Senior 4 practice book · Unit 3 of 10

Moments and Equilibrium of Bodies

50 questions that complete the Senior 4 quiz for this unit: 22 core and 28 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

The moment of a force is force × the distance along the bar to where it acts.Only the PERPENDICULAR distance from the pivot to the line of action counts; a force pulling along the bar towards the pivot has no moment at all.
If the forces on a body add up to zero, the body must be in equilibrium.A couple has zero resultant force but still makes the body rotate; equilibrium also needs the sum of the moments about any point to be zero.
The centre of gravity is always inside the material of an object.It can lie in empty space, as for a ring, a horseshoe or a boomerang; it is just the point where the whole weight can be taken to act.
A stiffer wire of the same metal has a bigger Young modulus.Young’s modulus belongs to the material, not to the wire; a thicker or shorter wire stretches less under a load, but stress ÷ strain is the same.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Scalars and vectors: adding vectors and resolving a force into components
Scalar (dot) and vector (cross) products
Moment of a force: perpendicular distance and everyday turning effects
Principle of moments: beams, see-saws and balances
Couples and torque
Conditions of equilibrium and free-body diagrams
Three non-parallel forces in equilibrium: triangle of forces and Lami's theorem
Locating the centre of gravity: plumb-line method and composite bodies
States of equilibrium, stability and toppling
Levers and simple machines: classes of lever and mechanical advantage
Hooke's law, force–extension graphs, springs in series and parallel, elastic and plastic behaviour
Stress, strain and Young's modulus
Energy stored in a stretched wire or spring (area under the force–extension graph)

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 28 harder ones are marked ★ Challenge.

1True or false
Lowering the centre of gravity and widening the base both make an object more stable.
Show answer
Answer: True
Both changes mean the object must be tilted through a larger angle before the vertical through its centre of gravity leaves the base.
Common mistakeSome think only a heavier object is more stable; extra weight helps only if it lowers the centre of gravity.
2True or false · ★ Challenge
A bus moving along a straight, level road at a constant velocity is in equilibrium.
Show answer
Answer: True
Constant velocity means zero acceleration, so the resultant force is zero (the driving force balances friction and drag): this is dynamic equilibrium.
Common mistakeMany learners think equilibrium means "at rest"; a body moving steadily in a straight line is also in equilibrium.
3True or false · ★ Challenge
If a copper wire is replaced by a longer, thicker copper wire, the Young modulus measured for it will be different.
Show answer
Answer: False
Young’s modulus is stress ÷ strain, which depends only on the material; the longer, thicker wire stretches by a different amount, but stress/strain is the same.
Common mistakeAnswering "true" confuses the stiffness of a particular wire (which changes with its size) with the Young modulus of the material (which does not).
4True or false
For any two vectors, the vector product A × B is equal to B × A.
Show answer
Answer: False
Swapping the order reverses the direction given by the right-hand rule: B × A = −(A × B). The scalar product, by contrast, does not depend on order.
Common mistakeThe trap is to think products always commute as in ordinary arithmetic; the cross product changes sign when the order is swapped.
5True or false · ★ Challenge
The centre of gravity of a body can lie outside the material of the body, as it does for a bicycle tyre or a boomerang.
Show answer
Answer: True
For a ring the centre of gravity is at the empty centre; it is simply the point where the total weight may be taken to act.
Common mistakeThe trap is to think the centre of gravity must be a point you can touch; for hollow or bent shapes it is often in empty space.
6True or false
A force whose line of action passes through the pivot has no moment about that pivot, however large the force is.
Show answer
Answer: True
The perpendicular distance from the pivot to the line of action is zero, so the moment F × 0 = 0.
Common mistakeSome learners think a very large force must always turn something; without a perpendicular distance, it cannot produce turning.
7True or false · ★ Challenge
If a copper wire is stretched beyond its elastic limit and then released, all the work done in stretching it is given back as it shortens.
Show answer
Answer: False
Beyond the elastic limit the wire is permanently stretched; part of the work goes into rearranging the metal and is lost as heat, so less energy is returned.
Common mistakeAnswering "true" applies the elastic rule to plastic behaviour; energy is fully recovered only when the wire returns to its original length.
