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A rod AB rests with the end A on rough ground and the end B against a smooth vertical wall. The rod is uniform and of weight W. If the rod is in equilibrium in the position shown in figure, the normal reaction at B is

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BPSC AE Paper IV General Engineering 19 Dec 2024 Official Paper
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  1. √3 W
  2. \(\rm \frac{\sqrt 3}{2}W\)
  3. \(\rm \frac{W}{2}\)
  4. W

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac{\sqrt 3}{2}W\)
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Detailed Solution

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Concept:

The rod is in equilibrium under the action of three forces: weight W acting at center, normal reaction at A (vertical), and horizontal reaction at B from the smooth wall. Friction at A balances the horizontal reaction.

Equilibrium Equations:

  • Vertical: \( R_A = W \)
  • Horizontal: \( F_A = R_B \)
  • Moment about A: \( R_B \times L \times \sin(30^\circ) = W \times \frac{L}{2} \times \cos(30^\circ) \)

Solving:

\( R_B \times \frac{L}{2} = \frac{W L \cos(30^\circ)}{2} \Rightarrow R_B = W \times \cos(30^\circ) = W \times \frac{\sqrt{3}}{2} \)

Final Answer:

\( \frac{\sqrt{3}}{2} W \)

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