A block of base width b and height h rests on a plane whose inclination is slowly increased. The coefficient of friction is μ. The block will overturn before it slides if

It depends on the weight of the block

The principle of virtual work is applied to a simply supported beam by releasing the roller at B and giving it a unit virtual displacement. The chief advantage over taking moments about A is that

The reaction at A does no work and never enters the equation

It gives a different, more accurate value of the reaction

The reaction at A does no work and never enters the equation

It applies to indeterminate structures where statics fails

It removes the need to know where the loads act

On a beam carrying both a UDL and a concentrated load, at the section where the concentrated load acts,

the shear force is discontinuous and the bending moment is continuous with a change of slope

both the shear force and the bending moment are discontinuous

the shear force is discontinuous and the bending moment is continuous with a change of slope

the bending moment is discontinuous and the shear force is continuous

both are continuous, since the beam itself is continuous

A pile hammer of mass 500 kg is dropped freely through 2 m onto a pile and is brought to rest in 0.05 s. Which of the following are correct? (One or more)

The impact velocity is about 6.26 m/s

The average resisting force is about 62.6 kN

The average force is about 12.8 times the weight of the hammer

The impact velocity is about 6.26 m/s

The average resisting force is about 62.6 kN

The average force is about 12.8 times the weight of the hammer

Halving the stopping time would leave the average force unchanged

Engineering Mechanics: Equilibrium, Internal Forces and Simple Stresses

Structural Engineering is the heaviest section of the Civil paper, and it opens with the two skills every later chapter spends: reading the internal forces at a section, and knowing which material constant to reach for. The syllabus words the first as internal forces in structures, and that phrase is worth taking literally. Cut a member anywhere and the part you throw away is replaced by exactly three quantities in the plane — an axial force, a shear force and a bending moment — and every diagram you will ever draw is those three plotted along the length. The relations between them are not three separate facts but one: dV/dx = −w and dM/dx = V, so the moment is stationary where the shear vanishes, load intensity is the slope of the shear diagram, and a point load puts a step in the shear and a kink in the moment. That last case is where candidates lose the mark, because a shear diagram that jumps through zero at a point load still locates the maximum moment even though V = 0 has no solution there. Friction is the other place marks go, and for a structure the question is rarely what is μ but which failure comes first — a body on a slope slides when tan θ > μ and tips when tan θ > b/h, and whichever threshold is lower decides. The principle of virtual work is the section's quiet workhorse: give the structure a displacement it is allowed to have, set the total work to zero, and any reaction or member force falls out without the reactions you did not ask about ever appearing. Then stress and strain, where the whole elastic story is two independent constants: pick any two of E, G, K and ν and the other two follow from G = E/2(1+ν) and K = E/3(1−2ν).

Free Bodies, Equilibrium, and the Three Internal Forces

Start with the free-body diagram, and treat it as the answer rather than the preamble. Draw a closed boundary round the part you are keeping, put back every force that crosses that boundary, and resolve nothing until the sketch is complete. A system of forces acting on a rigid body in the plane then gives you exactly three independent equations — ΣF_x = 0, ΣF_y = 0 and ΣM = 0 about any point whatever — and the freedom in that last phrase is the arithmetic you save: take moments about the point where two unknowns intersect and both drop out of the equation. A support is nothing more than a promise about which of those three a connection can supply. A roller supplies one force normal to its surface; a pin supplies two; a fixed end supplies two forces and a moment. Count the reactions against the equations before computing anything: three reactions and three equations means statically determinate and statics alone will finish the job, while four reactions means one equation short and the answer now depends on how stiff the members are.

Now cut the member. Whatever you removed is replaced by three quantities acting on the face you kept: an axial force N along the member, a shear force V across it, and a bending moment M about the section. Plot each along the length and you have the three diagrams the rest of this section lives on. What makes them quick is that they are not independent. Differentiate along the beam and dV/dx = −w and dM/dx = V, from which three working rules follow immediately: a length carrying no load has constant shear and therefore a straight moment diagram; a uniformly distributed load makes the shear straight and the moment parabolic; and the moment is stationary exactly where the shear passes through zero. Read that last one as a search instruction, because it turns "where is the maximum moment?" into "where does the shear change sign?", and the second question is answered by inspection of a diagram you have already drawn.

