Chapter 4, Problem 31P

(II) Christian is making a Tyrolean traverse as shown in Fig. 4-35. 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, 25 m away. The rope must sag sufficiently so it won’t break. Assume the rope can provide a tension force of up to 29 kN before breaking, and use a “safety factor” of 10 (that is, the rope should only be required to undergo a tension force of 2.9 kN) at the center of the Tyrolean traverse. (a) Determine the distance x that the rope must sag if it is to be within its recommended safety range and Christian’s mass is 72.0kg. (b) If the Tyrolean traverse is incorrectly set up so that the rope sags by only one-fourth the distance found in (a), determine the tension force in the rope. Will the rope break?

FIGURE 4-35.

Problem 31.

(a) We draw a free-body diagram for the piece of the rope that is directly above the person. That piece of rope should be in equilibrium. The person’s weight will be pulling down on that spot, and the rope tension will be pulling away from that spot towards the points of attachment. Write Newton’s second law for that small piece of the rope.

${\displaystyle \sum F_y=2F_T\sin \theta -mg=0\to \theta ={\sin }^{-1}\frac{mg}{2F_T}={\sin }^{-1}\frac{(72.0\text{kg})(9.80{\text{m/s}}^2)}{2(2900\text{N})}={6.988}^{\circ }}$ $\tan \theta =\frac{x}{12.5\text{m}}\to x=(12.5\text{m})\tan {6.988}^{\circ }=1.532\text{m}\approx \boxed{1.5\text{m}}$

(b) Use the same equation to solve for the tension force with a sag of only $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ that found above.

$F_{\text{T}}=\frac{mg}{2\sin \theta }=\frac{(72.0\text{kg})(9.80{\text{m/s}}^2)}{2(\sin {1.755}^{\circ })}=\boxed{11.5\text{ kN}}$

The rope will not break, but it exceeds the recommended tension by a factor of about 4.