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NEET 1998 (AIPMT) · Torque, Couple & Equilibrium of Rigid Bodies

Torque, Couple & Equilibrium of Rigid Bodies question from NEET 1998 (AIPMT)

Q48NEET 1998 · AIPMTMedium

Torque, Couple & Equilibrium of Rigid Bodies

A weightless ladder 20 ft20\text{ ft} long rests against a frictionless wall at an angle of 60∘60^\circ from the horizontal. A 150150 pound man is 4 ft4\text{ ft} from the top of the ladder. A horizontal force is needed to keep it from slipping. Choose the correct magnitude of force from the following:

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Answer:

  1. The horizontal force needed at the foot equals the wall's horizontal reaction NwN_w (smooth wall, weightless ladder).
  2. Man is 20−4=1620 - 4 = 16 ft from the foot. Moments about the foot:
  3. Nw×20sin⁡60∘=150×16cos⁡60∘N_w \times 20\sin 60^\circ = 150 \times 16\cos 60^\circ
  4. Nw=150×16×0.520×0.866=120017.32≈69.3N_w = \frac{150 \times 16 \times 0.5}{20 \times 0.866} = \frac{1200}{17.32} \approx 69.3 pound
  5. This value (40340\sqrt{3} pound) does not match any option.
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