⚠️ The beam this chapter keeps returning to, and the zero that is not there

Take a simply supported beam of span 8 m carrying a UDL of 12 kN/m over its whole length and a 40 kN point load 3 m from A. The UDL totals 96 kN acting at 4 m, so moments about A give R_B × 8 = 96(4) + 40(3) = 504, hence R_B = 63 kN and R_A = 73 kN — and the check 73 + 63 = 136 = 96 + 40 costs a second and catches most slips. Now hunt for the maximum moment the mechanical way, by setting the shear to zero. Left of the load the shear is 73 − 12x, which vanishes at x = 6.083 m — but that is not left of the load, so it is not a real root. Right of the load the shear is 73 − 12x − 40, which vanishes at x = 2.75 m — and that is not right of the load either. Neither equation has an admissible root, and a candidate who trusts the algebra over the diagram concludes there is no maximum. What actually happens is visible the moment you plot it: the shear runs +73 at A, falls to +37 just left of the load, drops by the full 40 kN to −3 just right of it, and continues to −63 at B. It changes sign by jumping over zero, at x = 3, and that is where the moment peaks: M = 73(3) − 12(3)(1.5) = 165 kN·m, confirmed from the other end as 63(5) − 12(5)(2.5) = 165 ✓. Midspan, where an unwary candidate looks first, carries only 156 kN·m. The rule to carry away is the honest form of the one in every textbook: the maximum moment is where the shear changes sign, which is not always where it equals zero.

The 8 m beam, read at five sections (sagging moment positive)
Section	Shear V (kN)	Moment M (kN·m)
Just right of A	+73	0
x = 2 m	+49	122
x = 3 m, just left of the load	+37	165 — the maximum
x = 3 m, just right of the load	−3	165 — no jump in M
x = 4 m (midspan)	−15	156
Just left of B	−63	0

Two features of that table are worth naming because they are what the examiner tests. First, the shear jumps by exactly the point load and the moment does not jump at all — a concentrated force is a step in V and only a change of slope in M, which is why the moment reads 165 on both sides of the section. A concentrated moment behaves the other way round: it steps the moment diagram and leaves the shear alone. Second, the moment at the load is larger than at midspan even though midspan is where the UDL alone would peak, because the two loadings put their maxima in different places and the sum peaks between them, at the discontinuity. Superposition is legitimate here — the beam is linear elastic and the loads are independent — but superposing the two maximum values is not, because they occur at different sections. Add the diagrams, then find the maximum; never add the maxima.

Friction: Which Failure Comes First — and Virtual Work

F = μN is the most over-applied formula in the section. It gives the friction force only at impending motion or while sliding; below that threshold friction is whatever equilibrium demands and no more, so the first question is always whether the body is moving at all. But for civil structures there is a second question that matters more, and the syllabus phrase friction and its applications is pointing at it: a body resting on a slope, a retaining wall pushed by soil, a gravity dam pushed by water — each can fail in two independent ways, and the design is governed by whichever comes first. It can slide, which happens when the driving force exceeds μN; or it can overturn, which happens when the resultant leaves the base. Neither implies the other, and checking only one is how a structure gets designed against the wrong failure.

🎯 The same block slides at one height and tips at another

A rectangular block of base b = 0.6 m and height h = 1.2 m sits on a plane whose angle is slowly raised, with μ = 0.4. Two thresholds, and both are pure geometry. It slides when the component of weight down the slope beats friction: W sin θ > μ W cos θ, that is tan θ > μ = 0.4, so at θ = 21.80°. It tips when the weight vector, which acts through the centroid at height h/2, falls outside the lower corner: that happens when tan θ > b/h = 0.5, so at θ = 26.57°. Since 0.4 < 0.5, this block slides first and never reaches the tipping angle. Now change one number and nothing else — make it slender at h = 2.0 m, keeping the same base and the same μ. Now b/h = 0.3, which is below μ, and the block tips first, at 16.70°, on a slope far too shallow to slide it. Notice what did the deciding: not the weight, which cancels out of both criteria, and not the material of the block, but the ratio b/h against μ. Slender bodies overturn, squat bodies slide — and that single comparison, μ against b/h, is the whole of it. It is also why a retaining wall is checked twice and why widening the base of a gravity dam is the standard cure for overturning while roughening its foundation is the cure for sliding: the two failures have two different remedies.

