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4 December, 20:49

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 10 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 = 9.8 m/s2 for the acceleration due to gravity. Round your answers to two decimal places.)

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  1. 4 December, 20:53
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    T₁ = 75.25 N : Wire tension forming angle of 52° with horizontal

    T₂ = 60.49 N : Wire tension forming angle of 40° with horizontal

    Explanation:

    We apply Newton's first law to the holiday decoration in equilibrium

    Forces acting on holiday decoration:

    T₁ : Wire tension forming angle of 52° with horizontal

    T₂ : Wire tension forming angle of 40° with horizontal

    W = m*g = 10 kg*9.8 m/s² = 98 N : weight of the decoration

    ∑Fx=0

    T₁x - T₂x = 0

    T₁x = T₂x

    T₁*cos52° = T₂*cos40°

    T₁ = T₂ * (cos40°) / (cos52°)

    T₁ = 1.244T₂ Equation (1)

    ∑Fy=0

    T₁y+T₂y - W = 0

    T₁*sin52° + T₂*sin40° - 98 = 0 Equation (2)

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

    1.244T₂*sin52° + T₂*sin40° - 98 = 0

    0.98T₂ + 0.643T₂ = 98

    1.62T₂ = 98

    T₂ = 98 / 1.62

    T₂ = 60.49 N

    We replace T₂ = 60.49 N in the Equation (1)

    T₁ = 1.244*60.49 N

    T₁ = 75.25 N
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