Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 3 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Find the tension in each wire and the magnitude of each tension. (Use g

T₁ = 19.39N with an angle of 52° with the horizontal.

T₂ = 15.64N with an angle of 40° with the horizontal.

Explanation:

1) We calculate the weight of holiday decoration

W = m*g

W: holiday decoration weight in Newtons (N)

m: holiday decoration weight = 3 kg

g: acceleration due to gravity = 9.8 m/s²

W = m*g = 3 kg * 9.8 m/s² = 29.4 N

2) We apply Newton's first law to the system in equilibrium:

ΣFx = 0

T₁Cos52° - T₂Cos40°=0

T₁Cos52° = T₂Cos40°

T₁ = T₂ (Cos40° / T₁Cos52°)

T₁ = 1,24*T₂ Equation (1)

ΣFy = 0

T₁Sin52° + T₂Sin40° - 29.4 = 0 Equation (2)

We replace T₁ of Equation (1) in the Equation (2)

1.24*T₂*Sin52° + T₂Sin40° - 29.4 = 0

1.88*T₂ = 29.4

T₂ = 29.4/1.88

T₂ = 15.64N

We replace T₂ = 15.64N in the Equation (1) to calculate T₁:

T₁ = 1.24*15.64N

T₁ = 19.39N