The principle of virtual work says that a structure in equilibrium does zero net work through any small displacement the supports permit. Used forwards it finds a reaction, and its advantage is entirely practical: the quantities you did not ask about do no work and never enter the equation. Return to the 8 m beam. Release the roller at B, let the beam rotate rigidly about the pin at A, and give B a virtual downward displacement of 1. Every point rides down in proportion to its distance from A, so the 40 kN load at 3 m moves 3/8 and the 96 kN UDL resultant at 4 m moves 4/8. Setting the total work to zero, R_B(1) = 40(0.375) + 96(0.5) = 15 + 48, so R_B = 63 kN — the same answer as the moment equation, reached without writing down either component of the pin reaction at A, because a pin that does not move does no work.

🧠 The unit load method, which is the same principle run backwards

Swap the roles and virtual work computes a deflection instead of a reaction. Apply a fictitious unit load at the point and in the direction whose movement you want, work out the internal forces m it causes, and equate its external work to the internal work done by those forces riding the real deformations: 1 × Δ = ∫ (m M / EI) dx for bending, or Σ (n N L / AE) for a truss. The unit load is a bookkeeping device and carries no units of its own; all it does is pick out the displacement you asked for. Take a cantilever of 3 m with 20 kN at the tip, E = 200 GPa, I = 9 × 10⁷ mm⁴. Real moment at distance x from the tip is M = −20x; the unit load gives m = −x; and ∫₀³ (20x²/EI) dx = 20L³/3EI, which is the familiar PL³/3EI and evaluates to exactly 10.0 mm. Two things this buys you that the standard formula does not. It handles a load the tables do not list, because the integral does not care how odd the loading is. And it gives a deflection anywhere, including at a point carrying no load at all — put the unit load there and the method still works, which is precisely the case a formula sheet cannot cover.

Centroids, Centre of Mass, and Impulse–Momentum

The centroid of a composite area is a weighted average and nothing more: x̄ = Σ(A_i x_i)/ΣA_i, with each part measured to its own centroid from a common origin. Two habits make it reliable. Choose the origin on a corner of the figure so no coordinate is negative unless it genuinely is, and treat a hole as an area with a minus sign rather than trying to reason about the remaining shape. For the section properties of the next chapter the centroid is not decoration — the neutral axis of a beam in pure bending passes through it, so every bending stress you compute depends on having located it correctly first.

ℹ️ An L-section whose centroid is in the air

Take an angle with a vertical leg 150 mm × 10 mm and a horizontal leg 100 mm × 10 mm, sharing the corner at the origin. Split it so nothing is counted twice: the vertical leg is 10 × 150 = 1500 mm² centred at (5, 75), and the horizontal leg is the remaining 90 × 10 = 900 mm² centred at (55, 5). Total area 2400 mm². Then x̄ = [1500(5) + 900(55)]/2400 = 57000/2400 = 23.75 mm and ȳ = [1500(75) + 900(5)]/2400 = 117000/2400 = 48.75 mm. Now look at where that point is. The vertical leg is only 10 mm wide and the horizontal leg only 10 mm thick, so the centroid at (23.75, 48.75) lies in the empty space inside the angle, on no metal at all. That is not a mistake and it is not unusual — a channel, a horseshoe and a ring all do the same. The centroid is a property of the distribution of area, not a point of the material, and the only consequence that matters is the one to remember: you cannot support a body at its centroid unless there is something there to hold.

For dynamics the Civil syllabus asks for kinematics of a point mass and a rigid body, energy methods, and impulse–momentum — and the choice between the last two is decided by what the question gives you. If it mentions a distance, use work–energy. If it mentions a time, use impulse–momentum: F Δt = m Δv, which rearranges to the fact that makes it worth knowing. A given change in momentum can be produced by a small force over a long time or an enormous force over a short one, and structures are routinely governed by the second. A pile hammer of 500 kg dropped 2 m arrives at v = √(2gh) = 6.264 m/s carrying 3132 kg·m/s of momentum. Bring it to rest in 0.05 s and the average force is 3132/0.05 = 62.6 kN — 12.8 times the hammer weight, and it is that multiplier, not the mass, that the pile and the surrounding soil actually feel. Halve the stopping time and you double the force for exactly the same drop.

