Explanation:
(i) Sum the forces at point M in the x direction.
∑F = ma
-T sin 30° − T sin 30° + 5 N = 0
2T sin 30° = 5 N
T = 5 N
(ii) Sum the forces on the ring in the x direction.
∑F = ma
T sin 30° − N = 0
N = 2.5 N
Sum the forces on the ring in the y direction.
∑F = ma
T cos 30° − mg − Nμ = 0
Nμ = T cos 30° − mg
2.5 μ = 5 cos 30° − 2
μ = 0.932
(iii) Sum the forces on the ring in the y direction.
∑F = ma
T cos 30° − m₁g − m₂g + Nμ = 0
m₂g = T cos 30° − m₁g + Nμ
m (10) = 5 cos 30° − 2 + (2.5)(0.932)
m = 0.466 kg