Question

A mass of \(10\) kg is suspended vertically by a rope of length \(5\) m 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, \(g = 10\; m/s^2\))

Answer 1

Correct Answer: 3

To find the value of \(x\), let's analyze the forces acting on the system. The mass of 10 kg is subject to gravitational force \(F_g = mg = 10 \times 10 = 100\) N vertically downward. The horizontal force applied at the midpoint of the rope is 30 N. Since the rope length is 5 m, the midpoint divides it into two segments of 2.5 m each. When the force is applied horizontally, the vertical component of tension \(T\) must balance the gravitational force, and the horizontal component must balance the applied force. Using trigonometry, let the angle \(θ\) be formed by the upper half with the vertical, such that \(tan\,θ = \frac{\text{horizontal force}}{\text{vertical component}}\).\(
The horizontal component \(T_h = T \sin θ\) and vertical \(T_v = T \cos θ\). From equilibrium \(T_v = 100\) N and \(T_h = 30\) N.
\(tan\,θ = \frac{T_h}{T_v} = \frac{30}{100} = 0.3\), therefore \(θ = tan^{-1}(0.3)\).
Given that \(θ = tan^{-1}(x \times 10^{-1})\), equating we find \(x \times 10^{-1} = 0.3\). Solving for \(x\), we get \(x = 3\). This value lies in the range 3, 3, as required. Thus, \(x\) is correctly determined to be 3.