A plane pin-jointed truss has 6 joints and 9 members, supported by a pin and a roller. The truss is

statically indeterminate to degree 1

determinate only if the loads are symmetric

Which of the following statements about zero-force members in a plane truss are correct?

Whether a given member carries zero force can depend on where the loads are applied

At a joint where three members meet, two of them collinear, and no load acts, the third carries zero force

Whether a given member carries zero force can depend on where the loads are applied

A zero-force member may be removed without affecting the truss under any loading

At a joint where three members meet, two of them collinear, and no load acts, the third carries zero force

A block of weight W rests on a plane inclined at 30°, with μ_s = μ_k = 0.3 between block and plane. No other force acts. The friction force on the block is

`0.500 W`, up the plane — it balances the weight component

zero, because the block is not in equilibrium

A solid cylinder and a thin hoop of the same mass and radius are released together from rest at the top of an incline and roll down without slipping. Which reaches the bottom first, and why?

The cylinder, because its smaller `k²/r²` leaves more of `g sin θ` for translation

The hoop, because its larger moment of inertia stores more energy

The cylinder, because its smaller `k²/r²` leaves more of `g sin θ` for translation

They arrive together, because both have the same mass and radius

The cylinder, but only if the incline is steeper than the angle of repose

A rod of length L has one end sliding along the floor and the other along a vertical wall, making angle θ with the wall. If the floor end moves at speed `v_A`, the wall end moves at speed

`v_A`, because the rod is rigid

Engineering Mechanics: Free Bodies, Trusses, Friction and Rigid-Body Dynamics

Section 2 of the Mechanical Engineering paper opens with the subject that everything after it borrows from, and it is worth being blunt about what the examinable skill actually is: almost every mark in engineering mechanics is won or lost when you draw the free-body diagram, not when you solve the equations. A free body is a closed boundary drawn around some part of the structure, with every force that crosses that boundary put back on the sketch — and the two habitual errors are forgetting a reaction and inventing one. Once the sketch is right, statics gives you exactly three equations in the plane, ΣF_x = 0, ΣF_y = 0 and ΣM = 0 about any point you like, and choosing the point well is most of the arithmetic saved: take moments about the intersection of two unknowns and both vanish. For a pin-jointed truss the count comes first — m = 2j − 3 for a determinate plane truss — and then you have a choice of two tools that answer different questions. The method of joints walks the whole structure and gives you every member; the method of sections cuts through at most three unknowns and gives you one member in a single moment equation, which is what an exam actually asks for. Friction is where candidates lose marks to a definite-sounding formula: F = μN is the friction force only at impending motion or during sliding; below that threshold friction is whatever equilibrium requires and no more, so the first question is always is it moving? Then dynamics, which is three conservation statements applied to whichever unknown you were given: work–energy when the question mentions distance, impulse–momentum when it mentions time, and both together with a coefficient of restitution when two bodies collide. Rolling without slipping is the one place where translation and rotation are locked together — v = ωr — and the acceleration down an incline depends on nothing but the shape, through k²/r².

Free Bodies, Determinacy, and the Two Ways to Read a Truss

Before any of that, the free-body diagram. Draw a closed boundary round the part you are analysing, put back every force that crosses it, and resolve nothing until the sketch is complete — a forgotten reaction and an invented one fail in exactly the same way and neither shows up later as an inconsistency. A pin-jointed truss is then an idealisation with exactly one useful consequence: every member carries only an axial force, because a frictionless pin cannot transmit a moment and loads are applied only at joints. So each member is either in tension (pulling its two joints towards each other, counted positive here) or in compression (pushing them apart, negative), and each joint gives you two scalar equations. Count before you compute. With m members, j joints and three support reactions, the unknowns number m + 3 and the equations 2j, so a statically determinate plane truss satisfies m = 2j − 3. Fewer members and it is a mechanism — it will collapse, and no set of member forces exists. More and it is indeterminate — the forces depend on how stiff each member is, which statics alone cannot tell you.

