Study Companion: S01 — What Statics Is: Quantities, Laws and Units

Statics answers one kind of question: a body is not accelerating, so which forces must be acting on it, and how large are they? The answer rests on very little. There are Newton's three laws, a few idealised models (the particle, the rigid body, the vector), one law of gravitation and one system of units. Then there is a disciplined method for turning a physical situation into equations. This lesson sets out each of those pieces and the reason it is there. Every later lesson in the book uses them without re-deriving them.

The sources are Meriam, Kraige and Bolton, Engineering Mechanics: Statics, articles 1/1–1/9. The vector operations that article 1/3 starts are developed fully in S02. Meriam mixes SI and US customary units; here every calculation is done in SI (see Nugget 04 for the translation).

How to use this companion

Each nugget is one idea. It opens with a claim. A picture shows why the claim is true, a number checks it, and the equation follows, labelled with what kind of statement it is. After that come a worked example with units, a real place it shows up, the wrong answer people give most often, and a question to test yourself. A short bridge says why the next nugget is needed. The Narration spine in each nugget summarises what its video beat shows. Traps and a self-check with full solutions close the lesson.

The puzzle

A 1.2\,\mathrm{kg} book lies on a desk. Nothing moves. Is any force acting on the book at all? If so, how hard does the desk push on it, and how could you know without a force sensor? By the end of Nugget 08 you will be able to answer: 11.8\,\mathrm N upwards, found from one law, one model, one unit and one method.

Nugget 01: Statics Is Mechanics With Zero Acceleration

Claim. A body at rest still has forces on it. Statics is the study of how those forces cancel.

The picture. Isolate the book: draw it alone and replace everything that touches or attracts it by a force.

Two forces act on the book. The Earth pulls it down with its weight W, and the desk pushes it up with a force N. Notice that the two forces on the book are not an action–reaction pair. Each force's partner acts on a different body (listed on the right).

The idea. Newton's three laws, in modern words (Meriam 1/4):

1. A particle stays at rest, or keeps moving in a straight line at constant speed, if no unbalanced force acts on it.
2. Its acceleration is proportional to the vector sum of the forces on it, and points the same way as that sum.
3. Two bodies act on each other with forces equal in size, opposite in direction and on the same line.

Mechanics divides into statics (bodies in equilibrium) and dynamics (bodies that accelerate). The first law is the second law with zero acceleration, so statics is the special case \mathbf a = \mathbf 0 of one law. That is why so few principles cover the whole subject.

The equation (physical law: Newton's second law, verified by experiment, not derived):

Here \mathbf F is the vector sum of all forces on the body (N), m its mass (kg) and \mathbf a its acceleration (\mathrm{m/s^2}). Both sides are vectors, so they agree in direction as well as size. Units: \mathrm{kg}\cdot\mathrm{m/s^2} = \mathrm N. Assumptions: an inertial frame of reference (the Earth is close enough for statics) and speeds far below the speed of light.

Worked number. The book has m = 1.2\,\mathrm{kg}, so W = mg = 1.2 \times 9.81 = 11.772\,\mathrm N. It does not accelerate, so the forces sum to zero: N - W = 0, giving N = 11.772\,\mathrm N \approx 11.8\,\mathrm N. Check: take the desk away and the only force left is W, so a = W/m = 11.772/1.2 = 9.81\,\mathrm{m/s^2}. That is free fall, as it should be.

Real example. Lean on a wall and you feel it push back on your hands. Push harder and it pushes harder. A wall supplies whatever force keeps the acceleration zero, up to the point where it breaks. Every support in this book behaves the same way.

Plausible wrong answer. "The desk's push and the book's weight are the action–reaction pair of the third law." They are equal here only because the book is not accelerating. Put the desk in a lift that is accelerating upwards and N grows while W does not. The true partner of W is the book's pull on the Earth.

Check. A 5 kg toolbox hangs at rest from a rope. Name the two forces on the toolbox and the partner of each. (Weight 49.05\,\mathrm N down, partner: the toolbox pulls the Earth up. Rope tension 49.05\,\mathrm N up, partner: the toolbox pulls the rope down.)

Narration spine. Show the book on the desk, then lift it out alone. The weight arrow appears from above and the push arrow from below, equal in length. Slide each partner force onto its own body (the desk, the Earth) to show they never act on the same object. Write \mathbf F = m\mathbf a, then set \mathbf a = \mathbf 0.

Bridge. We just treated the book as one point that either accelerates or does not. Real objects bend, rotate and have size. Statics only works because we choose, deliberately, which of those features to ignore.

Nugget 02: Particles and Rigid Bodies Are Models, Chosen on Purpose

Claim. A particle and a rigid body are idealisations. You choose one because the detail it ignores is too small to change the answer you want.

The picture.

