(1 point) A horizontal clothesline is tied between 2 poles, 20 meters apart. When a mass of 5 kilograms is tied to the middle of the clothesline, it sags a distance of 2 meters. What is the magnitude of the tension on the ends of the clothesline

Ft = 50.25 N

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.

multiply 2kg by latent heat of vap