A m = 2.15 m=2.15 kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a F W = 12.3 FW=12.3 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 g = 9.81 g=9.81 m/s2.

T = 24.4 N

β = 30.25°, angle from the vertical

Explanation:

Forces acting on the object

W = m*g = 2.15 kg * 9.81 m/s² = 21.0915 N : Weight of the object, vertical force

FW = 12.3 N : horizontal force of the wind

T : tension of the rope, angle (α) from the horizontal

Equilibrium of forces to the object

∑Fx=0

FW - Tcosα = 0

12.3 - Tcosα = 0

Tcosα = 12.3 Equation (1)

∑Fy=0

Tsinα - W = 0

Tsinα - 21.0915 = 0

Tsinα = 21.0915 Equation (2)

Equation (2) : Equation (1)

Tsinα / Tcosα = 21.0915 / 12.3

sinα / cosα = 1.7147

tanα = 1.7147

α = tan⁻¹ (1.7147)

α = 59.75°

We replace α = 59.75° in the equation (1)

Tcosα = 12.3

Tcos59.75° = 12.3

T = 12.3/cos59.75°

T = 24.4 N

α = 59.75°, angle from the horizontal

β = 90 - 59.75°

β = 30.25°, angle from the vertical