🎯 The truss this chapter keeps returning to, solved twice

Take a three-panel Pratt truss of span 9 m in panels of 3 m, height 4 m, pinned at A and on a roller at D. Bottom chord A(0,0)–B(3,0)–C(6,0)–D(9,0), top chord E(3,4)–F(6,4), members AB, BC, CD, EF, AE, BE, BF, CF, DF. Check first: m = 9, j = 6, 2j − 3 = 9 ✓, so it is determinate and the diagonals are length 5 with direction cosines 3/5 and 4/5 — which is why the arithmetic below stays in halves and thirds. Load 10 kN at B and 10 kN at C. Symmetry gives Ay = Dy = 10 kN. Joint A has only AB and AE, so 10 + (4/5)F_AE = 0 gives F_AE = −12.5 kN (compression) and then F_AB = −(3/5)(−12.5) = +7.5 kN (tension). Joint E has AE, BE and EF: the vertical component of AE arriving at E is 10 kN up, so F_BE = +10 kN and F_EF = −7.5 kN. Now joint B, and this is the part worth slowing down for: the load there is 10 kN down, BE brings exactly 10 kN up, so the diagonal has nothing left — F_BF = 0. Now change only the load: 20 kN at B and nothing at C. The reactions become 13.333 and 6.667, and the same joint walk gives F_BF = +8.333 kN while F_CF drops to zero. The member did not change. Its length, its angle and its neighbours did not change. A zero-force member is a property of the loading, not only of the geometry — and a candidate who has memorised the two geometric shortcuts without that will confidently zero out a member that is carrying 8 kN.

The 9 m Pratt truss: every member force, for two loadings (+ tension, − compression, kN)
Member	10 kN at B, 10 kN at C	20 kN at B only
AB	+7.5	+10
BC	+7.5	+5
CD	+7.5	+5
EF	−7.5	−10
AE	−12.5	−16.667
BE	+10	+13.333
BF	0 — zero-force	+8.333
CF	+10	0 — zero-force
DF	−12.5	−8.333
Reactions Ay, Dy	10, 10	13.333, 6.667

The method of sections answers the question an exam actually asks, which is never give me all nine members but what is the force in BC. Cut the truss with a line crossing BC, BF and EF — three unknowns, which is the most a single cut can resolve — throw away the right-hand part, and take moments about the point where two of the three unknowns meet. BF and EF both pass through E(3, 4), so moments about E leave BC alone: the left part carries Ay = 10 kN up at A(0,0) and the 10 kN load down at B(3,0). The load acts through a point directly below E and contributes nothing; the reaction contributes 10 × 3 = 30 kN·m; and F_BC acts horizontally 4 m below E, so 4 F_BC = 30 and F_BC = +7.5 kN, matching the full solve. One equation instead of walking four joints. The rule for choosing the cut is mechanical: cut no more than three unknown members, then take moments about their point of concurrency.

🧠 Virtual work: the method that never draws a reaction it does not need

Virtual work is the alternative to free bodies, and it earns its place on exactly one kind of problem: a mechanism with one degree of freedom where you want one force and nothing else. The statement is that for a system in equilibrium the total work done by all forces through any small kinematically admissible displacement is zero — and the phrase in bold is what makes it useful, because a displacement that respects the constraints does no work against the constraint forces, so every pin reaction, every smooth-surface normal and every internal force in a rigid link drops out of the equation before you write it. Work one. A toggle press has two equal links pinned at an apex, each at angle θ to the horizontal, carrying a load W downwards at the apex while a horizontal force P is applied at a sliding end. Give θ a virtual increase δθ. The apex rises by L cos θ · δθ and the slider moves in by 2L sin θ · δθ, so W L cos θ · δθ = P · 2L sin θ · δθ and P = W/(2 tan θ) — the link length cancels, and no pin force was ever computed. Read the numbers to see why toggles exist: at θ = 45° you need 0.500 W, at 30° 0.866 W, and at 15° 1.866 W. The mechanical advantage runs the other way as the links flatten towards θ → 0, where P → 0 holds any load at all, which is the principle of a toggle clamp and a rock crusher. The honest limit: virtual work gives you one equation per degree of freedom, so it cannot deliver a full set of member forces and is the wrong tool for a truss. Use free bodies when you want everything and virtual work when you want one thing cheaply.