8Fill in the blank
Strain has no unit because it is the ratio of two ______.
Show answer
Answer: lengths (extension and original length)
Strain = extension ÷ original length; the metres cancel, leaving a pure number.
Common mistakeGiving strain the unit N/m² confuses it with stress; only stress (and Young’s modulus) are measured in pascals.
9True or false · ★ Challenge
When a beam is in equilibrium you may take moments about any point you like, even a point where there is no pivot or support, and the equation will still be correct.
Show answer
Answer: True
In equilibrium the resultant moment is zero about EVERY point, so any point may be chosen; a clever choice simply removes an unknown force from the equation.
Common mistakeMany learners believe moments may only be taken about the actual pivot; that is just the most convenient point, not the only correct one.
10Fill in the blank · ★ Challenge
Torque is a vector product, τ = r × F, with magnitude rF sin θ. For a given force applied at a given distance, the torque is greatest when θ = ______.
Show answer
Answer: 90°
sin θ has its largest value, 1, at θ = 90°, i.e. when the force is perpendicular to the line from the pivot.
Common mistakeAnswering 0° confuses it with the scalar product; a force along r (θ = 0°) gives no torque at all.
11Fill in the blank
If three non-parallel forces keep a body in equilibrium, their lines of action must all pass through ______ point.
Show answer
Answer: the same (one common)
If two of the lines met at a point and the third missed it, the third force would have a moment about that point and the body would turn.
Common mistakeSome learners think the three forces only need to add to zero; they must also be concurrent, or there would be a resultant moment.
12Fill in the blank · ★ Challenge
For a lever with no friction, the mechanical advantage (load ÷ effort) is equal to the distance from the fulcrum to the effort divided by the distance from the fulcrum to the ______.
Show answer
Answer: load
Taking moments about the fulcrum: effort × effort arm = load × load arm, so load/effort = effort arm/load arm.
Common mistakeTurning the ratio upside down gives a mechanical advantage below 1 for a crowbar, which contradicts what a crowbar is used for.
13Multiple choice
Which list contains ONLY vector quantities?
1. AMass, speed, time
2. BDistance, velocity, force
3. CDisplacement, energy, momentum
4. DDisplacement, velocity, weight
Show answer
Answer: D. Displacement, velocity, weight
Displacement, velocity and weight all need a direction to be fully described; mass, speed, time, distance and energy are scalars.
Common mistakeChoosing the list with energy and momentum is the usual trap: momentum is a vector, but energy has no direction and is a scalar.
14Multiple choice · ★ Challenge
A goat is tied by two ropes; one pulls it with 5 N and the other with 12 N, and the angle between the ropes can be anything. Which total pull on the goat is impossible?
1. A7 N
2. B13 N
3. C20 N
4. D17 N
Show answer
Answer: C. 20 N
The resultant lies between the difference (12 − 5 = 7 N, forces opposite) and the sum (12 + 5 = 17 N, forces together); 20 N is outside this range.
Common mistakeChoosing 13 N assumes only the right-angle case is impossible; 13 N is exactly the result when the forces are perpendicular (√(25 + 144) = 13).
15Multiple choice
Which of these is an example of a couple acting on an object?
1. AOne hand pushing a classroom door near the handle
2. BOne hand turning a nut with a long spanner
3. CTwo hands turning the T-shaped handle of a water tap
4. DA child sitting on one end of a see-saw
Show answer
Answer: C. Two hands turning the T-shaped handle of a water tap
On a T-handle the two hands push with equal, opposite, parallel forces on either side of the axis: that is a couple.
Common mistakeChoosing the spanner is tempting, but one hand gives a single force; the nut must then push back, so it is not a pure couple.
16Multiple choice · ★ Challenge
Two identical springs each have k = 200 N/m. A 10 N load is hung from them joined end to end (in series), and then from them side by side (in parallel). What is the total extension in each case?
1. A2.5 cm in series, 10 cm in parallel
2. B5.0 cm in series, 5.0 cm in parallel
3. C10 cm in series, 5.0 cm in parallel
4. D10 cm in series, 2.5 cm in parallel
Show answer
Answer: D. 10 cm in series, 2.5 cm in parallel
In series each spring carries 10 N and stretches 5.0 cm, total 10 cm (k = 100 N/m). In parallel each carries 5 N and stretches 2.5 cm (k = 400 N/m).
