Question

A 1.55-kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a 14.3-N horizontal force. Find the magnitude of the tension in the rope, and the rope\'s angle from the vertical. The acceleration due to gravity is 9.81 m/s2.

Answer 1

Solution:

Let us go to the basics first.

The rope makes and angle with the vertical, θ.

Sum of the vertical forces (up is positive) = 0
0 = T*cosθ - m*g
=>T*cosθ = m*g

=>T = m*g / cosθ ...................Eqn.1

Sum of the horizontal forces (direction of the force is positive) = 0
0 = F - T*sinθ
T*sinθ = F ..................Eqn.2


Putting value of T from eqn.1 into eqn.2, we have:
m*g*sinθ/cosθ = F
=> m*g*tanθ = F
=> θ = tan^-1[F / (m*g)]
=> θ = tan^-1(14.3 N / (1.55 kg * 9.81 m/s^2)) = 43.24° [ANGLE (Answer)]

Using Eqn.1, we have:
T = m*g/cosθ = (1.55 kg * 9.81 m/s^2) / cos(43.24°) = 20.872 N [TENSION (Answer)]

Thanks!!!