🧠 The two geometric zero-force rules, and their exact scope

Both rules are statements about an unloaded joint, and both are worth stating with the qualifier attached, because the qualifier is what the previous note showed being violated. Rule one: at an unloaded joint where exactly two non-collinear members meet, both are zero — resolve along a direction perpendicular to one and the other must vanish. Rule two: at an unloaded joint where three members meet and two of them are collinear, the third is zero, and the two collinear ones carry equal force. Attach the words unloaded joint to both and they never mislead. Detach them and you get the standard error: applying rule two at a joint that carries an external load, where the third member carries precisely that load resolved along its own direction. In the truss above, joint B has four members and a load, so neither rule applies at all — F_BF = 0 there had to be computed, and it changed the moment the load did.

Friction: Why the First Question Is Always “Is It Moving?”

Coulomb friction is stated as an inequality, and treating it as an equation is the most reliable way to lose a mark in this section. The static law is F ≤ μ_s N: friction supplies whatever tangential force equilibrium demands, up to a ceiling proportional to the normal reaction, and only at impending motion does it sit exactly at μ_s N. Once sliding starts the law does become an equation, F = μ_k N, directed against the relative motion. So the procedure has a branch in it. Compute the friction force equilibrium requires, compare it with μ_s N, and only then decide which case you are in. A block of weight W on a 30° incline with μ_s = 0.3 requires W sin 30° = 0.5 W along the slope while the ceiling is 0.3 W cos 30° = 0.2598 W — the requirement exceeds the ceiling, so it slides, and any question that asks for "the friction force" wants μ_k N, not W sin θ. Turn it around and the same numbers answer the design question: to hold the block you must supply the shortfall, W(sin 30° − 0.3 cos 30°) = 0.2402 W up the slope; to push it steadily up, you fight both gravity and friction, W(sin 30° + 0.3 cos 30°) = 0.7598 W. Three times the force, for the same block, because friction reversed direction.

The friction branch, worked on one block: W on 30°, μ = 0.3
Question asked	What to compute	Answer
Does it stay put?	Compare `tan θ` with `μ`	`tan 30° = 0.5774 > 0.3` — no, it slides
Friction force while sliding	`μ N = μ W cos θ`, not `W sin θ`	`0.2598 W`
Force up the slope to just hold it	`W(sin θ − μ cos θ)`	`0.2402 W`
Force to drag it steadily up	`W(sin θ + μ cos θ)`	`0.7598 W` — 3.16× the holding force
Angle of repose	`θ = arctan μ`, independent of W	`arctan 0.3 = 16.70°` — well below 30°

🎯 The ladder, and where the wall being smooth actually enters

A uniform ladder of length L and weight W leans against a smooth vertical wall with its foot on a floor of coefficient μ. Find the smallest angle θ from the horizontal at which it does not slip. Three unknowns — the wall reaction N_w (horizontal only, because the wall is smooth), the floor normal N_f and the floor friction F. Vertical equilibrium gives N_f = W immediately, since the wall pushes only sideways. Horizontal equilibrium gives F = N_w. Moments about the foot, with the weight at mid-span: W (L/2) cos θ = N_w L sin θ, so N_w = (W/2) cot θ. At impending slip F = μ N_f = μW, so μW = (W/2) cot θ and tan θ = 1/(2μ) — the weight and the length both cancel, which is the result worth remembering: a heavier ladder is no more likely to slip, and neither is a longer one. With μ = 0.25 the answer is arctan 4 = 63.43°. Two things are worth noticing about the shape of this. First, W cancelling is what lets a person climb the ladder and change the answer: the climber adds weight away from mid-span, which changes the moment arm and not just the total, so the safe angle rises. Second, the wall being smooth is not a simplification you can drop casually — the moment a wall carries friction, vertical equilibrium no longer hands you N_f = W, the problem gains an unknown, and it becomes statically indeterminate unless the wall too is at impending slip.