Common mistakeChoosing "5.0 cm in parallel" forgets that the load is shared: each of the two parallel springs carries only half of it.
17Multiple choice
In which class of lever is the effort applied between the fulcrum and the load?
1. AFirst class, like a crowbar
2. BSecond class, like a wheelbarrow
3. CSecond class, like a bottle opener
4. DThird class, like tweezers
Show answer
Answer: D. Third class, like tweezers
In tweezers (and the human forearm) the fingers push between the hinge and the tips that grip the load, which makes a third-class lever.
Common mistakeChoosing the wheelbarrow mixes up the classes: there the LOAD is in the middle (second class), and the effort is at the handles.
18Multiple choice · ★ Challenge
A 20 kg sack of maize rests without sliding on a ramp inclined at 30° to the horizontal (g = 10 N/kg). What is the friction force acting on the sack?
1. A173 N up the slope
2. B200 N up the slope
3. C100 N up the slope
4. D0 N, since it is at rest
Show answer
Answer: C. 100 N up the slope
For equilibrium along the slope, friction balances the component of weight down the slope: F = mg sin 30° = 20 × 10 × 0.5 = 100 N.
Common mistakeChoosing 173 N uses mg cos 30°, which is the component pressing into the ramp (balanced by the normal reaction), not the one along it.
19Short answer · ★ Challenge
Concrete beams in Kigali buildings have steel bars set near the BOTTOM of the beam. Explain why, using the ideas of tensile and compressive stress.
Show answer
Model answer: When a beam carries a load it bends: the top surface is squashed (compressive stress) and the bottom surface is stretched (tensile stress). Concrete is strong in compression but cracks easily in tension, while steel is strong in tension. The steel bars are put where the tension is, near the bottom, so the beam can carry the load without cracking.
Common mistakePutting the steel "in the middle for strength" misses the idea; the middle of a bent beam has almost no stress, and the bottom is where tension is greatest.
20Multiple choice
A tall flask is clamped to a laboratory retort stand. Which change makes the stand LESS likely to topple over?
1. AClamping the flask at the top of the rod
2. BUsing a narrower base plate for the stand
3. CClamping the flask lower down the rod
4. DRemoving the heavy base plate altogether
Show answer
Answer: C. Clamping the flask lower down the rod
Lowering the flask lowers the centre of gravity, so the stand must tilt further before the line of action of the weight falls outside the base.
Common mistakeChoosing "remove the heavy base" may seem to make the stand lighter and safer, but the heavy base keeps the centre of gravity low and over a wide area.
21Short answer · ★ Challenge
Explain, using moments, why a minibus tilted sideways on a test platform falls over only when the vertical line through its centre of gravity passes outside its wheels.
Show answer
Model answer: While the vertical line through the centre of gravity falls inside the base (between the wheels), the weight has a moment about the lower wheels that turns the bus back to its upright position. Once that line passes outside the lower wheels, the moment of the weight about them acts in the other direction and turns the bus further over, so it topples.
Common mistakeSaying the bus topples "because it is heavy" misses the point; it is the position of the weight’s line of action relative to the base that decides.
22Multiple choice
When solving a beam problem, why is it wise to take moments about a support whose force you do NOT yet know?
1. AThat force has zero moment there, so it drops out of the equation
2. BThat force then becomes equal to the total weight of the beam
3. CThe moments of all the other forces become zero at that point
4. DThe beam can only turn about the support with the largest force
Show answer
Answer: A. That force has zero moment there, so it drops out of the equation
A force acting through the chosen point has zero perpendicular distance, so the equation contains one unknown fewer.
Common mistakeChoosing "all other moments become zero" is backwards: only forces passing through the chosen point lose their moment.
23Multiple choice · ★ Challenge
Two fishermen on Lake Kivu turn a capstan to pull up their boat. Each pushes a bar with a force of 150 N, at 1.2 m from the centre, on opposite sides and in opposite directions. What torque do they apply?
