Tension in a cable and pulley

Tension, two unequal weights m₁ and m₂
with m₁>m₂ and pulley
  T is tension in rope
  for m₁, F = m₁g – T
          m₁a = m₁g – T
  for m₂, F = T – m₂g
          m₂a = T – m₂g
  adding, g(m₁ – m₂) = a(m₁ + m₂)
  accel:  a = g(m₁ – m₂)/(m₁ + m₂)

  substituting:
          T – m₂g = m₂a
          T – m₂g = m₂g(m₁ – m₂)/(m₁ + m₂)
          T = m₂g(m₁ – m₂)/(m₁ + m₂) + m₂g
          T = 2gm₁m₂/(m₁ + m₂)

  tension in rope from pulley to ceiling:
    net force is tension and is sum of two weights minus
    F=ma where a is net acceleeration of masses
          T₁ = (m₁ + m₂)g – g(m₁ + m₂)(m₁ – m₂)/(m₁ + m₂)
          T₁ = (m₁ + m₂)g – g(m₁ – m₂)
          T₁ = 2gm₂


Angled rope pulling an a box. Rope is angled θ degrees
  above horizontal, Friction force is F, µ is coef of friction.
      T = Fcosθ
      T = mgµcosθ


Tension, two equal weights m and pulley
   acceleration is zero
   tension T = 2mg
   tension T₁ = 2mg


Tension, one weight, m, supported by two angled ropes.
   rope 1 makes angle θ with vertical
   rope 2 makes angle φ with vertical
   horizontal: T₁sinθ = T₂sinφ
   vertical:   T₁cosθ + T₂cosφ = mg
   (two eq in 2 unknowns, use example to solve numerically)
     example, θ=5º, φ=5º, mg = 100 N
        T₁sinθ = T₂sinφ
        T₁cosθ + T₂cosφ = mg
        T₁0.087 = T₂0.996
        T₁0.996 + T₂0.087 = 100
        T₁ = T₂ = 1018 N