Dynamics: Pick the Conservation Law the Question Already Named

Every particle-dynamics question in this paper is answerable by F = ma and enough patience, and almost none of them should be. The three integrated forms exist precisely so you can skip the acceleration, and the question tells you which one to use by naming its own variable. Distance appears → work–energy: ½mv₂² − ½mv₁² = ΣW, and work done by a constant force is F·d cos θ, by a spring ½kx², by friction −μNd and always negative. Time appears → impulse–momentum: m v₂ − m v₁ = ∫F dt, and when no external impulse acts on a system, momentum is conserved whatever happens internally. Two bodies collide → both, plus restitution: momentum is conserved through any collision, but energy is not, and the coefficient of restitution e = (v₂ − v₁)/(u₁ − u₂) supplies the second equation. e = 1 is perfectly elastic and conserves kinetic energy; e = 0 is perfectly plastic and the bodies leave together. Nothing in between conserves energy, which is the point of the coefficient.

🎯 A collision solved, and the energy audit that goes with it

Take m₁ = 2 kg at +6 m/s meeting m₂ = 3 kg at −2 m/s head on, with e = 0.6. Momentum before is 2(6) + 3(−2) = +6 kg·m/s. Restitution says the separation speed is 0.6 × the approach speed, and the approach speed is 6 − (−2) = 8 m/s, so v₂ − v₁ = 4.8 m/s. Two equations, two unknowns: v₁ = −1.68 m/s and v₂ = +3.12 m/s. Check both — momentum after is 2(−1.68) + 3(3.12) = +6.00 ✓ and (3.12 − (−1.68))/8 = 0.6 ✓. The light body has bounced back, which is worth noticing: it arrived at +6 and leaves at −1.68. Now the energy audit, because that is what a 2-mark question asks. Before: ½(2)(36) + ½(3)(4) = 42 J. After: ½(2)(1.68²) + ½(3)(3.12²) = 17.424 J. So 24.576 J — 58.5 per cent of the kinetic energy — is gone, into deformation, heat and sound. Two habits to take from the numbers. First, momentum is a vector and its sign carries information, so a collision problem done with speeds instead of velocities will silently give the wrong answer. Second, the energy loss is not a small correction here: at e = 0.6 most of the kinetic energy has left the system, and the only value of e at which none does is exactly 1.

Rigid bodies add rotation, and the useful simplification is that plane motion is a translation of the centre of mass plus a rotation about it: ΣF = m a_G and ΣM_G = I_G α, and the two are independent unless a constraint ties them. Rolling without slipping is exactly that constraint, v_G = ωr and a_G = αr, and it turns two unknowns into one. Roll a body of mass m, radius r and radius of gyration k down an incline θ: taking moments about the contact point, which is the instantaneous centre and therefore kills the friction force, gives a_G = g sin θ / (1 + k²/r²) — a result in which the mass and the radius have both cancelled. Only the shape survives, through k²/r², and that is why a solid cylinder always beats a hoop down a slope regardless of how heavy either one is. The friction needed to sustain the roll is μ ≥ tan θ/(1 + r²/k²), which for a solid cylinder is tan θ/3 — just 0.1925 at 30°, so rolling is easy to arrange and slipping is the special case.

Rolling down an incline: acceleration depends on shape alone
Body	`k²/r²`	`a_G` as a fraction of `g sin θ`	Minimum μ to roll at 30°
Thin hoop	1	1/2 = 0.5000	0.2887
Thin spherical shell	2/3	3/5 = 0.6000	0.2309
Solid cylinder	1/2	2/3 = 0.6667	0.1925
Solid sphere	2/5	5/7 = 0.7143	0.1650
Sliding block, μ = 0	0	1 — nothing to spin up	n/a

🧠 The instantaneous centre, and a two-line answer it gives away

For a rigid body in plane motion there is at every instant one point — possibly outside the body — whose velocity is zero, and about which the whole body is momentarily rotating. Find it and every velocity in the body follows from v = ω × r with no simultaneous equations at all. Locating it is geometric: draw perpendiculars to two known velocity directions and they meet at the instantaneous centre. A rod of length 2 m has one end sliding on the floor and the other on a vertical wall, at 30° from the wall, and the floor end moves at 1 m/s. The floor end moves horizontally, the wall end vertically, so the perpendiculars are a vertical and a horizontal line through those ends and the instantaneous centre sits at the corner of the rectangle they form. Then ω = v_A/(L cos 30°) = 1/(2 × 0.866) = 0.5774 rad/s, and the wall end has speed L sin 30° × ω = 0.5774 m/s, which is exactly v_A tan 30° — the length has cancelled again. Sanity-check the limits, because they are free marks: as the rod approaches the wall the top end slows to nothing, and as it flattens onto the floor the top end runs away to infinity. Any algebra that does not reproduce those two limits has a sine and a cosine swapped.

