Finding Tension of a Massless Rope

In summary, the archaeologist's rope will break if the tension in it exceeds 2.25*10^4N. If the angle between the rope and the horizontal is θ=11.7°, find the tension, T, in the rope.
  • #1
Yosty22
185
4

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
ƩFx=max
ƩFy=may

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.
Thanks again in advance.
 

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  • #2
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
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
Yosty22 said:
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?
Yes, obviously, since there is no acceleration.
If so, in the equation ƩFy=may, does ay=-g?
No. Again, there is no acceleration, so the vertical forces must cancel.
 
  • #5
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
 

What is "Finding Tension of a Massless Rope"?

"Finding Tension of a Massless Rope" is a scientific concept used to calculate the force or tension exerted by a rope that has negligible mass. It is often used in physics and engineering to analyze the forces acting on objects connected by ropes or strings.

Why is the mass of the rope considered negligible?

In order to simplify the calculations, the mass of the rope is often assumed to be negligible or zero. This is because the mass of the rope is significantly smaller compared to the objects it is connected to, and therefore its contribution to the overall force is considered to be insignificant.

How is the tension of a massless rope calculated?

The tension of a massless rope is calculated using the formula T = m x g, where T represents tension, m represents the mass of the object connected to the rope, and g represents the acceleration due to gravity. This formula assumes that the rope is in a state of equilibrium and all forces acting on it are balanced.

Can the tension of a massless rope be greater than the weight of the object it is supporting?

Yes, the tension of a massless rope can be greater than the weight of the object it is supporting. This is because tension is a force, while weight is a measure of the gravitational pull on an object. In certain situations, the tension in a rope can be greater than the weight of the object it is supporting, such as when the object is accelerating upwards.

What factors can affect the tension of a massless rope?

The tension of a massless rope can be affected by various factors such as the mass of the objects connected by the rope, the angle at which the rope is pulled, and the acceleration of the objects. Other factors like the type and condition of the rope, and external forces acting on it, can also affect the tension.

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