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9702 · 4.2

Equilibrium of Forces — practice questions

Practice and worked examples for 9702 Equilibrium of Forces. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A uniform beam of length 4.0 m and weight 80 N rests on supports at each end. A 120 N load is placed 1.5 m from the left support. Find the support forces.

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  1. Forces: Let left support = RLR_L, right support = RRR_R. Up = down: RL+RR=80+120=200R_L + R_R = 80 + 120 = 200 N.
  2. Pivot at left support (eliminates RLR_L from moments).
  3. Clockwise moments: load 120×1.5=180120 \times 1.5 = 180 N m; beam weight 80×2.0=16080 \times 2.0 = 160 N m. Total CW = 340 N m.
  4. Anticlockwise moment: RR×4.0R_R \times 4.0.
  5. Principle of moments: 4.0 RR=3404.0\,R_R = 340 → RR=85R_R = 85 N.
  6. Substitute: RL=200−85=115R_L = 200 - 85 = 115 N.

Worked example 2

A uniform ladder of length 5.0 m and weight 200 N leans against a smooth vertical wall at an angle of 60° to the rough horizontal ground. Find the reaction forces from the wall and ground, and the frictional force.

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  1. Diagram & Forces: Draw the ladder. Forces are: Weight (W=200 N) acting down from the centre (2.5 m), ground normal reaction (RgR_g) up, ground friction (FfF_f) horizontally inwards, wall normal reaction (RwR_w) horizontally outwards.
  2. Vertical Equilibrium (ΣF_y = 0): The only upward force is RgR_g and the only downward force is W. Thus, Rg=W=200R_g = W = 200 N.
  3. Horizontal Equilibrium (ΣF_x = 0): The only inward force is FfF_f and the only outward force is RwR_w. Thus, Ff=RwF_f = R_w.
  4. Rotational Equilibrium (ΣM = 0): Take moments about the base of the ladder to eliminate RgR_g and FfF_f.
    • Clockwise moment (from weight): Mcw=W×(perpendicular distance)=200×(2.5cos⁡60°)=200×1.25=250M_{cw} = W \times (\text{perpendicular distance}) = 200 \times (2.5 \cos 60°) = 200 \times 1.25 = 250 N m.
    • Anticlockwise moment (from wall reaction): Macw=Rw×(perpendicular distance)=Rw×(5.0sin⁡60°)=Rw×(5.0×0.866)=4.33RwM_{acw} = R_w \times (\text{perpendicular distance}) = R_w \times (5.0 \sin 60°) = R_w \times (5.0 \times 0.866) = 4.33 R_w.
  5. Apply Principle of Moments: Mcw=MacwM_{cw} = M_{acw} → 250=4.33Rw250 = 4.33 R_w.
    • Solving for RwR_w: Rw=250/4.33=57.7R_w = 250 / 4.33 = 57.7 N.
  6. Find Friction: From step 3, Ff=RwF_f = R_w. Therefore, Ff=57.7F_f = 57.7 N. Answer: Ground reaction Rg=200R_g = 200 N, Wall reaction Rw=57.7R_w = 57.7 N, Frictional force Ff=57.7F_f = 57.7 N.
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