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?

Alright, so i have this a little bit started but not sure if I'm going in the right direction. Here's what i have:
sin 2/sqrt(104)
cos 10/sqrt(104)
T1 = -|T1| cos 10/sqrt(104)i + |T1|sin 2/sqrt(104)j

T2 = |T2| cos 10/sqrt(104)i + |T2|sin 2/sqrt(104)j

T1 + T2 + F = 0
Force: -5(9.8)

Do i have this set up right? And if i do, why are certain things positive and negative? What's the next step?

The problem is much simpler than you are doing. Due to symettry, each side is supporting half the weight (1/2 mg= 2.5g)

Now, using your angles

SinTheta=2.5g/Tension
Now substitute for sintheta, 2/sqrt(104) and solve for tension

To find the magnitude of the tension on the ends of the clothesline, let's break down the problem step by step.

First, let's consider the symmetry of the problem. Since the mass is tied to the middle of the clothesline, we can assume that the tension forces at the ends of the clothesline are equal in magnitude and opposite in direction. Let's call the tension on one end T (without any subscript).

Now, let's draw a free body diagram for the mass in the middle. We have the force of gravity acting downward with a magnitude of -5 * 9.8 N (since it acts in the opposite direction of the positive y-axis).

Since the clothesline sags 2 meters, we can determine that the vertical component of the tension force on either end is 2 N (equal and opposite). However, the horizontal component of the tension force should cancel out since the forces on both sides are equal.

So, the tension force can be represented as T x cosθ i + (-2) j, where θ is the angle between the tension force and the positive x-axis.

Since the tension forces at both ends have the same magnitude, we have:

T x cosθ = T x cosθ

Now, let's look at the x and y-components of the tension forces:

x-component: T x cosθ = -T x cosθ (since the horizontal component cancels out)
y-component: -2 = 2

From the y-component equation, we can see that -2 = 2, which is not true. This implies that there must be an error in our setup.

Upon revisiting the problem, we realize that the sagging of the clothesline is not due to the weight of the mass but rather the horizontal component of the tension force. Therefore, the x-component should be 5g (since the mass is acting as a force pulling down). This means the equation for the x-component should be:

T x cosθ = 5 x 9.8

Now, we can proceed to solve for T.

T x cosθ = 5 x 9.8
T x cosθ = 49

To find the value of cosθ, we can use the Pythagorean theorem:

cosθ = adjacent / hypotenuse
cosθ = 20 / √(20^2 + 2^2)
cosθ = 20 / √(400 + 4)
cosθ = 20 / √404
cosθ ≈ 20 / 20.099
cosθ ≈ 0.995

Substituting the value of cosθ back into the equation:

T x 0.995 ≈ 49
T ≈ 49 / 0.995
T ≈ 49.25 N

So, the magnitude of the tension on the ends of the clothesline is approximately 49.25 Newtons.