and draw it as an arrow whose length is proportional to the magnitude.

Because forces are vectors, two forces acting at the same point can be replaced by a single force with the same effect. That single force is the resultant. Conversely, any force can be replaced by two or more forces whose resultant is the original; those are its components. Replacing several forces by one is composition; replacing one by several is resolution. Both are used constantly, and §21.2 treats them in detail.

Forces acting at a single point are called concurrent; forces all lying in one plane are coplanar. Everything in this unit deals with coplanar concurrent forces, which is what the syllabus requires.

The parallelogram law of forces

Two forces acting at a point are combined by the same construction used for vectors.

Put the law into symbols. Let two forces of magnitudes P and Q act at the point O with angle \theta between them, and let the resultant R make angle \alpha with P.

Drop a perpendicular from the tip of the resultant to the line of P extended. The horizontal side of the right triangle so formed is P + Q\cos\theta and the vertical side is Q\sin\theta. Pythagoras and the tangent ratio give the two working formulas.

Three special cases are worth memorising because they appear in multiple-choice questions.

Angle between the forces	Resultant R	Direction
\theta = 0^\circ (same direction)	P + Q — the maximum possible	along both forces
\theta = 90^\circ (perpendicular)	\sqrt{P^2+Q^2}	\tan\alpha = Q/P
\theta = 180^\circ (opposite)	|P - Q| — the minimum possible	along the larger force

So the resultant of two forces always lies between |P-Q| and P+Q. If a question claims two forces of 3 N and 4 N have a resultant of 9 N, it is wrong: the largest possible resultant is 7 N.

Triangle law and Lami's theorem

The parallelogram law has an equivalent form that is often quicker to use.

Turning that around gives the condition for three forces to be in equilibrium: if three concurrent forces are in equilibrium, they can be represented by the three sides of a triangle taken in order. Applying the sine rule to that triangle gives a result that solves most statics problems in one line.

21.2 Composition and resolution of forces; resultant of coplanar forces acting on a point

Resolving a force into components

Composition replaces many forces by one. Resolution is the reverse operation and is the more useful of the two, because once every force is written in components the whole problem becomes arithmetic with no geometry at all.

Two remarks make resolution safe in the examination.

• The direction matters for the sign. A force pointing left has a negative x-component; one pointing down has a negative y-component. Choose east and north as positive once, at the start, and keep that choice.
• You may resolve along any two perpendicular directions, not only horizontal and vertical. On an inclined plane it is almost always better to resolve along and perpendicular to the plane, as §21.4 does.

Resultant of any number of coplanar concurrent forces

Because components simply add, the method extends to as many forces as you like.

21.3 Dynamics: motion of a particle in a straight line; motion with uniform acceleration

The three quantities, and how calculus links them

A particle is a body small enough that its size can be ignored. Restrict it to a straight line and take a fixed point O on that line as origin. Then a single number describes where the particle is.

Distance and displacement are not the same. A particle that goes 5 m forward and 5 m back has travelled a distance of 10 m but has displacement 0. Likewise speed is the magnitude of velocity: velocity can be negative, speed never is.

Because v = ds/dt, everything from Unit 18 transfers directly. The particle is momentarily at rest when v = 0; it reverses direction where v changes sign; its velocity is greatest where a = 0 and changes from positive to negative.

The equations of uniformly accelerated motion

If the acceleration is constant, the calculus can be done once and for all. Let u be the velocity at t = 0, v the velocity at time t, a the constant acceleration, and s the displacement in that time.

The velocity-time graph makes the equations visible. For constant acceleration it is a straight line of gradient a and intercept u; the area under it between two times is the displacement, and for a trapezium that area is \tfrac{1}{2}(u+v)t — the fourth equation.

21.4 Motion under gravity; motion down a smooth inclined plane

Free vertical motion

Near the surface of the earth every body, whatever its mass, falls with the same constant acceleration g = 9.8\ \text{m/s}^2 directed downwards, provided air resistance is ignored. Constant acceleration means the four equations of §21.3 apply with a = \pm g; only the sign needs care.

Two standard results follow from taking upwards as positive with a = -g. At the greatest height v = 0, so 0 = u^2 - 2gH gives the maximum height H = u^2/(2g); and putting s = 0 in s = ut - \tfrac{1}{2}gt^2 gives t(u - \tfrac{1}{2}gt) = 0, so the total time of flight back to the starting level is t = 2u/g, exactly twice the time to rise.

