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Equating the forces along and perpendicular to AB and AC, for equilibrium of the beads;T cos α = F cos α + mg sin 300 . . . . (1)F sin α + N1 = mg cos 300 + T sin α . . . . . (2)T sin α = F sin α + mg cos 300 . . . . . . (3)N2 + F cos α = T cos α + mg cos 600 . . . . . (4)From equation (1) and (3);(T – F) cos α = mg sin 300and (T – F) sin α = mg cos 300Dividing, cot α = tan 300 = cot 600Therefore, α = 600.
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