(a) A governor of Hartnell type has each ball of weight 20 N and the lengths of vertical and horizontal arms of the bell crank…

(a)

Given: ball weight W = 20 N, so m = 20/9.81 = 2.0387 kg. The ball (vertical) arm is x = 125 mm and the sleeve (horizontal) arm is y = 65 mm. The fulcrum is 100 mm from the spindle axis. At r₁ = 80 mm, N₁ = 300 rpm; at r₂ = 125 mm, N₂ = 325 rpm.

Assumptions (standard): The sleeve weight is not given and is neglected. Arm obliquity and the moment of the ball weight are also neglected. With the fulcrum at 100 mm, the arms are exactly vertical and horizontal at r = 100 mm.

Moment equilibrium about the fulcrum. Each lever carries half of the total spring force S:

F_c · x = (S/2) · y, so S = 2F_c x/y = 2F_c(125/65) = 3.846 F_c

Minimum radius (r₁ = 0.08 m, N₁ = 300 rpm): ω₁ = 2π × 300/60 = 31.416 rad/s F_c1 = mω₁²r₁ = 2.0387 × 31.416² × 0.08 = 160.97 N S₁ = 3.846 × 160.97 = 619.12 N

Maximum radius (r₂ = 0.125 m, N₂ = 325 rpm): ω₂ = 2π × 325/60 = 34.034 rad/s F_c2 = 2.0387 × 34.034² × 0.125 = 295.19 N S₂ = 3.846 × 295.19 = 1135.33 N

Sleeve lift (ball travel × arm ratio): h = (r₂ – r₁)(y/x) = (125 – 80)(65/125) = 45 × 0.52 = 23.4 mm

Spring stiffness: k = (S₂ – S₁)/h = (1135.33 – 619.12)/23.4 = 516.21/23.4 = 22.06 N/mm (22.06 kN/m)

Speed at r = 100 mm. Lift from the lowest position: h' = (100 – 80) × 0.52 = 10.4 mm Spring force: S = S₁ + k h' = 619.12 + 22.06 × 10.4 = 848.55 N F_c = S y/(2x) = 848.55 × 65/(2 × 125) = 220.62 N ω = √(F_c/(m r)) = √(220.62/(2.0387 × 0.1)) = 32.896 rad/s N = 60ω/2π = 314.1 rpm

Interpretation. The spring stiffness is about 22 N/mm. The speed at mid-radius is 314 rpm, which is not the mean of 300 and 325 rpm because F_c varies as ω²r. Including ball weight and obliquity changes k by only about 5 per cent, and it leaves N at r = 100 mm unchanged because the arms are exactly vertical and horizontal there.

(b)

Take x from the left end, in kN and m. The UDL resultant is 24 kN at x = 1.2 m, and EI = 0.6 MN·m² = 600 kN·m².

Reactions. Moments about B: R_A × 2.4 = 15 × 3.0 + 24 × 1.8 + 15 × 1.2 = 45 + 43.2 + 18 = 106.2, so R_A = 44.25 kN. ΣF_y = 0: R_B = 15 + 24 + 15 – 44.25 = 9.75 kN.

Bending moment (Macaulay form, sagging positive). The UDL stops at x = 1.8 m, so it is continued to the right end and cancelled by an equal upward load:

EI d²y/dx² = M = –15x + 44.25⟨x – 0.6⟩ – 10⟨x – 0.6⟩² + 10⟨x – 1.8⟩² – 15⟨x – 1.8⟩

This gives M_A = –9 kN·m, M at 1.8 m = +11.7 kN·m, and M_B = 0 ✓.

First integration (slope): EI dy/dx = –7.5x² + 22.125⟨x – 0.6⟩² – (10/3)⟨x – 0.6⟩³ + (10/3)⟨x – 1.8⟩³ – 7.5⟨x – 1.8⟩² + C₁

Second integration (deflection): EI y = –2.5x³ + 7.375⟨x – 0.6⟩³ – (5/6)⟨x – 0.6⟩⁴ + (5/6)⟨x – 1.8⟩⁴ – 2.5⟨x – 1.8⟩³ + C₁x + C₂

Boundary conditions. At x = 0.6 m, y = 0: –2.5(0.216) + 0.6C₁ + C₂ = 0, so 0.6C₁ + C₂ = 0.54 … (1) At x = 3.0 m, y = 0: –67.5 + 7.375(13.824) – (5/6)(33.1776) + (5/6)(2.0736) – 2.5(1.728) + 3C₁ + C₂ = 0 –67.5 + 101.952 – 27.648 + 1.728 – 4.32 + 3C₁ + C₂ = 0, so 3C₁ + C₂ = –4.212 … (2) Subtracting (1) from (2): 2.4C₁ = –4.752, so C₁ = –1.98 kN·m² and C₂ = 0.54 + 1.188 = 1.728 kN·m³.

Deflection at x = 1 m (only the ⟨x – 0.6⟩ terms are active): EI y = –2.5(1)³ + 7.375(0.4)³ – (5/6)(0.4)⁴ – 1.98(1) + 1.728 = –2.5 + 0.472 – 0.0213 – 1.98 + 1.728 = –2.3013 kN·m³

y = –2.3013/600 = –3.836 × 10⁻³ m

Deflection at 1 m from the left end = 3.84 mm, downward.

At x = 0 the same equation gives EI y = +1.728 kN·m³, so the overhang tip rises by 2.88 mm while the span sags.

(c)

A cam converts rotation into a prescribed follower motion. Cams are classified by shape; followers by contact surface, motion and path.

Types of cams

• Disc / radial (plate) cam: merits — Simple, compact, cheap; the most widely used (IC-engine valves); demerits — Needs a spring or gravity to keep contact; large rise needs a large cam.
• Wedge / translating cam: merits — Simple to make; converts one reciprocation into another; demerits — Limited stroke; reciprocating inertia.
• Cylindrical (drum/barrel) cam: merits — Positive drive through a groove; follower moves parallel to the axis (machine-tool feeds); demerits — Harder to machine; backlash in the groove.
• Spiral, conical, globoidal and spherical cams: merits — Special motions, large oscillation angles (indexing); demerits — Costly, complex manufacture.

Types of followers

• Knife-edge: merits — Follows any profile precisely; demerits — Rapid wear, high contact stress; only for slow, light duty.
• Roller: merits — Rolling contact, low friction and wear; suits high speed (engines, textile machines); demerits — Bulkier; pin wear; cannot follow a concave profile of radius smaller than the roller.
• Flat-faced (mushroom): merits — Compact; takes large side thrust; offsetting it spreads wear; demerits — Cannot follow concave profiles; surface stress and sliding friction.
• Spherical-faced: merits — Avoids edge stress when misaligned (automobile tappets); demerits — More sliding friction than a roller.

By motion, followers are reciprocating or oscillating; by path, radial or offset. The choice balances load, speed, accuracy, wear and cost.