The boom is steel. Under load it shortens by about 2\,\mathrm{mm} over 10\,\mathrm m, a change of 0.002/10 = 2\times10^{-4}, or 0.02\,\%. Every distance on the boom, and so every lever arm used to find the cable tension, is wrong by at most 0.02\,\% if we ignore that shortening. That is below the three-figure accuracy we work to (Nugget 06).

The idea. Meriam 1/2 defines the basic quantities:

• Space: the region bodies occupy, located by measurements from a coordinate system. Three coordinates are needed in 3D, two in a planar problem.
• Time: the order of events. It does not enter statics directly.
• Mass: a body's resistance to a change of velocity (its inertia). It also sets how strongly gravity attracts the body.
• Force: the action of one body on another, given by its size, its direction and its point of application.
• Particle: a body whose dimensions do not matter to the question, treated as mass concentrated at a point.
• Rigid body: a body whose distances between points do not change noticeably under the loads considered.

Statics finds the external forces on rigid bodies in equilibrium. How a body deforms internally is the subject of the mechanics of deformable bodies, which usually follows statics.

The equation (definition of a rigid body): for every pair of points A, B of the body, the distance |\mathbf r_B - \mathbf r_A| stays constant. Its test is an approximation: a body may be treated as rigid when the relative change in length is well below the accuracy required, \Delta L/L well under 10^{-3} for three-figure answers. The boom's 2\times10^{-4} passes.

Worked number. A 5\,\mathrm{kg} box rests on a level floor, and the question is how hard the floor pushes on it. The answer is 5 \times 9.81 = 49.05\,\mathrm N whatever the box's size, so the box is a particle. Ask instead whether a 0.4\,\mathrm m tall box tips over when pushed near its top. Its size is now the whole question, so it must be a rigid body.

Real example. A mobile crane's load chart gives the safe load at each boom angle. It is computed with the boom as a rigid body. The boom's deflection is checked separately, in a strength calculation, not in the statics.

Plausible wrong answer. "Rigid means it does not deform at all." Nothing is perfectly rigid. Rigid means its deformation is negligible for this question. The same steel cable is a rigid straight member when you find its tension, and must be treated as flexible when you find its sag.

Check. Under your weight a 3\,\mathrm m steel ladder shortens by 0.6\,\mathrm{mm}, and a 1\,\mathrm m bungee cord holding a bag stretches by 0.4\,\mathrm m. Which can be treated as rigid when you find the forces? (The ladder: 0.0006/3 = 0.02\,\%. Not the cord: 0.4/1 = 40\,\% changes the geometry itself.)

Narration spine. Show the crane with the boom drawn as a rigid bar. Zoom in on its 2 mm shortening, put the ratio 0.002/10 beside it, then zoom back out until the change is invisible. Fade in the list of basic quantities, one at a time, each beside the part of the crane it describes.

Bridge. The model tells us what is acted on. Next we need a way to say how a force acts, and it takes more than one number.

Nugget 03: A Force Needs a Line of Action, Not Just a Size

Claim. A force is a vector: it has a size and a direction. For a rigid body its line of action matters too. For a deformable body, so does its exact point of application.

The picture.

Push the cart from behind with 30\,\mathrm N, or pull it from the front with 30\,\mathrm N along the same line. A rigid cart responds identically. The point where the force is applied can slide anywhere along its line of action. Move the force off that line, say to one corner, and the cart also starts to turn. So the line matters.

The idea. A scalar has a size only: time, volume, density, speed, energy, mass. A vector has a size and a direction, and adds by the parallelogram law (S02): displacement, velocity, acceleration, force, moment, momentum. Speed is a scalar; it is the magnitude of the velocity vector. Physical vectors come in three kinds (Meriam 1/3):

• Free vector: only its size and direction matter. Example: the displacement of a body that moves without rotating; every point moves by the same vector.
• Sliding vector: it has a definite line of action but no definite point on it. Example: an external force on a rigid body (the principle of transmissibility, taken up in S03).
• Fixed vector: its point of application matters too. Example: a force on a deformable body. Squeezing a sponge at its edge or at its centre deforms it differently.

Notation. In print a vector is bold, \mathbf V, and its magnitude is light italic, V or |\mathbf V|. By hand, mark every vector with an arrow or an underline, \vec V or \underline V. An equation that mixes up \mathbf V and V is a different equation.

The equation (definition): a force is written \mathbf F = F\,\mathbf n. Here F \ge 0 is its magnitude in newtons, and \mathbf n is a dimensionless unit vector giving its direction. For a sliding vector we also record one point on its line of action.

Worked number. The cart's push is \mathbf F = 30\,\mathbf i\ \mathrm N, with F = 30\,\mathrm N and \mathbf n = \mathbf i (along +x). Its line of action is the horizontal line through the cart's centre, y = 0.25\,\mathrm m above the floor. Push at the back or pull at the front: the same vector on the same line.

