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Homework Statement
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
Homework Equations
$$\vec{F}=m\vec{a}$$
The Attempt at a Solution
Assuming [itex]\theta=[/itex] 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?
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