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Question
A mass of 10 kg is suspended vertically by a rope of length 5m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is θ = tan-1(x × 10-1). The value of x is ______.
(Given g = 10 m/s2)
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MCQ
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Solution
A mass of 10 kg is suspended vertically by a rope of length 5m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is θ = tan-1(x × 10-1). The value of x is 3.
(Given g = 10 m/s2)
Explanation:
FBD of middle point is as shown below.
From FBD
T sin θ = 30
T cos θ = 100
So, tan θ = 0.3
= 3 × 10-1
So, θ = tan-1(3 × 10-1)
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Motion of Connected Bodies, Pulley and Equilibrium of Forces
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