Key takeaways

Draw the free body before you write an equation. Three planar equations exist; taking moments about the intersection of two unknowns removes both. Determinacy first: m = 2j − 3 for a plane truss.
A zero-force member is a property of the loading, not just the geometry. In the 9 m Pratt truss, BF carries 0 under 10 + 10 kN and +8.333 kN under 20 kN at B alone. Both zero-force rules apply only at an unloaded joint.
Use sections, not joints, when one member is asked for. Cut at most three unknowns and take moments about their point of concurrency — F_BC = +7.5 kN in one equation instead of four joints.
F = μN holds only at impending motion or during sliding. Compare what equilibrium requires with μ_s N first. On 30° with μ = 0.3 the block slides, holding needs 0.2402 W and dragging it up needs 0.7598 W.
Let the question name the conservation law: distance → work–energy, time → impulse–momentum, collision → momentum plus restitution. Momentum survives every collision; kinetic energy survives only e = 1.
Rolling down an incline forgets mass and radius. a_G = g sin θ/(1 + k²/r²) gives 1/2, 3/5, 2/3 and 5/7 of g sin θ for hoop, shell, cylinder and sphere — shape is the only variable that survives.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A plane pin-jointed truss has 6 joints and 9 members, supported by a pin and a roller. The truss is
1. statically determinate
2. statically indeterminate to degree 1
3. a mechanism — it will collapse
4. determinate only if the loads are symmetric
Show answer
Answer: A — statically determinate
Statically determinate. A pin gives two reaction components and a roller one, so there are three support unknowns and nine member unknowns, twelve in all. Each of the six joints supplies two equilibrium equations, so there are twelve equations — and m = 2j − 3 reads 9 = 12 − 3 ✓. Option D confuses two independent things: determinacy is a property of the structure, and the loading never enters it. The count is of members, joints and reactions only, which is why you can settle it before reading what the loads are. Option B would need a tenth member or a second roller; option C a member removed, after which some joint could move without stretching anything and no equilibrium solution would exist. One caveat the count cannot catch: a truss can satisfy m = 2j − 3 and still be unstable if the members are badly arranged — three collinear members at a joint, say — so the count is necessary and not sufficient.
In the three-panel Pratt truss of span 9 m (panels 3 m, height 4 m, pin at A, roller at D), vertical loads of 10 kN act at B and at C. The force in the bottom-chord member BC, in kN, is ______ (positive for tension)
Numerical answer — type the value.
Show answer
Answer: 7.5
+7.5 kN, tension. Use the method of sections and it is one equation. Symmetry gives Ay = Dy = 10 kN. Cut through BC, BF and EF and keep the left part. Take moments about E(3, 4), the point where the other two unknowns cross: F_EF passes through it and F_BF passes through it, so both drop out. The reaction acts 3 m to the left of E and contributes 10 × 3 = 30 kN·m; the 10 kN load at B(3, 0) lies directly below E, so its moment arm is zero and it contributes nothing at all; and F_BC acts horizontally along y = 0, 4 m below E, with arm 4. Hence 4 F_BC = 30 and F_BC = 7.5 kN, positive so tension — as a bottom chord in sagging should be. The full 12-unknown matrix solve agrees to fourteen decimal places. Two lessons for the exam. First, choosing E rather than any other point is the whole method — pick B and you carry F_EF around for nothing. Second, notice how much the awkward-looking load did not cost you: it sat on the line of action through the moment centre, so it never entered the equation.
Which of the following statements about zero-force members in a plane truss are correct?
1. At an unloaded joint where exactly two non-collinear members meet, both carry zero force
2. Whether a given member carries zero force can depend on where the loads are applied
3. A zero-force member may be removed without affecting the truss under any loading
4. At a joint where three members meet, two of them collinear, and no load acts, the third carries zero force
Show answer
