# Homework Help: Tension on a Tyrolean Traverse

1. Sep 5, 2014

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

Christian is making a Tyrolean traverse as shown in the figure. That is, he traverses a chasm by stringing a rope between a tree on one side of the chasm and a tree on the opposite side, 25m away. The rope must sag sufficiently so it won't break. Assume the rope can provide a tension force of up to 30kN before breaking, and use a "safety factor" of 10 (that is, the rope should only be required to undergo a tension force of 3.0kN ) at the center of the Tyrolean traverse.

Determine the distance x that the rope must sag if it is to be within its recommended safety range and Christian's mass is 75.0kg

2. Relevant equations

$$\vec{F}=m\vec{a}$$

3. The attempt at a solution

Assuming $\theta=$ the angle between one hypotenuse of one half of the rope and $x$, I know that
$$T_y=mg\rightarrow 75.0kg\cdot 9.8\frac{m}{s^2}=735N$$
Now, I assume (although I don't know if this correct) that the tension $\vec{T}$ of one side of the rope will equal half that of the entire rope so that $$\|\vec{T}\|\leq\frac{1}{2}\cdot 3000N\Rightarrow 1500N$$ assuming $3000N$ is the maximum amount of tension we will allow on rope.
$$arcos(\frac{735N} {.5\cdot3000N})\approx 60.66^{\circ}$$
If the length opposite of $\theta$ is half the total distance between the trees $(\frac{25m}{2m})$, we have
$$tan(\theta)=\frac{25m}{2m\cdot x}\Rightarrow x=\frac{25}{2tan(60.66^{\circ})}\approx {7.02m}$$

The system that generated this practice problem says that this is not the correct answer. Where did I go wrong?

Last edited: Sep 5, 2014
2. Sep 5, 2014

### SteamKing

Staff Emeritus
It's always essential to draw a free body diagram of the problem.

What you should have done is figure out the vertical tension component in each rope which is required to support Christian over the chasm. Then, knowing the vertical component and the maximum allowable tension in the rope, you can find the angle of the sag, and hence the sag in the rope.

Remember, the maximum allowable tension in each rope is 3000 N, not 1500 N.

3. Sep 5, 2014

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Thanks for the feedback, SteamKing.

Looking back over my work, I believe I found the vertical component in the rope as

$$T_y=mg\rightarrow 75.0kg\cdot 9.8\frac{m}{s^2}=735N$$

Is this incorrect?

If not, knowing that the total tension $\|\vec{T}\|$ must be

$$\|\vec{T}\|\leq3000N$$

for one side of the rope, I can substitute $3000~N$ for the $1500~N$ I mistakenly had:

$$arcos(\frac{735N} {3000N})\approx 75.82^{\circ}\Rightarrow tan(\theta)=\frac{25m}{2m\cdot x}\Rightarrow x=\frac{25}{2\cdot tan(75.82^{\circ})}\approx {3.2m}$$

This doesn't appear to be the correct answer either.

4. Sep 5, 2014

### SteamKing

Staff Emeritus
Still trying to wing it without drawing the free body diagram, I see.

The vertical force due to gravity is OK, but remember, there are two parts of rope supporting Herr Christian as he crosses the Chasm. How is that accounted for in your calculations?

You're just writing down numbers without trying to understand what's going on. This is a bad approach for analyzing a simple problem; trying to work a more complex problem this way will leave you even more frustrated.

5. Sep 5, 2014

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I think I see where you're going. Should the vertical component of one side of the rope be halved because the total weight $T_y=735N$ is being split/shared between the two sides of the rope?

If so, I get the following:

$$arcos(\frac{367.5N} {3000N})\approx 82.96^{\circ}$$
$$\Rightarrow \frac{25}{2\cdot tan(82.96^{\circ})}\approx 1.54~\mathrm{m}$$

Two questions regarding this problem:

1.In problems similar to this, is it useful to think of each side of the rope as an individual rope?

2.Why is it that we do not split the tension between the ropes? (i.e., when looking at one side of the total rope, we assume that $\|\vec{T}\|\leq3000~\mathrm{N}$ rather than $1500~\mathrm{N}$?

As for the free body diagrams, I assumed it was less common for members to post pictures of their work, but I'll remember to do so next time if that's the case.

6. Sep 5, 2014

### SteamKing

Staff Emeritus
That's why you draw the FBD; to help you identify all of the forces acting to keep the given situation static and in equilibrium.

In effect, you are splitting the tension, as a FBD would have shown you. At the point where the knabe was suspended, a FBD would have shown that there is tension in both parts of the rope leading from the boy to the trees on each side of the chasm. If there is tension in only one of the ropes, you will have an unbalanced situation, where the boy cannot possibly remain static.

The wrinkle here is, you were initially told the rope could withstand a certain maximum tension, after modification due to safety factors, of 3000 N. The problem wants to know by how much the rope can sag while holding Christian over the chasm and not exceed this maximum tension. Thus, the tension in both parts of the line is established. If this sag in the rope is less than what you calculated, then statics tells us that the rope tension of 3000 N will be exceeded in order to hold the boy over the chasm, and the possibility increases of the rope breaking and sending the boy plunging into the abyss.

In problems like these, symmetry often is useful in telling us that in static situations, we are dealing with forces which are equal on both sides of the axis of symmetry; in effect, symmetry reduces the amount of calculation required.

Whether or not you post your FBD, it is still essential to prepare one in order to help you analyze and solve the problem.