1. A180 N m
2. B125 N m
3. C300 N m
4. D360 N m
Show answer
Answer: D. 360 N m
The two forces form a couple with arm 2 × 1.2 = 2.4 m: τ = F × d = 150 × 2.4 = 360 N m.
Common mistakeChoosing 180 N m counts only one fisherman (150 × 1.2); in a couple the distance used is the full separation of the two forces.
24True or false
If three forces keep a body in equilibrium, arrows drawn to scale for them, placed head to tail, form a closed triangle.
Show answer
Answer: True
A closed triangle means the vector sum is zero, which is the first condition for equilibrium (the triangle of forces).
Common mistakeLearners sometimes draw the arrows from the same point and look for a triangle; the arrows must be joined head to tail.
25Multiple choice · ★ Challenge
Vector A has magnitude 4.0 units and vector B has magnitude 5.0 units; the angle between them is 30°. What is the magnitude of the vector product A × B?
1. A17.3 units
2. B20 units
3. C9.0 units
4. D10 units
Show answer
Answer: D. 10 units
|A × B| = |A||B| sin θ = 4.0 × 5.0 × sin 30° = 20 × 0.5 = 10 units.
Common mistakeChoosing 17.3 uses cos 30°, which belongs to the scalar (dot) product; the vector product uses sin θ.
26Multiple choice
Why is the handle of a classroom door fixed at the edge furthest from the hinges?
1. AA larger distance from the hinges reduces the weight of the door that must be lifted
2. BA larger perpendicular distance gives the same moment with less force
3. CThe door is thickest at that edge, so the handle is less likely to break off
4. DForces applied far from the hinges are always larger forces
Show answer
Answer: B. A larger perpendicular distance gives the same moment with less force
Moment = force × perpendicular distance; with the largest distance from the hinge, a small push gives enough turning effect to open the door.
Common mistakeChoosing "reduces the weight" mixes up turning and lifting: the hinges carry the door’s weight wherever the handle is.
27Multiple choice · ★ Challenge
A light rod 100 cm long carries a 2.0 kg mass at the 0 cm end and a 3.0 kg mass at the 100 cm end. Where is the centre of gravity of the system?
1. AAt the 60 cm mark
2. BAt the 40 cm mark
3. CAt the 50 cm mark
4. DAt the 75 cm mark
Show answer
Answer: A. At the 60 cm mark
x = (m₁x₁ + m₂x₂)/(m₁ + m₂) = (2.0 × 0 + 3.0 × 100) ÷ 5.0 = 60 cm, nearer the heavier mass.
Common mistakeChoosing 40 cm puts the centre of gravity nearer the LIGHTER mass; it always lies closer to the heavier end.
28Multiple choice · ★ Challenge
A gardener pushes on a gate 1.2 m from its hinges with a force of 50 N at right angles to the gate. She then pushes with the same force at the same point but along the gate, straight towards the hinges. What are the two moments about the hinges?
1. A60 N m, then 60 N m
2. B60 N m, then 0 N m
3. C41.7 N m, then 0 N m
4. D60 N m, then 30 N m
Show answer
Answer: B. 60 N m, then 0 N m
First: M = F × d = 50 × 1.2 = 60 N m. Second: the line of action passes through the hinge, so the perpendicular distance is zero and M = 0.
Common mistakeChoosing "60 N m both times" uses the distance along the gate instead of the perpendicular distance to the line of action.
29Multiple choice
A nylon rope with a cross-sectional area of 2.0 × 10⁻⁵ m² holds a 400 N goat-feed sack. What is the tensile stress in the rope?
1. A5.0 × 10⁻⁸ Pa
2. B2.0 × 10⁷ Pa
3. C2.0 × 10⁹ Pa
4. D8.0 × 10² Pa
Show answer
Answer: B. 2.0 × 10⁷ Pa
Stress = F/A = 400 ÷ 2.0 × 10⁻⁵ = 2.0 × 10⁷ Pa (N/m²).
Common mistakeChoosing 5.0 × 10⁻⁸ Pa divides the area by the force; stress is force per unit area, F ÷ A.
30Multiple choice · ★ Challenge
A 300 N load hangs from a crane rope. A worker pulls the load sideways with a horizontal rope until the crane rope makes 30° with the vertical, and holds it still. What is the pull in the horizontal rope?
