(a) A car is moving in a straight line with a velocity of v = (0·6t² + 2t) m/s for a short duration, where t is in seconds. Take…

(a)

Given: v = 0.6t² + 2t (m/s), with s = 0 at t = 0.

(i) Distance in 4 s. Since v = ds/dt, integrate: s = ∫v dt = 0.2t³ + t² + C₁ At t = 0, s = 0, so C₁ = 0 and s = 0.2t³ + t². v is positive for every positive t, so the car never reverses. The distance travelled therefore equals the displacement: s(4) = 0.2 × 4³ + 4² = 12.8 + 16 = 28.8 m

(ii) Acceleration at 4 s. Since a = dv/dt: a = 1.2t + 2 a(4) = 1.2 × 4 + 2 = 6.8 m/s²

Interpretation. At t = 4 s the velocity is 0.6 × 16 + 8 = 17.6 m/s, and the acceleration is still rising linearly with time. The car is accelerating ever harder over this short interval, which is why the problem restricts the law to "a short duration".

(b)

Reading the figure. The motor pinion drives the large gear D on shaft AB, which puts 20 kW into the shaft at D. The gear near A, meshing with F, takes off 8 kW, and gear E near B, meshing with G, takes off 12 kW. The shaft therefore carries different torques on either side of D.

Angular speed: ω = 2πN/60 = 2π × 450/60 = 47.12 rad/s

Torques (T = P/ω):

• Input at D: T_D = 20 000/47.12 = 424.4 N·m
• Take-off at F: T_F = 8000/47.12 = 169.8 N·m
• Take-off at G: T_G = 12 000/47.12 = 254.6 N·m

Internal torque in each segment (from a free body of the shaft cut in that segment):

• A to gear F: 0
• F to D (225 mm): 169.8 N·m
• D to E/G (225 mm): 254.6 N·m
• E/G to B: 0

As a check, 169.8 + 254.6 = 424.4 N·m, which equals the input torque at D. The shaft is never subjected to the full 20 kW torque in any segment. The critical segment is DE, with T_max = 254.6 N·m.

Design for torsion. τ_max = 16T/(πd³) ≤ 55 MPa, so d³ = 16T/(πτ_allow) = 16 × 254.6/(π × 55 × 10⁶) = 2.358 × 10⁻⁵ m³ d = 0.02868 m, so d_min = 28.7 mm. In practice, the next standard size of 30 mm would be adopted.

Remark. Designing for the total 20 kW (424.4 N·m) would give 34.0 mm, an oversized shaft. The problem specifies only an allowable shear stress, with no gear pressure angles or force directions, so the shaft is sized for torsion, as intended. If the gear-tooth loads were known, bending would be included through Guest's equivalent torque T_e = √(M² + T²).

(c)(i)

Let T_S, T_P and T_A be the numbers of teeth on the sun, planet and annulus.

Tabular method (arm C fixed first, S given +1 revolution):

• C fixed, S +1 rev: C — 0; S — +1; P — –T_S/T_P; A — –T_S/T_A
• × x: C — 0; S — x; P — –x T_S/T_P; A — –x T_S/T_A
• Add y to all: C — y; S — x + y; P — y – x T_S/T_P; A — y – x T_S/T_A

Setting A stationary gives y = x T_S/T_A. The required ratio is N_S/N_C = (x + y)/y = 1 + T_A/T_S = 5, so T_A = 4T_S.

Pitch-circle (meshing) condition: T_A = T_S + 2T_P, so 4T_S = T_S + 2T_P and T_P = 1.5T_S.

The smallest wheel is the sun. With T_S = 14 (even, so T_P is an integer): T_S = 14, T_P = 21, T_A = 56.

Strictly, three equally spaced planets also need (T_A + T_S)/3 to be an integer. Here 70/3 is not, so exact 120° spacing would need T_S = 18, T_P = 27, T_A = 72.

Fixing torque. With no losses, the external torques balance and the power balances: T_S + T_C + T_A = 0 and T_S ω_S + T_C ω_C = 0 (because ω_A = 0) T_C = –200 × 5 = –1000 N·m (the resisting torque on the spider) T_A = –(T_S + T_C) = –(200 – 1000) = 800 N·m

The equivalent tooth-force route gives the same result: 200 × T_A/T_S = 200 × 56/14 = 800 N·m. A holding torque of 800 N·m, acting in the same sense as the input torque, must be applied to the annulus.

(c)(ii)

Data: m₂ = m₃ = 500 kg, with cranks 2 and 3 at 90°. Cylinder planes measured from the centre are at –900, –300, +300 and +900 mm. r = 0.2 m, l = 0.8 m, n = l/r = 4. By symmetry m₁ = m₄ = m.

Primary balance. Place crank 2 at 0° and crank 3 at 90°, and write each mass-crank product as a vector u = m₁∠θ₁ and v = m₄∠θ₄. Force balance: u + v + 500∠0° + 500∠90° = 0, so u + v = –500 – 500j Couple balance about the central plane: 900(v – u) = 300 × 500(∠0° – ∠90°), so v – u = 166.7 – 166.7j

Solving: u = –333.3 – 166.7j and v = –166.7 – 333.3j.

m₁ = m₄ = √(333.3² + 166.7²) = 372.7 kg θ₁ = 206.57° and θ₄ = 243.43°, measured from crank 2 in the direction of crank 3. The engine is mirror-symmetric: crank 1 is 206.57° from crank 2, and crank 4 is 153.43° from crank 3. The two extreme cranks are 36.87° apart.

Secondary forces. The secondary cranks lie at 2θ: 0°, 180°, 53.13° and 126.87°, so Σm∠2θ = 500∠0° + 500∠180° + 372.7(∠53.13° + ∠126.87°) = 372.7 × 1.6 j = 596.3 kg, directed at 90°.

ω = 2π × 150/60 = 15.71 rad/s Maximum secondary unbalanced force = (Σm∠2θ) ω² r/n = 596.3 × 246.74 × 0.2/4 = 7.36 kN

There is also an unbalanced secondary couple of about 8.67 kN·m about the central plane. The engine is therefore fully balanced for primary effects, but it transmits secondary forces at twice engine speed to its mountings.