Real example. A tow truck pulls a car by a chain hooked to the front bumper. To move the car the same way, you could push the rear bumper on the same line instead.

Plausible wrong answer. "Two equal force vectors always have the same effect." Equal vectors on different lines turn a body differently; that difference is the moment of a force (S04). Marghitu puts it as: equality of vectors does not imply physical equivalence.

Check. Classify each as free, sliding or fixed: (a) the weight of a rigid steel beam, for finding its supports; (b) a finger pressing a rubber balloon; (c) the displacement of a drawer sliding straight out. (Sliding; fixed; free.)

Narration spine. Show the cart with a push arrow on its line of action. Slide the arrow along the dashed line, through the cart, to the front, where it becomes a pull. The cart's response stays the same. Then shift the arrow sideways off the line and the cart rotates. Label the three kinds of vector.

Bridge. A force's size is a number with a unit. Before we can write any equation with forces in it, we need to know what a newton is and how it is tied to the kilogram.

Nugget 04: The Newton Is Built From Kilograms, Metres and Seconds

Claim. Mass, length, time and force cannot all have independent units, because \mathbf F = m\mathbf a ties them together. SI picks mass, length and time, and derives force.

The picture.

The kilogram is the same lump of matter in both places. The spring stretches less on the Moon, because the force on the lump is smaller there (g_\text{Moon} \approx 1.62\,\mathrm{m/s^2}). So mass and force are different quantities, and a spring scale measures force.

The idea. Mechanics uses four dimensions: mass M, length L, time T and force F. Their units cannot all be chosen freely, because the second law fixes one of them once the other three are set.

• SI (Système International) chooses kilogram (kg), metre (m) and second (s) as base units. The newton (N) is derived: 1\,\mathrm N gives 1\,\mathrm{kg} an acceleration of 1\,\mathrm{m/s^2}. SI is an absolute system, because its base unit of mass does not depend on where you are.
• US customary (foot–pound–second) chooses the foot, the second and the pound-force as base units, and derives the slug: 1\,\mathrm{slug} is the mass that 1\,\mathrm{lbf} accelerates at 1\,\mathrm{ft/s^2}. It is a gravitational system, because its base unit is a force defined by gravity on a standard mass. Meriam writes "lb" for pound-force, "lbm" for pound-mass and "sec" for second. The kip (1000\,\mathrm{lbf}) and the ton (2000\,\mathrm{lbf}) are also used.
• In SI the kilogram is only a mass, never a force. The old MKS gravitational system used a "kilogram-force", 1\,\mathrm{kgf} = 9.806\,65\,\mathrm N, which is not part of SI.

Translation note for this book. Meriam gives some data in US units. Each is converted to SI where it first appears, and all working is in SI:

Primary standards. Since 20 May 2019 every SI base unit is defined by a fixed constant of nature. The second is set by the caesium-133 transition frequency, 9\,192\,631\,770\,\mathrm{Hz}. The metre is set by the speed of light, 299\,792\,458\,\mathrm{m/s}. The kilogram is set by the Planck constant, h = 6.626\,070\,15\times10^{-34}\,\mathrm{J\,s}. Meriam still describes the kilogram as a platinum–iridium cylinder kept near Paris; that artefact was retired in 2019. The standard gravitational acceleration is g = 9.806\,65\,\mathrm{m/s^2} (sea level, 45^\circ latitude). We use 9.81\,\mathrm{m/s^2}, which is accurate enough for engineering work.

The equation (definition of the newton, from Newton's second law):

Dimension check of W = mg: [\mathrm{kg}][\mathrm{m/s^2}] = [\mathrm N].

Worked number. A car of mass 1500\,\mathrm{kg} has weight W = 1500 \times 9.81 = 14\,715\,\mathrm N \approx 14.72\,\mathrm{kN}. In US units that is 14\,715/4.448 = 3308\,\mathrm{lbf}. Going the other way, a 3000\,\mathrm{lbf} car weighs 3000 \times 4.448 = 13\,345\,\mathrm N, so its mass is 13\,345/9.81 = 1360\,\mathrm{kg}.

Real example (one you can try). A kitchen scale displays grams, but its sensor measures force. It divides the force by g = 9.81\,\mathrm{m/s^2} before showing it. Taken to the Moon, it would show a 1\,\mathrm{kg} bag of sugar as 1000 \times 1.62/9.81 = 165\,\mathrm g, although the sugar has not changed.

Plausible wrong answer. "I weigh 70 kg." That is your mass. Your weight is 70 \times 9.81 = 687\,\mathrm N. Mixing the two puts a factor of about 10 into every force you calculate.

