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3 years ago

5

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.)

1 answer:

Katyanochek1 [597]3 years ago

6 0

Answer:

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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