Applying equilibrium equations, Av + Bv = 200 N … (i) AH = BH … (ii) From FBD of block B,
BH + FB cos 60° - NB sin 60° = 0 NB cos 60° - Bv - 300 + FB sin 60° = 0 FB = 0.25 NB BH - 0.74 NB = 0 ... (iii) - BV + 0.71 NB = 300 ... (iv) FBD of block A
FA−AH = 0 NA−AV = 400 ... (v) FA = µANA ∴ µANA = 0 ... (vi) On solving above equations, we get NA = 650 N , FA = 260 N , FA = µANA ∴ µA =