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

Answer 1

Answer:

Explanation:

Let the tension in ropes be T₁ and T₂ . Ropes are making angle of 52 and 40 degree with the horizontal . The vertical component of tension will add up together to balance the weight of the decoration

T₁ sin52 + T₂ sin40 = 3 x 9.8

.788 T₁ + .643 T₂ = 29.4

The horizontal component of tension will add up to zero because the decoration piece is at rest .

T₁ cos52 + T₂ cos40 = 0

.615 T₁ -  .766 T₂ = 0

T₁ =  1.24 T₂

Substituting this value of T₁ in earlier equation , we have

.788 x 1.24 T₂ + .643 T₂ = 29.4

1.62 T₂ = 29.4

T₂ = 18.15 N

T₁ = 1.24 x 18.15 = 22.51 N .