Finding Tension of a Massless Rope

[Yosty22]

Homework Statement

An adventurous archaeologist crosses between two rock cliffs by slowly going hand-over-hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope. The rope will break if the tension in it exceeds 2.25*10^4N, and our hero's mass is 86.9kg.

If the angle between the rope and the horizontal is θ=11.7°, find the tension, T, in the rope.

Homework Equations

ƩF=ma
ƩF_x=ma_x
ƩF_y=ma_y

The Attempt at a Solution

All I could think of doing was breaking everything down into components, but I still can't get it right. Since θ is 11.7° below the horizontal (from the x-axis down 11.7°) [I will attach a picture], I took the components of the tension. Doing this, I got T_x=Tcos(11.7) and T_y=Tsin(11.7). Since the man is in the middle of the rope, it would mean that there is an equal vector of T_x and T_y on each side of the man. No matter how I break it up, I get: 2T_x=ma_x and 2T_y-mg=ma_y. No matter how I set up the equations, I am left with 2 unknowns, so I don't know if I am missing something or doing something wrong.
Thanks again in advance.

 

[tms]

The man remains at rest, so what is his acceleration? As for the x direction, are the signs of the tension on the left and right the same?

 

[Yosty22]

That is where I got confused. Taking another look at the problem, I realize that T_x in each direction are the same. Does this mean that: ƩF_x=0? If so, in the equation ƩF_y=ma_y, does a_y=-g?

 

[tms]

That is where I got confused. Taking another look at the problem, I realize that T_x in each direction are the same. Does this mean that: ƩF_x=0?

Yes, obviously, since there is no acceleration.

If so, in the equation ƩF_y=ma_y, does a_y=-g?

No. Again, there is no acceleration, so the vertical forces must cancel.

 

[Chestermiller]

Your original equation in the y-direction was correct, except that a_y=0. So solve for T_y, and then use the trigonometric relation in the y-direction to back out the value of T.

Chet