Check. What is the weight in newtons of a 250\,\mathrm g apple? (0.25 \times 9.81 = 2.45\,\mathrm N: roughly 1\,\mathrm N for every 100\,\mathrm g.)

Narration spine. Hang 1 kg on a spring scale reading 9.81 N. Swap the background to the Moon: the spring shortens and the reading drops to 1.62 N, while the "1 kg" label stays fixed. Build \mathrm N = \mathrm{kg\cdot m/s^2} from the three base units dropping into place. Finish on the conversion table.

Bridge. We wrote W = mg and simply used g = 9.81\,\mathrm{m/s^2}. Where does that number come from, and is the Earth's pull the only attraction worth counting?

Nugget 05: Weight Is Gravitation, Measured at the Surface

Claim. Every two masses attract each other. At the Earth's surface the Earth's pull is the only one large enough to matter, and g is that pull per kilogram.

The picture.

Moving out from radius R to R + h weakens the pull by the factor \left(R/(R+h)\right)^2. At 400\,\mathrm{km} up, where the International Space Station orbits, that factor is (6371/6771)^2 = 0.885, so gravity there is still 88.5\,\% of its surface value.

The idea. The law of gravitation (Meriam 1/6): any two particles attract each other along the line joining their centres, with equal and opposite forces, as the third law requires. For a body at the Earth's surface that attraction is its weight. Since the body falls with acceleration g, \mathbf F = m\mathbf a gives W = mg.

The equation (physical law, found by experiment; G is measured, not derived):

The terms: F is the mutual attraction (N), m_1 and m_2 are the masses (kg), and r is the distance between their centres (m). The constant is G = 6.674\,30\times10^{-11}\,\mathrm{m^3/(kg\cdot s^2)} (Meriam prints the older value 6.673\times10^{-11}). Units: \mathrm{m^3/(kg\cdot s^2)}\cdot\mathrm{kg}^2/\mathrm m^2 = \mathrm{kg\cdot m/s^2} = \mathrm N. It assumes particles, or uniform spheres (which attract as if all their mass were at the centre).

Check with numbers. Put the Earth (m_e = 5.972\times10^{24}\,\mathrm{kg}, R = 6.371\times10^{6}\,\mathrm m) in for m_1 and divide by m_2:

That is close to 9.81, but not equal to it. The difference (about 0.1\,\%) is the Earth's rotation. A scale measures the apparent weight, slightly less than the true gravitational pull. Statics neglects the difference and uses g = 9.81\,\mathrm{m/s^2}.

Worked number. A 70 kg person: from the law of gravitation, W = 70 \times 9.820 = 687.4\,\mathrm N. From W = mg, W = 70 \times 9.81 = 686.7\,\mathrm N. Aboard the station, W = 687.4 \times 0.8853 = 608.6\,\mathrm N.

Is any other attraction worth counting? Take two steel spheres of diameter 100\,\mathrm{mm} (density 7830\,\mathrm{kg/m^3}). Each has mass 7830 \times \frac43\pi(0.05)^3 = 4.10\,\mathrm{kg} and weight 40.2\,\mathrm N. Touching, their centres are 0.1\,\mathrm m apart, and they attract each other with 6.6743\times10^{-11}\times4.10^2/0.1^2 = 1.12\times10^{-7}\,\mathrm N. That is 360 million times smaller than either weight. On the Earth's surface, the Earth is the only attraction we count.

Real example. The Moon stays in orbit because of the same law. The Earth pulls it with about 1.98\times10^{20}\,\mathrm N (self-check Q7).

Plausible wrong answer. "Astronauts float because there is no gravity in orbit." There is: 88.5\,\% of surface gravity at the station's height. They float because they and the station fall around the Earth together, so nothing pushes on them. That is a dynamics effect, not an absence of weight.

Check. By what factor does your weight change if you stand on a 1\,\mathrm{km} high mountain? ((6371/6372)^2 = 0.99969, a 0.03\,\% drop, invisible at three figures.)

Narration spine. Draw two masses with equal and opposite arrows along the line between them. Grow one into the Earth and put a person on its surface. Move the person up to the station's orbit, showing the 1/r^2 curve and the reading falling from 687 N to 609 N. Finish with the two steel spheres and their tiny 10^{-7}\,\mathrm N arrows beside their 40\,\mathrm N weights.

Bridge. We just wrote 687.4\,\mathrm N and 686.7\,\mathrm N and called the difference small. Deciding what "small" means, and how many digits an answer deserves, needs its own rule.

Nugget 06: An Answer Is Only as Precise as Its Data

Claim. An answer cannot have more significant figures than its data justify. Subtracting nearly equal numbers loses figures quickly.

