Am = 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

v=ir = 3x15 =45 v