Engineering Mechanics: Solved Problems in Statics and Dynamics

Posted on Aug 12, 2026 in Physics

D’Alembert’s Principle: Elevator Dynamics

Given: Weight W = 1960 N, g = 9.81 m/s2. Mass m = 1960 / 9.81 = 199.8 kg (approx. 200 kg using g = 9.8 m/s2).

D’Alembert Equation: Tension T = m(g + a) for upward acceleration/downward deceleration, and T = m(g – a) for downward acceleration/upward deceleration.

  1. Moving UP with acceleration a = 2 m/s2: T = 1960 + (200 * 2) = 2360 N
  2. Moving UP with deceleration a = 1.5 m/s2: T = 1960 – (200 * 1.5) = 1660 N
  3. Moving DOWN with acceleration a = 2 m/s2: T = 1960 – (200 * 2) = 1560 N
  4. Moving DOWN with deceleration a = 1.5 m/s2: T = 1960 + (200 * 1.5) = 2260 N

Varignon’s Theorem and Vertical Post Systems

1. Theorem Statement

Varignon’s Theorem states that the moment of the resultant of a system of coplanar forces about any point is equal to the algebraic sum of the moments of all individual component forces about that same point. Equation: MR = ΣM

2. Vertical Post Problem

Given forces F1 = 100 N, F2 = 150 N, F3 = 200 N (all horizontal) and Couple = 150 N·m.

  • ΣFx = 100 + 150 + 200 = 450 N
  • ΣFy = 0; Resultant R = 450 N (Horizontal)
  • Moment about point B: ΣMB = Σ(Moments of forces) + Couple = 150 N·m
  • Position of Resultant: R * d = ΣMB; 450 * d = 150; d = 0.33 m from point B.

Answer: Resultant R = 450 N, Position d = 0.33 m from B.

Ladder Friction: Man Ascending

Given: Ladder length L = 5 m, WL = 100 N, θ = 60°, WM = 600 N, μfloor = 0.25, μwall = 0.

  1. Vertical Equilibrium (ΣFy = 0): Nfloor = WL + WM = 100 + 600 = 700 N
  2. Limiting Friction at Floor: Ffloor = μfloor * Nfloor = 0.25 * 700 = 175 N
  3. Horizontal Equilibrium (ΣFx = 0): Nwall = Ffloor = 175 N
  4. Moment about Base A (ΣMA = 0): Let x = distance man ascends.
    (100 * 2.5 * cos 60°) + (600 * x * cos 60°) – (Nwall * 5 * sin 60°) = 0
    125 + 300x – 757.75 = 0; 300x = 632.75; x = 2.11 m

Answer: Man can ascend 2.11 m along the ladder before slipping.

Ladder Friction: Horizontal Force P

Given: L = 4 m, WL = 200 N, θ = 60°, WM = 600 N at 3 m from base A, μwall = 0.2, μfloor = 0.3.

  1. Geometry: Height h = 4 sin 60° = 3.464 m; Horizontal distance = 4 cos 60° = 2.0 m
  2. Friction Relations: Ffloor = 0.3 * Nfloor; Fwall = 0.2 * Nwall
  3. Vertical Equilibrium (ΣFy = 0): Nfloor + Fwall = 800 N; Nfloor + 0.2 * Nwall = 800 [Eq 1]
  4. Moment about A (ΣMA = 0): (200 * 1) + (600 * 1.5) – (Nwall * 3.464) – (Fwall * 2.0) = 0; Nwall = 284.68 N
  5. Solve Nfloor and Ffloor: Nfloor = 743.06 N; Ffloor = 222.92 N
  6. Horizontal Equilibrium (ΣFx = 0): P + Ffloor = Nwall; P + 222.92 = 284.68; P = 61.76 N

Answer: Required force P = 61.76 N