Simple Stress and Strain, and the Two Independent Constants

Direct stress is σ = P/A and direct strain is ε = δ/L, and Hooke’s law ties them with σ = Eε, from which the elongation of a prismatic bar is δ = PL/AE. The formula is trivial; the examinable part is that A is whatever the area is at the section you are looking at, so a bar of changing section is a sum and not a single application. A stepped steel bar carrying 60 kN with E = 200 GPa makes the point: 1.5 m of 600 mm² runs at 100 MPa and stretches 0.75 mm, while 1.0 m of 300 mm² runs at 200 MPa and stretches 1.00 mm. Total 1.75 mm, and the shorter segment contributes more — halving the area doubles both the stress and the strain, so the thin part dominates twice over. The same reasoning covers a tapered bar, where the sum becomes an integral, and a bar of two materials in series, where E changes from term to term.

An isotropic elastic material has exactly two independent constants, and the four you meet are related by G = E/2(1+ν) and K = E/3(1−2ν). Know any two and the rest follow — which is why an examiner can give you E and K and ask for ν. The relations also carry the physics in their denominators. As ν → 0.5 the term (1−2ν) goes to zero and K diverges: a material with ν = 0.5 cannot change volume at all, which is what incompressible means, and rubber sits close to it. At the other end ν = 0 gives K = E/3, a material that thins not at all when stretched. Cork is near that, which is exactly why a cork goes into a bottle neck and a rubber bung fights you.

Elastic constants at E = 210 GPa: both derived values come out exact
Quantity	Relation	Value at ν = 0.25
Shear modulus G	`E/2(1+ν)`	210/2.5 = 84 GPa
Bulk modulus K	`E/3(1−2ν)`	210/1.5 = 140 GPa
Back-check on E	`9KG/(3K+G)`	210 GPa ✓
Limit ν → 0.5	`K → ∞`	incompressible
Limit ν = 0	`K = E/3`	70 GPa

Key takeaways

Cut it, and three quantities replace what you removed: axial force, shear and bending moment. dV/dx = −w and dM/dx = V make the three diagrams one object, not three.
The maximum moment sits where the shear changes sign — which at a point load is a jump across zero, not a root of V = 0. On the 8 m beam it is 165 kN·m at the load and only 156 at midspan.
Sliding and overturning are independent failures with independent remedies. Slides when tan θ > μ, tips when tan θ > b/h; the weight cancels out of both, so only the geometry and the friction decide.
Virtual work earns its keep by omission: a support that does not move does no work and never appears. Run forwards it gives a reaction; run backwards with a unit load it gives a deflection anywhere, including where no load acts.
An isotropic material has two independent elastic constants, no more: G = E/2(1+ν) and K = E/3(1−2ν). The denominators carry the physics — ν = 0.5 makes K infinite, which is what incompressible means.