Answer: A — At an unloaded joint where exactly two non-collinear members meet, both carry zero force; B — Whether a given member carries zero force can depend on where the loads are applied; D — At a joint where three members meet, two of them collinear, and no load acts, the third carries zero force
A, B and D. A and D are the two standard rules, and both carry the words unloaded joint for a reason — attach a load and neither survives. B is the statement this chapter is built around, and it is the one most often got wrong: in the 9 m Pratt truss, member BF carries 0 kN under 10 kN at B and 10 kN at C, and +8.333 kN under 20 kN at B alone. Same member, same geometry, different load, different answer. C is false, and dangerously so. Being zero-force under one loading says nothing about another, so removing BF would leave the truss with m = 8, 2j − 3 = 9 — a mechanism, which collapses under the very next asymmetric load. Real trusses keep such members precisely because the load case that makes them redundant is not the only one they will see; a diagonal that is idle under symmetric live load is what carries a one-sided wind load. If a question ever does ask you to remove a zero-force member, it is asking about that single load case, and the determinacy count is the check that tells you what you have destroyed.
A block of weight W rests on a plane inclined at 30°, with μ_s = μ_k = 0.3 between block and plane. No other force acts. The friction force on the block is
1. 0.500 W, up the plane — it balances the weight component
2. 0.260 W, up the plane
3. 0.300 W, up the plane
4. zero, because the block is not in equilibrium
Show answer
Answer: B — `0.260 W`, up the plane
0.260 W up the plane. Run the branch. Equilibrium would require W sin 30° = 0.500 W along the slope, but the static ceiling is μ_s N = 0.3 W cos 30° = 0.2598 W. The requirement exceeds the ceiling, so equilibrium is impossible — the block slides, and once it is sliding the friction force is μ_k N = 0.3 W cos 30° = 0.2598 W ≈ 0.260 W, directed up the plane against the motion. Option A is the trap, and it is the answer you get by assuming equilibrium instead of testing for it: W sin θ is what friction would have to be, not what it is. Option C forgets the cos 30° — friction is proportional to the normal reaction, which on an incline is W cos θ and not W. Option D gets the physics half right, correctly spotting that the block is not in equilibrium, and then draws the wrong conclusion: sliding friction is at its maximum, not zero. The cheap first check is the angle of repose, arctan 0.3 = 16.7°: any slope steeper than that and this block is going to move.
A uniform ladder rests against a smooth vertical wall, its foot on a floor with μ = 0.25. The minimum angle it can make with the horizontal without slipping, in degrees, is ______ (round to two decimals)
Numerical answer — type the value.
Show answer
Answer: 63.43
63.43°. Three unknowns and three equations. The wall is smooth, so it pushes horizontally only, which means vertical equilibrium hands you N_f = W at once. Horizontal equilibrium gives F = N_w. Moments about the foot, with the weight acting at mid-span: W (L/2) cos θ = N_w L sin θ, so N_w = (W/2) cot θ. At impending slip F = μ N_f = μW, and equating gives μW = (W/2) cot θ, that is tan θ = 1/(2μ). With μ = 0.25 that is tan θ = 2, so θ = arctan 2 = 63.4349°. The structure of the result is worth more than the number: both W and L cancel, so a heavier ladder is no safer and a longer one no worse — a genuinely counter-intuitive fact that this problem exists to teach. What does change the answer is a person on the ladder, because their weight acts away from mid-span and alters the moment arm rather than just the total load, pushing the safe angle up. And the smooth wall is not decoration: give the wall friction and vertical equilibrium no longer yields N_f = W, a fourth unknown appears, and the problem is statically indeterminate unless the wall is at impending slip too.
A 2 kg body moving at 6 m/s strikes a 3 kg body moving at 2 m/s in the opposite direction, head on. The coefficient of restitution is 0.6. The kinetic energy lost in the impact, in joules, is ______ (round to two decimals)
