Answer:
Ft = 50.25 N
Step-by-step explanation:
As you can see in the attached diagram, the angle θ is given by
tanθ = opp/adj
The opposite is the sag of distance 2 m and the adjacent is the half of the horizontal clothesline that is 20 m.
tanθ = 2/10
θ = tan⁻¹(2/10)
θ = 11.30°
There are two forces along the y axis, one is the F = mg acting downwards and the other is the Ftsinθ component of tension force in the string.
The sum of forces along y-axis is
5Ftsin(11.30) - mg = 0
Ft = mg/5sin(11.30)
Ft = 5*9.8/5sin(11.30)
Ft = 9.8/0.195
Ft = 50.25 N
Therefore, the magnitude of the tension on the ends of the clothesline is 50.25 N.