1. A173 N
2. B150 N
3. C346 N
4. D520 N
Show answer
Answer: A. 173 N
Resolving: vertically T cos 30° = 300, horizontally P = T sin 30°. So P = 300 tan 30° = 300 × 0.577 = 173 N (and T = 346 N).
Common mistakeChoosing 346 N gives the tension in the crane rope, not the horizontal pull; the triangle of forces gives both, so read the question carefully.
31Multiple choice
A farmer pulls a cart with a rope that makes 25° with the horizontal ground. The tension in the rope is 200 N. What is the horizontal part of the pull?
1. A181 N
2. B84.5 N
3. C200 N
4. D221 N
Show answer
Answer: A. 181 N
Horizontal component = F cos θ = 200 × cos 25° = 200 × 0.906 = 181 N.
Common mistakeChoosing 84.5 N uses sin 25°, which gives the vertical (lifting) part; the component next to the angle uses cos.
32Multiple choice · ★ Challenge
A steelyard balance in a market is a rod pivoted at its own centre of gravity. The load hook is 4.0 cm from the pivot, and a 5.0 N sliding weight balances a sack of beans when it is 32 cm from the pivot on the other side. What is the weight of the beans?
1. A0.63 N
2. B160 N
3. C37 N
4. D40 N
Show answer
Answer: D. 40 N
Moments about the pivot: W × 4.0 = 5.0 × 32, so W = 160 ÷ 4.0 = 40 N. The rod’s own weight acts at the pivot and has no moment.
Common mistakeChoosing 160 N stops at the moment (5.0 × 32 = 160 N cm) and forgets to divide by the 4.0 cm distance of the load.
33Short answer
Explain, using the principle of moments, why shears for cutting sheet metal have long handles and short blades.
Show answer
Model answer: The shears are first-class levers with the pivot between handles and blades. Long handles give a large distance from the pivot to the hand, and short blades put the metal close to the pivot. By the principle of moments, effort × long distance = load × short distance, so a small effort gives a large cutting force on the metal.
Common mistakeSaying long handles "add force" is vague; it is the ratio of the distances from the pivot that multiplies the force.
34Multiple choice · ★ Challenge
A builder uses a 1.5 m crowbar to lift the edge of a heavy rock. The fulcrum (a small stone) is 0.10 m from the end under the rock, which presses down with 1400 N. What effort is needed at the other end?
1. A93 N
2. B1400 N
3. C100 N
4. D140 N
Show answer
Answer: C. 100 N
Effort arm = 1.5 − 0.10 = 1.4 m. Moments about the fulcrum: E × 1.4 = 1400 × 0.10, so E = 140 ÷ 1.4 = 100 N (MA = 14).
Common mistakeChoosing 93 N divides by the whole 1.5 m bar; the effort arm is only the part from the fulcrum to the hand, 1.4 m.
35True or false
A spring that obeys Hooke’s law gives a straight-line graph of force against extension that passes through the origin.
Show answer
Answer: True
F = ke means force is directly proportional to extension: a straight line through (0, 0) with gradient k.
Common mistakeSome learners plot force against total LENGTH and expect a line through the origin; only extension is proportional to force.
36Multiple choice · ★ Challenge
Four bottles have these base diameters and centre-of-gravity heights: A 6 cm, 10 cm; B 8 cm, 16 cm; C 5 cm, 6 cm; D 10 cm, 25 cm. Which bottle can be tilted through the largest angle before it topples?
1. AA
2. BC
3. CB
4. DD
Show answer
Answer: B. C
Toppling angle: tan θ = (base/2) ÷ height. A: 3/10 = 0.30; B: 4/16 = 0.25; C: 2.5/6 = 0.42; D: 5/25 = 0.20. C has the largest ratio (θ ≈ 23°).
Common mistakeChoosing D because it has the widest base ignores how high its centre of gravity is; stability depends on the ratio of half-base to height.
37Short answer · ★ Challenge
Two equal and opposite parallel forces F act a perpendicular distance d apart. Show that their total moment is the same about any point between the two forces, and also about a point outside them.
