A mass of 6 kg is suspended by a rope of length 2 m from a ceiling. A force of 50N in the horizontal direction is applied at the midpoint of the rope as shown in figure. What is the angle the rope makes with the vertical in equilibrium? Take g = 10ms–2. Neglect mass of the rope.

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Shiksha Nation · Accepted Answer

T1 sinθ = T3 = 50N           ….(i)T1 cosθ = T2 = 6kg wt = 60NDividing (i) by (ii), we getDividing (i) by (ii), we get, tanθ = 5/6 or θ = tan–1(5/6) = 39.8°

A mass of 6 kg is suspended by a rope of length 2 m from a ceiling. A force of 50N in the horizontal direction is applied at the midpoint of the rope as shown in figure. What is the angle the rope makes with the vertical in equilibrium? Take g = 10ms^–2. Neglect mass of the rope.

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