Motion down a smooth inclined plane

Let a particle of mass m rest on a plane inclined at angle \theta to the horizontal. Two forces act on it: its weight mg vertically downwards, and the normal reaction N perpendicular to the plane. Smooth means there is no friction, so no force acts along the plane except the component of the weight.

Resolve the weight along and perpendicular to the plane — not horizontally and vertically. The component along the plane, pointing down the slope, is mg\sin\theta; the component pressing into the plane is mg\cos\theta.

Perpendicular to the plane the particle does not move, so N = mg\cos\theta. Along the plane the only force is mg\sin\theta, so by Newton's second law ma = mg\sin\theta, and the mass cancels.

Chapter summary

• A force is a vector: magnitude, direction, point of application. Forces at a point are concurrent; forces in one plane are coplanar.
• Parallelogram law: two forces P, Q at angle \theta have resultant R = \sqrt{P^2+Q^2+2PQ\cos\theta} making angle \alpha = \tan^{-1}\dfrac{Q\sin\theta}{P+Q\cos\theta} with P.
• The resultant always lies between |P-Q| and P+Q; it equals P+Q at \theta = 0^\circ, \sqrt{P^2+Q^2} at 90^\circ, |P-Q| at 180^\circ.
• Resolution: F_x = F\cos\theta, F_y = F\sin\theta. Any number of coplanar concurrent forces combine as X = \sum F\cos\theta, Y = \sum F\sin\theta, R = \sqrt{X^2+Y^2}, \tan\theta = Y/X; read the quadrant from the signs.
• Equilibrium \Leftrightarrow X = 0 and Y = 0. For three forces use Lami's theorem: each force divided by the sine of the angle between the other two is the same. The equilibrant is minus the resultant.
• Kinematics: v = ds/dt, a = dv/dt = d^2s/dt^2. Distance is not displacement; reverse of direction happens where v changes sign.
• Uniform acceleration only: v = u+at, s = ut+\tfrac12 at^2, v^2 = u^2+2as, s = \tfrac12(u+v)t, and s_n = u + \tfrac{a(2n-1)}{2}. On a v–t graph the gradient is a and the area is s.
• Gravity: a = -g upwards positive, a = +g downwards positive, with g = 9.8\ \text{m/s}^2. Maximum height u^2/2g, time of flight 2u/g.
• Smooth inclined plane: a = g\sin\theta down the slope (mass cancels) and N = mg\cos\theta. Speed after a vertical drop h is \sqrt{2gh}, the same as in free fall.

Practice questions

Group A — Multiple choice (1 mark each)

1. The greatest possible resultant of two forces of 8 N and 6 N is [1]a2 Nb10 Nc14 Nd48 N
2. Two forces, each of 5 N, act at a point at 120^\circ. Their resultant is [1]a0b5 Nc5\sqrt{3} Nd10 N
3. Coplanar concurrent forces are in equilibrium if and only if [1]aX = 0 onlybY = 0 onlycX = 0 and Y = 0dX = Y
4. In Lami's theorem each force is divided by the sine of the angle [1]ait makes with the verticalbbetween the other two forcescbetween it and the next forcedof 90^\circ
5. A body dropped from rest falls, in the first second, a distance of (take g = 9.8\ \text{m/s}^2) [1]a4.9 mb9.8 mc19.6 md2.45 m
6. If s = 4t^2, the velocity at t = 3 s is [1]a12 m/sb24 m/sc36 m/sd8 m/s
7. At the highest point of the path of a stone thrown vertically upwards [1]av = 0 and a = 0bv = 0 and a = g downwardscv = g and a = 0dboth v and a are upwards
8. The acceleration of a particle sliding freely down a smooth plane inclined at 30^\circ is [1]a9.8\ \text{m/s}^2b4.9\ \text{m/s}^2c8.49\ \text{m/s}^2dit depends on the mass

Answers to Group A

1. (c) — the resultant is greatest when \theta = 0^\circ, giving 8 + 6 = 14 N.

2. (b) — R = \sqrt{25+25+2(25)\cos 120^\circ} = \sqrt{50-25} = 5 N.

3. (c) — the resultant \sqrt{X^2+Y^2} vanishes only when both component sums vanish.

4. (b) — that is the statement of the theorem; the angles used sum to 360^\circ.

5. (a) — s = \tfrac12 gt^2 = \tfrac12(9.8)(1)^2 = 4.9 m.

6. (b) — v = ds/dt = 8t, so v(3) = 24 m/s.

7. (b) — the velocity is momentarily zero but gravity never switches off.

8. (b) — a = g\sin 30^\circ = 9.8 \times 0.5 = 4.9\ \text{m/s}^2, independent of mass.

