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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

हिन्दी

A uniform ladder of length 3 m and mass 20 kg rests against a smooth vertical wall with its foot 1 m away from the wall on a rough floor. The total reaction force exerted by the floor on the ladder has a magnitude of approximately: (Take \( g = 9.8\text{ m/s}^2 \))

  1. 196.0 N
  2. 228.2 N
  3. 205.5 N
  4. 199.0 N

Correct answer · Option 4

199.0 N

Explanation

Vertical normal force \( N = W = mg = 196.0\text{ N} \). Taking torque about foot of ladder gives horizontal wall force \( F_1 = \frac{W}{4\sqrt{2}} \approx 34.6\text{ N} \). Friction \( F = F_1 = 34.6\text{ N} \). Resultant floor force \( F_2 = \sqrt{N^2 + F^2} = \sqrt{196^2 + 34.6^2} \approx 199.0\text{ N} \).
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