The picture. A square bar has a side measured as 36\,\mathrm{mm}, to the nearest millimetre. So the true side lies between 35.5 and 36.5\,\mathrm{mm}. The true area therefore lies anywhere in the shaded band:

The band runs from 1260 to 1332\,\mathrm{mm^2}. Writing 36^2 = 1296\,\mathrm{mm^2} claims four figures, but the last two are noise. Writing 1300\,\mathrm{mm^2} (two figures, like the data) is honest.

The idea. The number of significant figures in an answer should be no greater than the accuracy of the data justifies (Meriam 1/7). Small differences of large quantities need especially accurate data. For example, 7.6382 - 7.6271 = 0.0111 needs five-figure inputs to give a three-figure difference. Round the inputs to three figures first and you get 7.64 - 7.63 = 0.01, only one figure. Meriam's working rule, used in this book: give answers to three significant figures, or four if the first digit is 1. Treat given data as exact. Store intermediate results at full precision and round only once, at the end.

The equation (approximation, a rule for reporting): if x has relative uncertainty \varepsilon, then x^2 has relative uncertainty about 2\varepsilon. Here \varepsilon = 0.5/36 = 1.4\,\%, so the area is uncertain by about 2.8\,\%, or \pm36\,\mathrm{mm^2}. That matches the band: 1332 - 1296 = 36.

Worked number. The 1500\,\mathrm{kg} car of Nugget 04: W = 14\,715\,\mathrm N. The first digit is 1, so keep four figures: 14\,720\,\mathrm N (14.72\,\mathrm{kN}). Divide by 4.448 to get 3308\,\mathrm{lbf}, starting from the stored 14\,715, not from the rounded 14\,720.

Real example. The tolerance on a machined shaft is written \varnothing\,25.00 \pm 0.02\,\mathrm{mm}. The machinist reads those two decimals as a promise. Writing 25.0000 would be a promise nobody can keep.

Plausible wrong answer. "More decimals is more accurate." A calculator's eight digits only repeat the input's error more precisely. Accuracy comes from the data, not from the arithmetic.

Check. Using the rule, report A + B, A - B, AB and A/B for the exact values A = 9.24 and B = 1.316 (self-check Q6).

Narration spine. Draw a 36 mm square, then a slightly smaller and a slightly larger outline at 35.5 and 36.5 mm. Unroll the three areas onto a number line as a shaded band. Drop 1296 and 1300 into the band to show both fit. Then flash 7.6382 - 7.6271 and watch the leading digits cancel.

Bridge. Knowing how small "small" is lets us do something stronger than round. We can drop whole terms from an expression, before the arithmetic starts.

Nugget 07: Small Things Can Be Dropped, If You Know How Small

Claim. In a limit, products of small quantities vanish faster than the quantities themselves. For small angles in radians, \sin\theta \approx \tan\theta \approx \theta and \cos\theta \approx 1.

The picture. Two pictures, one for each idea.

The slab of thickness \Delta x at distance x from the vertex is almost a flat disc of radius rx/h. The parts that make it not quite a disc are thin rims whose volume is a small quantity times another small quantity.

With hypotenuse 1, the vertical side is \sin\theta and the arc is \theta (in radians). For a small angle the two are almost the same length, and the base \cos\theta is almost the whole hypotenuse.

The idea, differentials. The volume from the vertex to x is a cone of radius rx/h, so V(x) = \frac13\pi\left(\frac{rx}{h}\right)^2x = \frac{\pi r^2x^3}{3h^2}. Step the source skips: expand (x + \Delta x)^3 = x^3 + 3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3. Then

As \Delta x \to dx the higher-order terms drop out, leaving dV = \frac{\pi r^2}{h^2}x^2\,dx. That is exact when integrated (it is how centroids and distributed loads are found in Part 5).

Check with numbers. Take r = h, x = 1\,\mathrm m and \Delta x = 0.01\,\mathrm m. The three bracket terms are 0.01, 0.0001 and 0.000\,000\,33\,\mathrm{m^3}. The second is \Delta x/x = 1\,\% of the first, and the third is smaller again. Halve \Delta x and the second term shrinks four times while the first only halves, so in the limit only the first survives.

The idea, small angles. For \theta = 1^\circ = 0.017\,453\,\mathrm{rad} the values are \sin1^\circ = 0.017\,452, \tan1^\circ = 0.017\,455 and \cos1^\circ = 0.999\,848. Replacing \sin\theta by \theta is wrong by 0.005\,\% at 1^\circ, 0.13\,\% at 5^\circ (0.0873\,\mathrm{rad}), 0.51\,\% at 10^\circ (0.1745\,\mathrm{rad}) and 2.1\,\% at 20^\circ. For more accuracy, keep the next terms of the series:

In the limit, \sin d\theta = \tan d\theta = d\theta and \cos d\theta = 1.

The equation (approximation; valid only with \theta in radians): \sin\theta \approx \tan\theta \approx \theta, \cos\theta \approx 1. The error is of order \theta^3 for the sine and tangent and \theta^2 for the cosine.

