Finding Tension of a Massless Rope

1. Feb 16, 2013

Yosty22

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

ƩF=ma
ƩFx=max
ƩFy=may

3. 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 Tx=Tcos(11.7) and Ty=Tsin(11.7). Since the man is in the middle of the rope, it would mean that there is an equal vector of Tx and Ty on each side of the man. No matter how I break it up, I get: 2Tx=max and 2Ty-mg=may. 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.

Attached Files:

• MP Tension.jpg
File size:
8.4 KB
Views:
225
2. Feb 16, 2013

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?

3. Feb 16, 2013

Yosty22

That is where I got confused. Taking another look at the problem, I realize that Tx in each direction are the same. Does this mean that: ƩFx=0? If so, in the equation ƩFy=may, does ay=-g?

4. Feb 16, 2013

tms

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

5. Feb 16, 2013

Staff: Mentor

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

Chet