(a) A heavy carriage wheel of weight W, and radius r is to be dragged over an obstacle of height h by a horizontal force P…

(a) Force to drag a wheel over an obstacle

Let C be the point where the wheel touches the edge of the obstacle and O the centre. At the verge of lifting, the ground reaction vanishes, so only W, P and the reaction at C act, and the reaction at C passes through C itself.

The vertical height of O above C is r − h. The horizontal distance of O from C is x, where x² + (r − h)² = r², so x = √(2rh − h²).

Moments about C: P acts horizontally at O, with arm r − h, and W acts vertically at O, with arm x. Hence P(r − h) = W√(2rh − h²), P = W√(2rh − h²)/(r − h).

This is the value at the verge of lifting. To actually drag the wheel over, P must exceed it slightly, so P is slightly greater than W√(2rh − h²)/(r − h), for 0 < h < r.

(b) Countermass for static balance

For static balance the vector sum of the mr products, including the countermass, is zero. The mr products are 6 × 70 = 420 kg·mm at 45°, 5 × 80 = 400 kg·mm at 135°, and 4 × 50 = 200 kg·mm at 240°.

Σ mr cosθ = 420 cos45° + 400 cos135° + 200 cos240° = 297.0 − 282.8 − 100 = −85.86 kg·mm Σ mr sinθ = 420 sin45° + 400 sin135° + 200 sin240° = 297.0 + 282.8 − 173.2 = 406.62 kg·mm

Resultant of the three masses = √(85.86² + 406.62²) = 415.59 kg·mm, at tan⁻¹(406.62/−85.86), which is 101.92° (second quadrant).

The countermass must balance it, so it lies at 101.92° + 180° = 281.92°, and m_c × 85 = 415.59, so m_c = 4.89 kg.

Countermass ≈ 4.89 kg at 85 mm radius, at 281.9° counter-clockwise from the reference line.

(c) Height of the balloon

Take upward as positive and g = 9.8 m/s². At release the stone has the balloon's velocity, 25 m/s upward. It first rises and then falls back to the release level after 2 × 25/9.8 = 5.10 s, before it can even start descending to the lake. But the splash is heard only 5 s after release, and that total time includes the sound's travel time as well.

Equations confirm this. If the stone takes t seconds to reach the water, H = 4.9t² − 25t, and the sound takes H/340 = 5 − t. So 4.9t² − 25t = 340(5 − t), which gives 4.9t² + 315t − 1700 = 0 and t = 5.007 s. This exceeds 5 s, so the sound time 5 − t would be negative, which is impossible.

The data are inconsistent for an ascending balloon: no positive height satisfies them. If the balloon were descending at 25 m/s, the same method gives 4.9t² + 365t − 1700 = 0, t = 4.40 s, and H = 340 × (5 − 4.40) ≈ 205 m (204.7 m), which is probably the intended answer.

(d) Atomic packing factor

Importance: the atomic packing factor (APF) is the fraction of a unit cell volume occupied by atoms, treated as hard spheres. It shows how closely a structure is packed, and so relates to density, number of slip systems, ductility, diffusion rates, solid solubility and phase stability, which guide the choice and design of alloys.

FCC: there are 4 atoms per unit cell. Atoms touch along the face diagonal, so 4r = a√2, giving a = 2√2 r.

V_cell = a³ = 16√2 r³ V_atoms = 4 × (4/3)πr³ = (16/3)πr³

APF = (16/3)πr³/(16√2 r³) = π/(3√2) = 0.7405, that is 74% of the volume.

(e) Critical speed of the shaft

Ends held in long bearings are effectively fixed. For a fixed-fixed shaft with a central disc, the static deflection is δ = WL³/(192EI).

W = 20 kgf, L = 150 cm, E = 2 × 10⁶ kgf/cm², I = πd⁴/64 = π(2)⁴/64 = π/4 = 0.7854 cm⁴.

δ = 20 × 150³/(192 × 2 × 10⁶ × 0.7854) = 0.2238 cm

Critical angular speed ω_c = √(g/δ) = √(981/0.2238) = 66.2 rad/s.

N_c = 60ω_c/(2π) = 632 cycles per minute (rpm). The shaft's own mass is neglected.