Worked number. A cable pulls a post with 5\,\mathrm{kN} at 4^\circ above the horizontal. Exactly: horizontal part 5\cos4^\circ = 4.988\,\mathrm{kN}, vertical part 5\sin4^\circ = 0.3488\,\mathrm{kN}. With small angles (4^\circ = 0.069\,81\,\mathrm{rad}): 5\,\mathrm{kN} and 5 \times 0.069\,81 = 0.3491\,\mathrm{kN}, which is within 0.25\,\% and 0.08\,\%.

Real example. A shelf bracket that sags 1^\circ moves its tip down by \theta L. For a 300\,\mathrm{mm} shelf that is 0.017\,45 \times 300 = 5.2\,\mathrm{mm}, and you need no trigonometry to see it.

Plausible wrong answer. "\sin 5 \approx 5." Only if 5 is in radians, and 5 rad is not small. In degrees the approximation reads \sin5^\circ \approx 5\times\pi/180 = 0.0873, not 5.

Check. Is \cos\theta \approx 1 good to 1 % at 10^\circ? (\cos10^\circ = 0.9848, a 1.5\,\% error, so no. Use 1 - \theta^2/2 = 0.9848.)

Narration spine. Show the cone and a thin slab at x. Lift the slab out and peel off its rim, which shrinks away as \Delta x shrinks. Then draw the unit-hypotenuse triangle and close the angle: the vertical side and the arc merge while the base stretches to 1. The table of errors at 1°, 5°, 10° and 20° ticks up beside it.

Bridge. Laws, models, vectors, units and approximations are the ingredients. Statics problems are solved by putting them together in a fixed order, every time.

Nugget 08: The Method Turns a Picture Into Equations

Claim. Every statics problem is solved the same way: formulate it, isolate the body in a free-body diagram, apply the laws, and check the result.

The picture. A 25\,\mathrm{kg} lamp hangs from two equal cables, each at \theta to the horizontal. Isolate the lamp.

The two tensions T pull along the cables. Their horizontal parts cancel by symmetry, and their vertical parts, T\sin\theta each, hold up the weight.

The idea. Meriam 1/8 sets the procedure. Formulate the problem: state the given data, the result wanted, and the assumptions and approximations. Develop the solution: draw the diagrams needed, state the governing principles, calculate, keep the accuracy the data justifies, keep the units consistent, check the answer is reasonable in size, direction and common sense, and draw conclusions. The central tool is the free-body diagram: the body isolated from everything else, with all external forces acting on it drawn in. Isolating the body separates cause from effect (S10 teaches how to draw one). Work in symbols as long as you can, for three reasons: the physics stays visible, the result is reusable, and every line can be checked for dimensional homogeneity (every term in an equation must have the same units). Problems are solved by hand, graphically, or by computer. The computer is the right tool for many equations, for a parameter study, or for an equation with no closed-form solution.

The equation (physical law applied to the model): with x to the right and y upwards, the vertical forces sum to zero:

Assumptions: the cables are massless and inextensible, the lamp is a particle, and the system is in static equilibrium. Dimensions: [\mathrm N]/[\,1\,] = [\mathrm N]. Limiting cases: when \theta = 90^\circ the cables hang vertically and T = W/2, each carrying half. When \theta \to 0, T \to \infty.

Worked number. W = 25 \times 9.81 = 245.25\,\mathrm N. At \theta = 30^\circ, T = 245.25/(2 \times 0.5) = 245.25\,\mathrm N, so each cable carries the full weight. At 10^\circ, T = 245.25/(2\times0.173\,65) = 706.2\,\mathrm N. At 5^\circ, T = 1407\,\mathrm N, nearly six times the weight.

Real example (one you can try). Hang a bag on a washing line and try to pull the line straight. However hard you pull, it still sags, because a straight line (\theta = 0) would need infinite tension. The symbolic result predicted this. The single number at 30^\circ could not have.

Plausible wrong answer. "Two cables share the load, so each carries half the weight, 122.6\,\mathrm N." That is true only when they hang vertically. At 30^\circ each carries 245\,\mathrm N, double the guess.

Check. Is T = W\sin\theta/2 dimensionally possible? Does it pass the limiting cases? (Dimensionally fine, but at \theta = 0 it gives T = 0, the opposite of the physics. Wrong.)

Narration spine. Show the lamp on its two cables. Cut it free and redraw it alone with the two tension arrows and the weight. Split each tension into horizontal and vertical parts: the horizontals cancel, the verticals stack up to match W. Then let \theta shrink while a readout of T climbs toward infinity, with a washing line beside it.

