Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square.

Question

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 weight of 9 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Give your answers correct to two decimal places. (a) Find the tension in 3 m rope. (ii) Find the magnitude of this tension.(b) Find the tension in 5 m rope. (ii) Find the magnitude of this tension.

Answer 1

In general, the length of these ropes does not affect the tension. So from there, we could simply draw a picture of the sign being hung from these two ropes; we'll label the tensions on these two ropes as T₃ and T₅ (based on their respective lengths). Each tension will have a horizontal element and a vertical element. The total vertical force would be Force = mass•gravity; this will shared and separated into each rope based on the sine of it's angle:

9kg • 9.8 m/s² = 88.2 N = T₃sin(52°) + T₅sin(40°) ≈ 0.788T₃ + 0.6428T₅

The horizontal forces are equal as there's no pull towards one rope or the other. This means the horizontal components are equal (horizontal will be based on the cosine of the angle):

T₃cos(52°) = T₅cos(40°)

0.61566T₃ = 0.766T₅

Now you have a system of equations that you can solve. I'm finding that T₃ = 67.61 N and T₅ = 56.73 N.