Forces and equilibrium

• The only numbers you bring into the room for this chapter are \(g = 10\) and the four trigonometric results: \(\sin(90^\circ - \theta) \equiv \cos\theta\), \(\cos(90^\circ - \theta) \equiv \sin\theta\), \(\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}\), \(\sin^2\theta + \cos^2\theta \equiv 1\).
• W down, R perpendicular, F against the tendency, T along the string; resolve in two directions and set each sum to zero; F = μR only when about to slip. If these come back instantly on day 30, the chapter has stuck.

• 4.1.1 I can identify — identify the forces acting in a given situation
• 4.1.2 I can understand — understand the vector nature of force, and find and use components and resultants
• 4.1.3 I can use — use the principle that, when a particle is in equilibrium, the vector sum of the forces acting is zero, or equivalently, that the sum of the components in any direction is zero
• 4.1.4 I can understand — understand that a contact force between two surfaces can be represented by two components, the normal component and the frictional component
• 4.1.5 I can use — use the model of a 'smooth' contact, and understand the limitations of this model
• 4.1.6 I can understand — understand the concepts of limiting friction and limiting equilibrium, recall the definition of coefficient of friction, and use the relationship F = μR or F ≤ μR, as appropriate
• 4.1.7 I can use — use Newton's third law

Accuracy and working are marked. The syllabus states that non-exact numerical answers are to be given correct to three significant figures, or one decimal place for angles in degrees, unless the question specifies otherwise, and that to earn accuracy marks you should avoid rounding until the final answer. It also states that no marks are given for unsupported answers from a calculator, and that graphic calculators and calculators with symbolic algebra or calculus are not permitted. Every worked example in this chapter is therefore written the way an answer must be written: the method line first, the substitution visible, full precision carried, one rounding at the end.

