Correct : 0.41

Block A rests against the vertical wall and block B rests on the horizontal floor. Both are connected by a rigid inclined rod at θ = 45°. Let T be the compressive force in the rod, W the weight of each block, N_A and N_B the normal reactions, and μ the coefficient of friction.
Equilibrium of Block A:
Horizontal: N_A = T cos45° = T/√2
Vertical: μN_A + T sin45° = W → (T/√2)(1 + μ) = W ... (i)
Equilibrium of Block B:
Vertical: N_B = W + T sin45° = W + T/√2
Horizontal: T cos45° = μN_B → T/√2 = μ(W + T/√2)
Rearranging: T/√2 = μW/(1 − μ) ... (ii)
Substituting (ii) into (i):
[μW/(1 − μ)] × (1 + μ) = W
μ(1 + μ) = 1 − μ
μ² + μ = 1 − μ
μ² + 2μ − 1 = 0
Solving the quadratic:
μ = (−2 + √8)/2 = √2 − 1 = 0.4142 ≈ 0.41
Correct answer: 0.41 ✓