Show answer
Model answer: Point between them, at distance x from one force: both forces turn the same way, total moment = Fx + F(d − x) = Fd. Point outside, at distance x from the nearer force: the forces turn opposite ways, total moment = F(x + d) − Fx = Fd. In every case the moment is Fd, independent of the point chosen.
Common mistakeSome learners take the moment of only one force; a couple must always be treated as a pair, and then the result never depends on the point.
38Short answer
Explain why a steering wheel acted on by a couple is not in equilibrium, even though the resultant force on it is zero.
Show answer
Model answer: A couple gives zero resultant force, so the first condition (ΣF = 0) is met and the wheel does not accelerate along a line. But the couple has a resultant moment Fd that is not zero, so the second condition (Σ moments = 0) fails and the wheel starts to rotate.
Common mistakeSaying "zero force means equilibrium" ignores the second condition; both the resultant force and the resultant moment must be zero.
39Short answer · ★ Challenge
A canoe on Lake Ruhondo is paddled due north at 2.4 m/s relative to the water, while a steady wind-driven current carries the water east at 0.70 m/s. Find the canoe’s resultant velocity.
Show answer
Model answer: The two velocities are perpendicular: v = √(2.4² + 0.70²) = √(5.76 + 0.49) = √6.25 = 2.5 m/s. Direction: tan θ = 0.70 ÷ 2.4 = 0.292, θ = 16.3° east of north.
Common mistakeAdding the speeds (2.4 + 0.70 = 3.1 m/s) treats the velocities as scalars; perpendicular vectors must be added with Pythagoras.
40Multiple choice
A spring in a kitchen scale stretches 2.0 cm under a load of 5.0 N. Assuming it obeys Hooke’s law, what load stretches it by 7.0 cm?
1. A17.5 N
2. B14 N
3. C12 N
4. D3.5 N
Show answer
Answer: A. 17.5 N
k = F/e = 5.0 ÷ 2.0 = 2.5 N/cm, so F = ke = 2.5 × 7.0 = 17.5 N.
Common mistakeChoosing 12 N adds the numbers (5 + 7); Hooke’s law says force is PROPORTIONAL to extension, so use the ratio.
41Short answer · ★ Challenge
A learner hangs loads on a spring: load 0, 2, 4, 6, 8, 10 N; extension 0, 1.0, 2.0, 3.0, 4.4, 6.5 cm. Up to what load does the spring obey Hooke’s law? Find its spring constant in N/m.
Show answer
Model answer: Up to 6 N the extension rises by 1.0 cm for every 2 N (proportional). After 6 N the steps are 1.4 cm and 2.1 cm, so the limit of proportionality is passed at about 6 N. k = 2 N ÷ 1.0 cm = 2.0 N/cm = 200 N/m.
Common mistakeUsing the last reading (10 N ÷ 6.5 cm) gives a wrong k because that point is beyond the limit of proportionality; use only the straight part.
42Fill in the blank
The energy stored in a stretched wire is equal to the ______ under its force–extension graph.
Show answer
Answer: area
Work done = force × distance; when the force changes, the work is the area under the force–extension graph, ½Fe for a straight line.
Common mistakeTaking the gradient of the graph gives the spring constant (or stiffness), not the energy stored.
43Short answer · ★ Challenge
A 50 N lamp hangs from two strings that are at right angles to each other; one string makes 30° with the vertical and the other 60° with the vertical. State Lami’s theorem and use it to find the tension in each string.
Show answer
Model answer: Lami: if three forces in equilibrium act at a point, each force is proportional to the sine of the angle between the other two. The angles are: between the strings 90°, between W and T₁ 180 − 30 = 150°, between W and T₂ 180 − 60 = 120°. So T₁/sin 120° = T₂/sin 150° = 50/sin 90°, giving T₁ = 50 × 0.866 = 43.3 N (string at 30°) and T₂ = 50 × 0.5 = 25 N (string at 60°).
Common mistakeA frequent slip is pairing each force with the angle next to it; in Lami’s theorem each force goes with the angle OPPOSITE it, between the other two forces.
44Short answer · ★ Challenge
A eucalyptus log lies on the ground. Lifting end A just off the ground (end B stays down) needs an upward force of 400 N; lifting end B instead needs 600 N. The log is 5.0 m long. Find its weight and the position of its centre of gravity.