Group B — Short answer (5 marks each)

1. Forces of 4 N and 7 N act at a point with 60^\circ between them. Find the magnitude and direction of the resultant. [5]
2. The resultant of two forces P and 2P acting at a point is perpendicular to P. Find the angle between the forces and the magnitude of the resultant. [5]
3. Three forces of 5 N, 10 N and 15 N act at a point in directions making 0^\circ, 120^\circ and 240^\circ with the x-axis. Find their resultant. [5]
4. A particle moves so that s = 2t^3 - 9t^2 + 12t metres. Find when it is at rest, and the total distance it travels in the first 3 seconds. [5]
5. A ball is thrown vertically upwards with velocity 24.5\ \text{m/s}. Find the greatest height reached and the total time of flight. (g = 9.8\ \text{m/s}^2) [5]
6. A particle slides from rest down a smooth plane of inclination 45^\circ and length 10 m. Find the acceleration, the speed at the bottom and the time taken. [5]

Solutions to Group B

1. With P = 4, Q = 7, \theta = 60^\circ:

R = \sqrt{4^2 + 7^2 + 2(4)(7)(0.5)} = \sqrt{16+49+28} = \sqrt{93} = 9.64\ \text{N}

For the direction from the 4 N force,

\tan\alpha = \frac{7\sin 60^\circ}{4 + 7\cos 60^\circ} = \frac{6.062}{7.5} = 0.8083

so \alpha = 38.95^\circ = 38^\circ 57'. The resultant is 9.64 N at 38^\circ 57' to the 4 N force.

2. Take P along the x-axis. The resultant is perpendicular to P exactly when its x-component is zero:

X = P + 2P\cos\theta = 0 \;\Longrightarrow\; \cos\theta = -\tfrac12 \;\Longrightarrow\; \theta = 120^\circ

The resultant is then purely the y-component,

R = 2P\sin 120^\circ = 2P\left(\tfrac{\sqrt{3}}{2}\right) = \sqrt{3}\,P

So the angle is 120^\circ and R = \sqrt{3}P.

3. Components, using \cos 120^\circ = \cos 240^\circ = -\tfrac12, \sin 120^\circ = \tfrac{\sqrt{3}}{2}, \sin 240^\circ = -\tfrac{\sqrt{3}}{2}:

X = 5 - 5 - 7.5 = -7.5, \qquad Y = 0 + 8.660 - 12.990 = -4.330

Hence

R = \sqrt{56.25 + 18.75} = \sqrt{75} = 5\sqrt{3} = 8.66\ \text{N}

and \tan\theta = (-4.330)/(-7.5) = 0.5774 with both components negative, so the resultant lies in the third quadrant: \theta = 180^\circ + 30^\circ = 210^\circ. The resultant is 5\sqrt{3} N at 210^\circ to the x-axis.

4. Differentiating, v = 6t^2 - 18t + 12 = 6(t-1)(t-2), so the particle is at rest at t = 1 s and t = 2 s. The displacements at the turning times and the end points are

s(0) = 0, \quad s(1) = 5, \quad s(2) = 4, \quad s(3) = 9

Adding the magnitudes of the three legs,

|5-0| + |4-5| + |9-4| = 5 + 1 + 5 = 11\ \text{m}

Total distance = 11 m (while the displacement is only 9 m).

5. Upwards positive: u = 24.5, a = -9.8. At the top v = 0, so from v^2 = u^2 + 2as,

H = \frac{u^2}{2g} = \frac{(24.5)^2}{2(9.8)} = \frac{600.25}{19.6} = 30.63\ \text{m}

The time of flight is

T = \frac{2u}{g} = \frac{2(24.5)}{9.8} = 5\ \text{s}

Greatest height 30.63 m, time of flight 5 s (rising for 2.5 s).

6. The acceleration down the slope is

a = g\sin 45^\circ = 9.8 \times 0.7071 = 6.93\ \text{m/s}^2

With u = 0 and s = 10 m, from v^2 = 2as,

v = \sqrt{2(6.93)(10)} = \sqrt{138.6} = 11.77\ \text{m/s}

and from v = at,

t = \frac{11.77}{6.93} = 1.70\ \text{s}

Group C — Long answer (8 marks each)

