A crazed armadillo with a mass of 60kg stands at the center of a rope which was initially strung horizontally between two poles 50m apart. Its weight causes the rope to sag 1.20m. What is the tension in the rope

Solution:

2Tsin(angle) - mg = 0

Tan-1() = 1.2/25 = 2.75°

2Tsin(2.75°) - mg = 0

T = mg/2Sin(2.75°)
T = 6200 N

My questions is regarding the first part of the solution and how would I know I have to set 2Tsin(angle) - mg equal to 0.
Please help me understand this problem.

the rope is evidently mass less.

total force up = total force down because nothing is accelerating.

Total force down = m g

Tension in rope is T on each side of mass
horizontal component of T = T cos angle above horizontal right and left

vertical component of T = T sin angle, twice because from both sides of mass

so
total up force = 2 T sin angle
so 2 T sin angle = m g

to get angle simple geometry
tan angle = 1.2/(.5*50)

In this problem, we are dealing with a rope that is sagging due to the weight of the armadillo. To find the tension in the rope, we need to analyze the forces acting on the rope.

First, let's consider the forces acting on the armadillo. The only force acting on the armadillo is its weight, which is given by mg, where m is the mass of the armadillo and g is the acceleration due to gravity.

Next, let's consider the forces acting on the rope. The tension in the rope is an upward force that opposes the weight of the armadillo. This tension force is acting in both directions because the rope is symmetric. Therefore, we can consider 2T, where T is the tension in the rope.

Since the rope is sagging, there is an angle formed between the sagging rope and the horizontal line connecting the two poles. Let's call this angle θ.

Now, using the equilibrium condition, we can write the forces acting on the rope as a vector sum. The vertical component of the tension force is 2Tsin(θ), which is the upward force. We subtract the downward force due to the weight of the armadillo, which is mg.

Therefore, we have 2Tsin(θ) - mg = 0.

To solve for θ, we can use the given information that the rope sags 1.20m over a distance of 50m. Dividing the sag height by the distance between the poles gives us the tangent of the angle: tan(θ) = 1.2/25.

Using the inverse tangent (arctan) function, we find that θ ≈ 2.75°.

Finally, substituting the value of θ back into the equation 2Tsin(θ) - mg = 0, we can solve for T:

2Tsin(2.75°) - mg = 0.

Simplifying this equation, we can solve for T:

T = mg / (2sin(2.75°)).

After substituting the given values for m, g, and solving the equation, we find that T ≈ 6200 N.

So, in summary, setting the equation 2Tsin(θ) - mg = 0 allows us to find the tension in the rope by accounting for the vertical force balance.