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 simply supported beam of span 8 m carries a UDL of 12 kN/m over its full length and a point load of 40 kN at 3 m from the left support A. The maximum bending moment, in kN·m, is ______
Numerical answer — type the value.
Show answer
Answer: 165
165 kN·m, at the point load. Reactions first: the UDL totals 12 × 8 = 96 kN acting at midspan, so R_B × 8 = 96(4) + 40(3) = 504, giving R_B = 63 and R_A = 136 − 63 = 73 kN. Now find where the shear changes sign, and do not stop at the algebra. Left of the load V = 73 − 12x vanishes at x = 6.083 m, which is not left of the load; right of it V = 73 − 12x − 40 vanishes at x = 2.75 m, which is not right of the load. Neither root is admissible — the shear falls to +37 just left of the load, is knocked down by the whole 40 kN to −3 just right of it, and so changes sign by jumping across zero at x = 3. That is where the moment peaks: M = 73(3) − 12(3)(1.5) = 219 − 54 = 165 kN·m, and computing it from the far end as 63(5) − 12(5)(2.5) = 165 confirms it. The common wrong answer is 156, obtained at midspan, which is where the UDL alone would peak; the point load drags the maximum towards itself. Note also that the moment does not jump at the load even though the shear does — a concentrated force puts a kink in the moment diagram, not a step.
A block of base width b and height h rests on a plane whose inclination is slowly increased. The coefficient of friction is μ. The block will overturn before it slides if
1. b/h < μ
2. b/h > μ
3. b/h < 1/μ
4. It depends on the weight of the block
Show answer
Answer: A — `b/h < μ`
b/h < μ. Two thresholds, each a pure ratio. Sliding begins when the driving component beats friction, W sin θ > μ W cos θ, that is at tan θ = μ. Overturning begins when the weight vector through the centroid falls outside the lower corner, which is at tan θ = b/h. Whichever threshold is reached first as θ rises is the one that governs, so the block tips first when b/h < μ. Check it on numbers: with b = 0.6 m, h = 1.2 m and μ = 0.4, b/h = 0.5 > 0.4, so this block slides first, at 21.80°, and never gets to its tipping angle of 26.57°. Make it slender at h = 2.0 m and now b/h = 0.3 < 0.4, so the same block on the same surface tips first, at 16.70°. Option B is the same comparison backwards and describes sliding. Option D is the trap worth understanding rather than just avoiding: W cancels out of both criteria, so a heavier block of identical proportions fails at exactly the same angle in exactly the same way — mass changes the forces but not which threshold arrives first. This is the reason a retaining wall must be checked separately against sliding and against overturning, and why widening its base cures one while roughening its foundation cures the other.
A steel bar carries an axial pull of 60 kN. It consists of a 1.5 m length of area 600 mm² followed by a 1.0 m length of area 300 mm². Taking E = 200 GPa, the total elongation in mm is ______
Numerical answer — type the value.
Show answer
Answer: 1.75
1.75 mm. The load is the same in both segments — a bar in series carries the same axial force throughout, whatever its shape — so the only thing that changes is A. First segment: σ = 60000/600 = 100 MPa, and δ = PL/AE = 60000(1500)/(600 × 200000) = 0.75 mm. Second: σ = 60000/300 = 200 MPa, and δ = 60000(1000)/(300 × 200000) = 1.00 mm. Total 1.75 mm. The instructive part is the split: the shorter segment contributes more, and by a wide margin. Halving the area doubles the stress, and since strain is proportional to stress it doubles the strain per unit length too — so the thin part is penalised twice and dominates the total even at two-thirds the length. The wrong answers are diagnostic. Using the average area of 450 mm² over 2.5 m gives 1.667 mm, which is not the same thing and never is. Using only the larger area gives 1.25 mm. And a candidate who assumes the stress is uniform through the bar has made the one error this question exists to catch: the force is uniform, the stress is not.
For a material with E = 210 GPa and Poisson’s ratio 0.25, the bulk modulus in GPa is ______
Numerical answer — type the value.
Show answer
Answer: 140
140 GPa. Straight from K = E/3(1−2ν) = 210/[3(1 − 0.5)] = 210/1.5 = 140 GPa, and for the same material G = E/2(1+ν) = 210/2.5 = 84 GPa. Both come out exact at ν = 0.25, which is why that value is a favourite in question papers. Confirm by going the other way: E = 9KG/(3K+G) = 9(140)(84)/(420 + 84) = 105840/504 = 210 ✓. The relations matter more than the arithmetic because they say an isotropic material has only two independent elastic constants — give any two and the other two are fixed, so a question can hand you K and G and ask for ν. The denominators also carry the physics. As ν → 0.5 the factor (1 − 2ν) → 0 and K → ∞: the material resists volume change absolutely, which is what incompressible means and why rubber, at about 0.4997, is treated that way. At ν = 0 the same relation gives K = E/3 = 70 GPa. Thermodynamic stability requires −1 < ν < 0.5, so a claimed Poisson ratio of 0.6 is not an unusual material but an arithmetic error.
The principle of virtual work is applied to a simply supported beam by releasing the roller at B and giving it a unit virtual displacement. The chief advantage over taking moments about A is that
1. It gives a different, more accurate value of the reaction
2. The reaction at A does no work and never enters the equation
3. It applies to indeterminate structures where statics fails
4. It removes the need to know where the loads act
Show answer