Synthesis

The book on the desk now has its answer. It is a particle (Nugget 02). The forces on it are vectors along vertical lines of action (Nugget 03). Its weight is the Earth's gravitational pull (Nugget 05), 1.2 \times 9.81 = 11.772\,\mathrm N in newtons (Nugget 04). Because it does not accelerate, the first law says the forces cancel (Nugget 01). The method, a free-body diagram and one equation, gives the desk's push as 11.8\,\mathrm N to the accuracy the data justify (Nuggets 06 and 08). No single nugget gives that number. Each one removes one thing you would otherwise have to guess.

• Central idea: statics is Newton's second law with zero acceleration, applied to an idealised body.
• Key relationship: \sum\mathbf F = \mathbf 0 on an isolated body, with forces in newtons, 1\,\mathrm N = 1\,\mathrm{kg\cdot m/s^2}, and weight W = mg.
• Most important assumption: the model (particle or rigid body, massless cable, small angle) must ignore only effects smaller than the accuracy you need.
• Open question: a force has a line of action, so two equal forces on different lines act differently. How do we add vectors and measure that difference? (S02 and S04.)

Traps

• Calling mass "weight". Mass is in kilograms, weight in newtons: 70\,\mathrm{kg} weighs 687\,\mathrm N. (From Nugget 04.)
• Pairing the wrong forces. The desk's push and the book's weight act on the same body, so they are not a third-law pair. The pair of the weight is the book's pull on the Earth. (From Nugget 01.)
• "No gravity in orbit." At 400\,\mathrm{km}, gravity is 88.5\,\% of its surface value; orbiting bodies are falling, not weightless. (From Nugget 05.)
• Rounding before the end. Store 14\,715\,\mathrm N, not 14\,720, if you will use it again. Round once, at the answer. (From Nugget 06.)
• Small angles in degrees. \sin\theta\approx\theta only with \theta in radians: \sin5^\circ \approx 0.0873, not 5. (From Nugget 07.)
• Too many figures from a difference. 7.64 - 7.63 = 0.01 has one significant figure; subtracting nearly equal numbers needs more accurate data. (From Nugget 06.)
• Pound of what? In Meriam "lb" is a force (lbf), and the US mass unit is the slug. Convert a weight in lbf to newtons (\times4.448), then divide by g: a 3000\,\mathrm{lbf} car has m = 13\,345/9.81 = 1360\,\mathrm{kg}. Putting a mass in lbm into F = ma with a in \mathrm{ft/s^2} gives a force 32.2 times too large. (From Nugget 04.)
• Equal vectors, equal effect. Two equal forces on different lines of action turn a body differently. Only a sliding along the same line changes nothing. (From Nugget 03.)
• "Each of two cables carries half." Only if they are vertical. At 30^\circ each carries the whole weight. (From Nugget 08.)

Self-check

1. A vector is \mathbf V = 12\mathbf i - 5\mathbf j (units). Find its magnitude, the angles it makes with the positive x- and y-axes, and the unit vector \mathbf n along it.
2. A force is \mathbf F = 200\mathbf i - 150\mathbf j + 100\mathbf k\ \mathrm N. Find the angles it makes with the positive x-, y- and z-axes.
3. A car's data sheet gives its weight as 3500\,\mathrm{lbf}. Find its weight in newtons and its mass in kilograms.
4. Use the law of gravitation to find the weight of an 80\,\mathrm{kg} astronaut (a) on the Earth's surface and (b) in a station orbiting 400\,\mathrm{km} up. Compare (a) with mg.
5. A person's weight is recorded as 130\,\mathrm{lbf}. Find it in newtons and find the person's mass in kilograms.
6. Two dimensionless quantities are exactly A = 9.24 and B = 1.316. Report A + B, A - B, AB and A/B by this book's rule for significant figures.
7. Find the force the Earth exerts on the Moon (m_e = 5.972\times10^{24}\,\mathrm{kg}, m_\text{Moon} = 7.342\times10^{22}\,\mathrm{kg}, centre distance 384\,400\,\mathrm{km}).
8. Two lead spheres of 10\,\mathrm{kg} each hang with their centres 0.2\,\mathrm m apart. Find their mutual attraction and compare it with the weight of one of them.
9. Evaluate E = 3\sin^2\theta\,\tan\theta\,\cos\theta exactly for \theta = 3^\circ, then with the small-angle approximations. What is the error?
10. A quantity is given by Q = k\,m\,b^2/(c\,t^2), where k is dimensionless, m is a mass, b and c are lengths and t is a time. Find its SI units, using base units, and name the quantity it could be.
11. A 40\,\mathrm{kg} shop sign hangs from two equal wires, each at 20^\circ to the horizontal. Find the tension in each wire, and check the result in the limit \theta \to 90^\circ.
12. Classify each as a scalar or a vector, and each vector as free, sliding or fixed: (a) the temperature of a beam; (b) the force of a rigid tie rod on a rigid bracket; (c) the pressure of a thumb on a soft eraser; (d) the displacement of a lift car.