1. State the parallelogram law of forces and derive expressions for the magnitude and the direction of the resultant of two forces P and Q inclined at an angle \theta. Hence find two forces whose magnitudes are in the ratio 3:5, act at 60^\circ to each other and have a resultant of 35 N. [8]
2. Starting from the definitions v = ds/dt and a = dv/dt, derive the equations v = u+at, s = ut+\tfrac12 at^2 and v^2 = u^2+2as for constant acceleration. A train starts from rest, accelerates uniformly at 1\ \text{m/s}^2 for 20 s, runs at constant speed for 60 s, and is then brought to rest by a uniform retardation of 2\ \text{m/s}^2. Find the total distance travelled. [8]
3. Show that a particle sliding freely on a smooth plane of inclination \theta has acceleration g\sin\theta down the plane and that the normal reaction is mg\cos\theta. A body slides from rest down a smooth plane of inclination 30^\circ whose vertical height is 10 m. Find the speed at the bottom and the time taken, and compare the speed with that of a body falling freely through 10 m. [8]

Solutions to Group C

1. Statement. If two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from that point, their resultant is represented by the diagonal of the parallelogram through that point.

Derivation. Put O at the origin with P along the x-axis, so \vec P = (P, 0) and \vec Q = (Q\cos\theta,\ Q\sin\theta). Adding,

\vec R = (P + Q\cos\theta,\ Q\sin\theta)

Squaring and adding the components,

R^2 = (P+Q\cos\theta)^2 + Q^2\sin^2\theta = P^2 + 2PQ\cos\theta + Q^2(\cos^2\theta+\sin^2\theta)

so, using \cos^2\theta+\sin^2\theta = 1,

R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}

and the angle \alpha the resultant makes with P satisfies

\tan\alpha = \frac{Q\sin\theta}{P+Q\cos\theta}

Application. Let the forces be 3k and 5k. Then with \theta = 60^\circ,

R^2 = 9k^2 + 25k^2 + 2(3k)(5k)\left(\tfrac12\right) = 9k^2+25k^2+15k^2 = 49k^2

so R = 7k. Given R = 35 N,

7k = 35 \;\Longrightarrow\; k = 5

The forces are 3(5) = 15 N and 5(5) = 25 N.

2. Derivation. With a constant, integrating a = dv/dt gives v = at + C, and v = u at t = 0 forces C = u, so

v = u + at

Integrating v = ds/dt next,

s = \int (u+at)\,dt = ut + \tfrac12 at^2 + C_1

and measuring s from the start gives C_1 = 0, so s = ut + \tfrac12 at^2. Eliminating t = (v-u)/a from this,

2as = 2u(v-u) + (v-u)^2 = 2uv - 2u^2 + v^2 - 2uv + u^2 = v^2 - u^2

hence v^2 = u^2 + 2as.

Application. Phase 1 (u = 0, a = 1, t = 20):

v = 0 + 1(20) = 20\ \text{m/s}, \qquad s_1 = 0 + \tfrac12(1)(20)^2 = 200\ \text{m}

Phase 2 (constant 20 m/s for 60 s):

s_2 = 20 \times 60 = 1200\ \text{m}

Phase 3 (u = 20, v = 0, a = -2): from v^2 = u^2+2as,

0 = 400 - 4s_3 \;\Longrightarrow\; s_3 = 100\ \text{m} \quad (\text{taking } t = 20/2 = 10\ \text{s})

Total distance

s = 200 + 1200 + 100 = 1500\ \text{m}

3. Derivation. The particle of mass m has only two forces on it: the weight mg vertically down and the normal reaction N perpendicular to the plane (the plane is smooth, so there is no friction). Resolve the weight along and perpendicular to the plane: the components are mg\sin\theta down the slope and mg\cos\theta into the plane.

Perpendicular to the plane there is no motion, so those forces balance:

N = mg\cos\theta

Along the plane the only force is mg\sin\theta, so Newton's second law gives

ma = mg\sin\theta \;\Longrightarrow\; a = g\sin\theta

The mass cancels, so the acceleration is independent of the mass.

Application. The length of the slope is

l = \frac{h}{\sin\theta} = \frac{10}{\sin 30^\circ} = 20\ \text{m}

and the acceleration is a = 9.8\sin 30^\circ = 4.9\ \text{m/s}^2. With u = 0,

v = \sqrt{2al} = \sqrt{2(4.9)(20)} = \sqrt{196} = 14\ \text{m/s}

t = \frac{v}{a} = \frac{14}{4.9} = 2.86\ \text{s}

For free fall through the same height,

v = \sqrt{2gh} = \sqrt{2(9.8)(10)} = \sqrt{196} = 14\ \text{m/s}

The two speeds are equal: the slope does not change the final speed, only the time taken to acquire it (2.86 s on the plane against 14/9.8 = 1.43 s in free fall).