Numerical answer — type the value.
Show answer
Answer: 24.58
24.58 J. Set the first body moving in the + direction, so u₁ = +6 and u₂ = −2 — and getting that sign right is most of the question. Momentum: 2(6) + 3(−2) = +6 kg·m/s, conserved. Restitution: the approach speed is u₁ − u₂ = 8 m/s, so the separation speed is 0.6 × 8 = 4.8, giving v₂ − v₁ = 4.8. Solving the pair: v₁ = −1.68 m/s (the lighter body bounces back) and v₂ = +3.12 m/s. Check momentum after: 2(−1.68) + 3(3.12) = +6.00 ✓. Energy before is ½(2)(36) + ½(3)(4) = 36 + 6 = 42 J; after, ½(2)(2.8224) + ½(3)(9.7344) = 2.8224 + 14.6016 = 17.424 J. The loss is 42 − 17.424 = 24.576 J, which is 58.5 per cent of what went in — not a small correction. Two traps. Using speeds instead of signed velocities gives an approach speed of 4 m/s and the whole answer collapses. And a candidate who reaches for e = 1 out of habit conserves energy and answers 0, which is what the coefficient of restitution exists to rule out: e = 1 is the only value at which no energy is lost, and any e < 1 means the collision has consumed some.
A solid cylinder and a thin hoop of the same mass and radius are released together from rest at the top of an incline and roll down without slipping. Which reaches the bottom first, and why?
1. The hoop, because its larger moment of inertia stores more energy
2. The cylinder, because its smaller k²/r² leaves more of g sin θ for translation
3. They arrive together, because both have the same mass and radius
4. The cylinder, but only if the incline is steeper than the angle of repose
Show answer
Answer: B — The cylinder, because its smaller `k²/r²` leaves more of `g sin θ` for translation
The cylinder, because of k²/r². Moments about the contact point give a_G = g sin θ/(1 + k²/r²), in which mass and radius have both cancelled. For a solid cylinder k²/r² = 1/2, so a_G = (2/3) g sin θ; for a thin hoop k²/r² = 1, so a_G = (1/2) g sin θ. The cylinder accelerates a third faster and wins by a margin that does not depend on how heavy either body is. Option A has the physics of energy partition right — the hoop does put a larger share of the released potential energy into rotation — but draws the opposite conclusion from it: energy going into spin is energy not going into forward speed, so a larger moment of inertia is a handicap, not an advantage. Option C is the intended trap, and it fails because mass and radius are exactly the two things that cancel; distribution of mass is what survives. Option D confuses the condition for rolling at all with the ranking once it rolls: rolling needs μ ≥ tan θ/3 for the cylinder and tan θ/2 for the hoop, which is a separate question, and above those thresholds the ordering never changes.
A rod of length L has one end sliding along the floor and the other along a vertical wall, making angle θ with the wall. If the floor end moves at speed v_A, the wall end moves at speed
1. v_A tan θ
2. v_A cot θ
3. v_A, because the rod is rigid
4. v_A L sin θ
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Answer: A — `v_A tan θ`
v_A tan θ. Locate the instantaneous centre: the floor end moves horizontally and the wall end vertically, so the perpendiculars to those two velocities are a vertical line through the floor end and a horizontal line through the wall end, meeting at the far corner of the rectangle the rod spans. With θ measured from the wall, the floor end sits L sin θ from the wall and the wall end L cos θ from the floor, so ω = v_A/(L cos θ) and the wall end has speed L sin θ × ω = v_A tan θ. The length cancels, which is the check worth running: velocity ratios in a rigid body depend on angles, not on size. Take L = 2 m, θ = 30°, v_A = 1 m/s: ω = 0.5774 rad/s and the wall end moves at 0.5774 m/s, matching tan 30°. Option C is the standard error — rigidity fixes the distance between the ends, not their speeds, and the two ends of a rotating body generally move at different rates. Option B is the same answer with θ measured from the floor instead of the wall, which is why reading the figure matters as much as the algebra. Check the limits: as the rod nears the wall (θ → 0) the wall end stops, and as it flattens (θ → 90°) it runs away without bound — tan θ does both, cot θ does neither.