Mechanical systems: statics, kinematics, dynamics and transmission
This domain rewards clean free-body diagrams and consistent units. The exam asks you to balance forces and moments, compute speed and torque through a transmission, and evaluate efficiency from input to output.
Statics
- Draw the free-body diagram, then apply ΣFx = 0, ΣFy = 0 and ΣM = 0 about a convenient point (a pin removes two unknowns, a roller one).
- Friction: F ≤ μ·N; at the limit of slipping F = μ·N.
- Levers: F1·d1 = F2·d2, with distances measured perpendicular to the force. First class (fulcrum between), second (load between) and third (effort between).
Kinematics of mechanisms
- ω = 2πn/60 (rad/s from rpm); v = ω·r at a pitch radius.
- Gears and pulleys: i = Z2/Z1 = n1/n2 = ω1/ω2 = d2/d1. Ideal torque ratio is the inverse: M2 = M1·i·η.
- Centre distance of a gear pair: a = m·(Z1+Z2)/2, with m the module in mm.
- Worm gear: i = Z2/Z1 (single-start worm counts as Z1 = 1). Movable pulley halves the force and doubles the rope travel. Crank–slider stroke = 2r.
Dynamics and energy
P = F·v = M·ω; η = Pu/Pa; W = F·d. A gearbox reduces speed and multiplies torque, minus losses.
Worked example
Motor at 1450 rpm drives Z1 = 20 into Z2 = 60; on the same shaft Z3 = 25 drives Z4 = 75. Total ratio i = (60/20)·(75/25) = 3·3 = 9. Output speed n4 = 1450/9 = 161.1 rpm. If the output must deliver 90 N·m with η = 0.846, input torque M1 = M3/(i·η) = 90/(9 × 0.846) = 11.8 N·m, and input power P = M·ω = 11.8 × (1450·2π/60 = 151.8) = 1791 W.
Traps that cost marks
- Mixing rpm into P = M·ω without converting to rad/s.
- Using the speed ratio for torque instead of its inverse, or forgetting η.
- Belt slip makes the real ratio lower than the geometric one.
- Mixing radius in mm with power in W; convert to metres.
- Ignoring that an external gear mesh reverses the direction of rotation.