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 |
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