Fully worked solutions

1. V = \sqrt{12^2 + (-5)^2} = \sqrt{169} = 13. With the x-axis: \cos\theta_x = 12/13 = 0.9231, \theta_x = 22.6^\circ. With the y-axis: \cos\theta_y = -5/13 = -0.3846, \theta_y = 112.6^\circ. Unit vector \mathbf n = \mathbf V/V = 0.923\mathbf i - 0.385\mathbf j (dimensionless). Check: 0.9231^2 + 0.3846^2 = 0.8521 + 0.1479 = 1.

2. F = \sqrt{200^2 + 150^2 + 100^2} = \sqrt{40\,000 + 22\,500 + 10\,000} = \sqrt{72\,500} = 269.3\,\mathrm N. \cos\theta_x = 200/269.3 = 0.7428, \theta_x = 42.0^\circ; \cos\theta_y = -150/269.3 = -0.5571, \theta_y = 123.9^\circ; \cos\theta_z = 100/269.3 = 0.3714, \theta_z = 68.2^\circ. Check: 0.7428^2 + 0.5571^2 + 0.3714^2 = 0.5517 + 0.3103 + 0.1379 = 1.000.

3. W = 3500 \times 4.448\,22 = 15\,569\,\mathrm N (four figures, leading 1: 15\,570\,\mathrm N). Mass m = W/g = 15\,569/9.81 = 1587\,\mathrm{kg}.

4. (a) W = Gm_em/R^2 = 6.6743\times10^{-11}\times5.972\times10^{24}\times80/(6.371\times10^{6})^2 = 785.6\,\mathrm N. Compare mg = 80 \times 9.81 = 784.8\,\mathrm N, about 0.1\,\% less because of the Earth's rotation. (b) r = 6371 + 400 = 6771\,\mathrm{km}: W = 785.6 \times (6371/6771)^2 = 785.6 \times 0.8853 = 695.5\,\mathrm N.

5. W = 130 \times 4.448\,22 = 578.3\,\mathrm N. m = 578.3/9.81 = 58.9\,\mathrm{kg}.

6. A + B = 10.556; the first digit is 1, so four figures: 10.56. A - B = 7.924 \to 7.92. AB = 12.1598 \to 12.16 (leading 1). A/B = 7.021\,28 \to 7.02.

7. F = \frac{6.6743\times10^{-11}\times5.972\times10^{24}\times7.342\times10^{22}}{(3.844\times10^{8})^2}. Numerator: 6.6743\times5.972\times7.342 = 292.6, times 10^{-11+24+22} = 10^{35}, so 2.926\times10^{37}. Denominator: 3.844^2\times10^{16} = 14.776\times10^{16}. F = 2.926\times10^{37}/1.4776\times10^{17} = 1.980\times10^{20}\,\mathrm N.

8. F = 6.6743\times10^{-11}\times10\times10/0.2^2 = 6.6743\times10^{-9}/0.04 = 1.669\times10^{-7}\,\mathrm N. Each weighs 10 \times 9.81 = 98.1\,\mathrm N, so the attraction is 98.1/1.669\times10^{-7} = 5.9\times10^{8} times smaller. It is negligible.

9. Since \tan\theta\cos\theta = \sin\theta, E = 3\sin^3\theta exactly. \sin3^\circ = 0.052\,336, so E = 3\times0.052\,336^3 = 3\times1.4335\times10^{-4} = 4.3005\times10^{-4}. Approximate: \theta = 3\pi/180 = 0.052\,360\,\mathrm{rad}, \sin\theta\approx\theta, \tan\theta\approx\theta, \cos\theta\approx1: E \approx 3\theta^3 = 3\times1.4355\times10^{-4} = 4.3064\times10^{-4}. Error: (4.3064 - 4.3005)/4.3005 = 0.14\,\%.

10. Units: \mathrm{kg}\cdot\mathrm m^2/(\mathrm m\cdot\mathrm s^2) = \mathrm{kg\cdot m/s^2}, which is the newton. Q could be a force. (In US base units it would be the pound-force, lbf.)

11. W = 40 \times 9.81 = 392.4\,\mathrm N. Vertical equilibrium: 2T\sin20^\circ = W, so T = 392.4/(2\times0.342\,02) = 573.7\,\mathrm N. In the limit \theta\to90^\circ, \sin\theta \to 1 and T \to W/2 = 196.2\,\mathrm N: two vertical wires share the load equally, as expected.

12. (a) Scalar. (b) Vector, sliding: the bracket is rigid, so only the line of action matters. (c) Vector, fixed: the eraser deforms, so the point of application matters. (d) Vector, free: every point of the car moves by the same displacement.