A 6 kg mass hangs from a light rope. A horizontal force of 50 N is applied at the rope's midpoint (g = 10 m s⁻²). In equilibrium the upper part of the rope makes angle θ with the vertical such that:
Atan θ = 1
Btan θ = 3/5
Ctan θ = 5/6
Dtan θ = 6/5
Answer & Solution
Correct answer: C. tan θ = 5/6
At the midpoint: T₁cosθ = weight side tension = 6×10 = 60 N (vertical) and T₁sinθ = 50 N (horizontal). Dividing gives tan θ = 50/60 = 5/6; the result is independent of where the force is applied.
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