Show answer
Model answer: Lifting A, moments about B: 400 × 5.0 = W × (distance of G from B). Lifting B, moments about A: 600 × 5.0 = W × (distance of G from A). Adding: (400 + 600) × 5.0 = W × 5.0, so W = 1000 N. Distance of G from A = 3000 ÷ 1000 = 3.0 m (2.0 m from B), nearer the heavier end B.
Common mistakeA common slip is to assume the log is uniform with G at 2.5 m; the two different lifting forces show that G is closer to the thicker end B.
45Multiple choice
A textbook rests on a table. Which pair of forces should appear on the free-body diagram of the BOOK?
1. AIts weight and the push of the book on the table
2. BIts weight and the normal reaction from the table
3. CThe normal reaction and the push of the book on the table
4. DIts weight and the pull of the book on the Earth
Show answer
Answer: B. Its weight and the normal reaction from the table
A free-body diagram shows only forces acting ON the chosen body: the Earth’s pull (weight) and the table’s upward push.
Common mistakeChoosing "push of the book on the table" adds a force the book exerts on something else; that belongs on the table’s diagram, not the book’s.
46Short answer · ★ Challenge
Two porters carry a uniform 2.0 m pole of weight 80 N on their shoulders, one at each end. A 120 N bunch of bananas hangs from the pole 0.50 m from the front end. How much force does each porter support?
Show answer
Model answer: Moments about the front porter: R_back × 2.0 = 80 × 1.0 + 120 × 0.50 = 80 + 60 = 140, so R_back = 70 N. Upward forces = downward forces: R_front = 80 + 120 − 70 = 130 N.
Common mistakeForgetting the pole’s own 80 N (acting at its centre) gives R_back = 30 N and R_front = 90 N; the weight of a uniform beam must be included.
47Short answer
Describe how to find the centre of gravity of a flat, irregular piece of cardboard cut in the shape of a map of Rwanda.
Show answer
Model answer: Make three small holes near the edge. Hang the card from a pin through one hole so that it swings freely, hang a plumb line from the same pin and draw a line along the thread when everything is still. Repeat from the second hole. The centre of gravity is where the lines cross; the third hole is used to check that its line passes through the same point.
Common mistakeBalancing the card on one finger and guessing gives only a rough point; the plumb lines work because the centre of gravity always hangs vertically below the point of suspension.
48Short answer · ★ Challenge
The force needed to draw a bow rises steadily from 0 to 240 N as the string is pulled back 0.50 m. Find the energy stored, and the speed of a 0.040 kg arrow if all this energy becomes its kinetic energy.
Show answer
Model answer: Energy = area under the force–draw graph = ½ × 240 × 0.50 = 60 J. ½mv² = 60, so v² = 2 × 60 ÷ 0.040 = 3000 and v = 54.8 m/s.
Common mistakeUsing 240 × 0.50 = 120 J forgets that the force builds up from zero; the energy is the triangular area, ½Fx.
49Multiple choice
A guitar string is stretched by 2.0 mm when the tension is raised steadily from 0 to 150 N, within its elastic limit. How much energy is stored in it?
1. A0.30 J
2. B0.15 J
3. C150 J
4. D0.075 J
Show answer
Answer: B. 0.15 J
E = ½Fe = ½ × 150 × 0.0020 = 0.15 J (the area of the triangle under the force–extension line).
Common mistakeChoosing 0.30 J uses F × e without the ½; the force grows from zero, so the average force is only half of the final value.
50Short answer · ★ Challenge
A wheel nut needs a moment of 90 N m to loosen it. A mechanic can push with at most 300 N at the end of his 0.25 m spanner. He slips a pipe over the handle so that he can push 0.75 m from the nut. Show whether each method works, and find the least force needed with the pipe.
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Model answer: Spanner alone: maximum moment = 300 × 0.25 = 75 N m < 90 N m, so the nut does not move. With the pipe: 300 × 0.75 = 225 N m > 90 N m, so it works. Least force with the pipe = 90 ÷ 0.75 = 120 N.
Common mistakeSome learners add the pipe length to the force instead of to the distance; the pipe increases the perpendicular distance, not the force.