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

b.

explanation:

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