• “The particle is at rest on a rough plane, so \(F = \mu R\).” Correct \(F = \mu R\) only when the particle is in limiting equilibrium (“about to slip”, “on the point of moving”, the least or greatest force). A particle that is merely at rest has whatever friction equilibrium needs, found by resolving, and \(F \le \mu R\). Ask “is it limiting?” before writing either (section E).
• “On a slope at \(\alpha\), the weight has component \(mg\cos\alpha\) down the slope.” Correct The weight makes \(90^\circ - \alpha\) with the slope, so its component down the slope is \(mg\sin\alpha\) and its component into the slope is \(mg\cos\alpha\). Check with the extremes: at \(\alpha = 0\) nothing pulls along a flat surface, and \(\sin 0 = 0\) (figure 5c).
• “The normal reaction equals the weight.” Correct \(R\) is whatever balances the other forces perpendicular to the surface. On a slope with nothing else acting, \(R = mg\cos\alpha\); a pull at an angle above the horizontal reduces \(R\) on level ground; a push downwards increases it. Find \(R\) by resolving every time.
• “Friction always acts down the slope.” Correct Friction opposes the motion or the tendency to move. A particle tending to slide down has \(F\) up the slope; a large enough force up the slope reverses the tendency and \(F\) acts down. That is why a range question has two cases.
• “Weight and normal reaction are a Newton's third law pair.” Correct A third-law pair acts on two different bodies. \(W\) and \(R\) both act on the particle. The partner of \(R\) is the push of the particle on the surface; the partner of \(W\) is the particle's pull on the Earth.
• A resultant given as “10.5 N”. Correct A force is a vector: give the magnitude and the direction, as an angle to 1 decimal place measured from a stated direction (“\(72.9^\circ\) above the positive \(x\)-direction”).
• Drawing “\(ma\)” as an arrow on a force diagram. Repair A force diagram shows forces only. In this chapter the resultant is zero; in chapter 21 \(ma\) is what the resultant equals, and it is still never drawn as a force.
• “The normal reaction on a slope is \(mg\).” Repair \(R\) is perpendicular to the slope and is found by resolving in that direction. It equals \(mg\cos\alpha\) only when no other force has a component perpendicular to the slope.
• “The component of \(W\) along a slope at \(\alpha\) is \(W\cos\alpha\).” Repair The angle between \(W\) and the slope is \(90^\circ - \alpha\), so the component along the slope is \(W\cos(90^\circ - \alpha) = W\sin\alpha\); perpendicular to it, \(W\cos\alpha\).
• “The direction of the resultant is \(\tan^{-1}(X/Y)\).” Repair \(\tan^{-1}(|Y|/|X|)\) is the angle to the \(X\)-direction. Say which direction the angle is measured from, and fix the quadrant from the signs of \(X\) and \(Y\).
• Writing \(F = \mu R\) for a particle that is merely at rest. Repair \(F = \mu R\) only in limiting equilibrium (or when sliding). Otherwise \(F\) comes from resolving and satisfies \(F \le \mu R\).
• Friction drawn down the slope for a particle that tends to slide down. Repair Friction opposes the tendency to move, so \(F\) is up the slope. When a force up the slope is large enough, the tendency reverses and so does \(F\): two cases.
• “\(\mu = 0.35\) N.” Repair \(\mu\) is a ratio of two forces, \(F_{\max}/R\), so it has no unit.
• “A rod in thrust pulls the particle.” Repair Thrust pushes: the arrow points into the particle along the rod. Tension in a string pulls: the arrow points from the particle along the string.
• “Weight and normal reaction are a third-law pair.” Repair Both act on the same body. The partner of \(R\) is the force the body exerts on the floor.
• A resultant given as a magnitude alone. Repair A force is a vector: give the angle to 1 decimal place and the direction it is measured from.
• For a pull \(P\) at \(\theta\) above the horizontal on rough ground: “\(R = mg\), so \(F = \mu mg\).” Repair The pull has an upward component \(P\sin\theta\), so resolving vertically gives \(R = mg - P\sin\theta\). A push at \(\theta\) below the horizontal gives \(R = mg + P\sin\theta\).
• Rounding \(R\) to 3 significant figures, then using the rounded value to find \(F\) and \(P\). Repair Keep full calculator precision (at least four significant figures, marked unrounded) and round only the final answers; an early rounding can change the third figure of the answer.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• What makes a force diagram right. Count the forces against the situation: one weight; one \(R\), and one \(F\) if rough, per surface; one \(T\) per string or rod; each applied force. Then check each direction: \(W\) vertical, \(R\) perpendicular to its surface, \(F\) along the surface against the tendency to move, a string's \(T\) pulling away from the particle, a rod in thrust pushing into it.
• Interleave with the chapters that use this one. When you reach chapter 21, redo worked example 3 with the pull increased to 25 N: the same diagram, but now the resultant is not zero and equals ma. When you reach chapter 22, re-answer retrieval question 11 and find the work done against friction as the particle moves 2 m up the plane. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: later chapters use these methods without re-teaching them, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Mechanics content, examined in Paper 4. Paper 4 (Mechanics) is 40% of an AS Level that includes it and 20% of an A Level that includes it. It assumes knowledge of the algebraic methods in the Paper 1 content. An A Level route that includes Paper 4 takes Papers 1, 3, 4 and 5; Paper 4 cannot be combined with Paper 6. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A question on this section describes a situation in words — a particle on rough ground pulled at an angle, a particle hanging from two strings, a particle on a rough inclined plane with a force along it — and asks for forces, an angle or \(\mu\). The reasoning is in three places: the force diagram (which forces, which way), the choice of resolving directions, and the decision whether friction is limiting. A “show that” part can give \(R\) or a tension so that a later part can use it; a final part may ask for the range of a force for which the particle stays at rest, which needs two limiting cases.
• MF19's Mechanics entry is the four constant-acceleration formulae (chapter 19), and none of them is used here. Everything in this chapter must be known: \(W = mg\), the components \(F\cos\theta\) and \(F\sin\theta\), the magnitude \(\sqrt{X^2 + Y^2}\), \(F = \mu R\) and \(F \le \mu R\), and Newton's third law. The value of \(g\) is not printed either: the syllabus says to use \(g = 10\ \mathrm{m\,s^{-2}}\).
• Write each resolving equation in full, naming the direction (“Resolving up the plane”), before any number: an answer with no equation behind it earns nothing. Carry unrounded values (at least four significant figures) and round once: forces to 3 significant figures with the unit N, angles to 1 decimal place, \(\mu\) to 3 significant figures with no unit. Exact values stay exact: \(R = 20\) N is written as 20, not 20.0.
• The model this whole component uses. The syllabus states it once for Mechanics, and this chapter obeys it throughout. Every body is a particle: an extended body in a realistic context (a crate, a car, a person) is treated as a particle, so every force on it acts at a single point. Vector notation is not used in the question papers: no i, j or column vectors, only magnitudes and directions in words or on a diagram. Calculations are always required, never approximate answers by scale drawing. Take \(g = 10\ \mathrm{m\,s^{-2}}\). Questions are mainly numerical and do not involve difficult algebra or trigonometry; the results \(\sin(90^\circ - \theta) \equiv \cos\theta\), \(\cos(90^\circ - \theta) \equiv \sin\theta\), \(\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}\) and \(\sin^2\theta + \cos^2\theta \equiv 1\) are assumed, as is knowledge of the algebraic methods in the content of Paper 1.