Answer: B — The reaction at A does no work and never enters the equation
The reaction at A does no work and never enters the equation. A pin that does not displace contributes nothing to the work equation, so the unknown you did not ask about is eliminated by the choice of virtual displacement rather than by algebra. On the 8 m beam, rotating rigidly about A with δ_B = 1 moves the 40 kN load at 3 m through 3/8 and the 96 kN UDL resultant at 4 m through 1/2, so R_B = 40(0.375) + 96(0.5) = 63 kN — identical to the moment equation, as it must be. Option A is the error worth naming: virtual work is the same statics, so it gives exactly the same number, never a better one; anyone who gets a different answer has made a mistake, not found a refinement. Option C overstates it — the principle does extend to indeterminate structures, but only once compatibility is brought in as well, and the version used here is pure equilibrium. Option D is the reverse of the truth: the method depends entirely on knowing where each load acts, because the work each does is its magnitude times the displacement at its own point, which is what makes 3/8 and 1/2 different numbers in the first place.
An angle section is made of a vertical leg 150 mm × 10 mm and a horizontal leg 100 mm × 10 mm meeting at a corner. Measuring from the outer face of the vertical leg, the distance of the centroid in mm is ______ (round to two decimals)
Numerical answer — type the value.
Show answer
Answer: 23.75
23.75 mm. The only real difficulty is not counting the corner twice. Split the angle into the full vertical leg, 10 × 150 = 1500 mm² with its centroid at x = 5 mm, and the remaining part of the horizontal leg, 90 × 10 = 900 mm² with its centroid at x = 10 + 45 = 55 mm. Total area 2400 mm², and x̄ = [1500(5) + 900(55)]/2400 = (7500 + 49500)/2400 = 57000/2400 = 23.75 mm. The same walk on the other axis gives ȳ = [1500(75) + 900(5)]/2400 = 48.75 mm. Now put that point on the sketch: the vertical leg occupies only 0 ≤ x ≤ 10 and the horizontal leg only 0 ≤ y ≤ 10, so the centroid at (23.75, 48.75) sits in the empty space inside the angle, on no material at all. That is normal for an angle, a channel or a ring, and it is worth having seen once, because the centroid is a property of how the area is distributed rather than a point of the body. The classic wrong answer, 27.5 mm, comes from taking the horizontal leg as the full 100 mm at x = 50 and so counting the 10 × 10 corner twice.
On a beam carrying both a UDL and a concentrated load, at the section where the concentrated load acts,
1. both the shear force and the bending moment are discontinuous
2. the shear force is discontinuous and the bending moment is continuous with a change of slope
3. the bending moment is discontinuous and the shear force is continuous
4. both are continuous, since the beam itself is continuous
Show answer
Answer: B — the shear force is discontinuous and the bending moment is continuous with a change of slope
Shear steps, moment kinks. Take a slice of vanishing width around the load. Vertical equilibrium of that slice must absorb the whole concentrated force, so V changes by exactly the load across it — a step. Moment equilibrium of the same vanishing slice involves the load times an arm that goes to zero, so M cannot change across it — it is continuous — but since dM/dx = V and V has just jumped, the slope of the moment diagram changes abruptly. That is a kink, not a break. The 8 m beam shows both: V runs +37 to −3 across the load while M reads 165 on both sides. Option C describes a different and equally examinable case — a concentrated couple applied to the beam steps the moment diagram by its magnitude and leaves the shear untouched, which is the mirror image and a favourite way of testing whether the two have been learned as one idea or two memorised pictures. Option D confuses continuity of the beam with continuity of the internal force diagrams; the beam is continuous in all these cases and the diagrams are not.
A pile hammer of mass 500 kg is dropped freely through 2 m onto a pile and is brought to rest in 0.05 s. Which of the following are correct? (One or more)
1. The impact velocity is about 6.26 m/s
2. The average resisting force is about 62.6 kN
3. The average force is about 12.8 times the weight of the hammer
4. Halving the stopping time would leave the average force unchanged
Show answer
Answer: A — The impact velocity is about 6.26 m/s; B — The average resisting force is about 62.6 kN; C — The average force is about 12.8 times the weight of the hammer
A, B and C. Free fall through 2 m gives v = √(2gh) = √(2 × 9.81 × 2) = 6.264 m/s, so A is right, and the momentum arriving is 500 × 6.264 = 3132 kg·m/s. Impulse–momentum over the stopping interval gives F Δt = m Δv, so F = 3132/0.05 = 62 642 N ≈ 62.6 kN — B is right. Against the hammer’s own weight of 500 × 9.81 = 4.905 kN, that is a factor of 62.642/4.905 = 12.77, so C is right too, and it is the number that matters in practice: the pile and the soil beneath it feel not the mass but the multiplier. D is false, and it is the point of the question. The impulse m Δv is fixed by the drop, so force and time trade off inversely — halve the stopping time and the force doubles, to about 125 kN. That is why a cushion or dolly is placed on the pile head: it cannot change the momentum the hammer arrives with, but by stretching Δt it lowers the peak force, which is what cracks concrete piles. The same trade is behind crumple zones and safety nets, and it is worth reading the relation in the direction F = m Δv/Δt rather than as a formula to be solved once.