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)

पर्याय

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Answer 1

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)