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hjr
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first time posting
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, 23 m away. The rope must sag sufficiently so it won't break. Assume the rope can provide a tension force of up to 28 kN before breaking, and use a "safety factor" of 10 (that is, the rope should only be required to undergo a tension force of 2.8 kN) 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 73.0 kg.
F=ma
So basically the x component is throwing me off. The tension in the rope shouldn't be greater then 2.8kN. I tried to add the two vectors of the tension force so I had one vector pointing straight up and the force of mg pointing down. I believe the acceleration should be zero since you don't want the guy moving. So the force pushing down shouldn't be greater then the upward force. But then isn't the downward force always the same? I thought maybe it had to do something with velocity but the acceleration is constant. Then tried making two triangles with the x component being the base for each. But they give no angles, so I gave up on that idea. Then I tried to relate the length of the rope(they give it for a reason) to the max force allowed but I can't see the connection. I can't jump from the force to the distance. If someone could point me in the right direction I would greatly appreciate it. You can send me a link to a concept I'm missing here, anything that helps.
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, 23 m away. The rope must sag sufficiently so it won't break. Assume the rope can provide a tension force of up to 28 kN before breaking, and use a "safety factor" of 10 (that is, the rope should only be required to undergo a tension force of 2.8 kN) 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 73.0 kg.
Homework Equations
F=ma
The Attempt at a Solution
So basically the x component is throwing me off. The tension in the rope shouldn't be greater then 2.8kN. I tried to add the two vectors of the tension force so I had one vector pointing straight up and the force of mg pointing down. I believe the acceleration should be zero since you don't want the guy moving. So the force pushing down shouldn't be greater then the upward force. But then isn't the downward force always the same? I thought maybe it had to do something with velocity but the acceleration is constant. Then tried making two triangles with the x component being the base for each. But they give no angles, so I gave up on that idea. Then I tried to relate the length of the rope(they give it for a reason) to the max force allowed but I can't see the connection. I can't jump from the force to the distance. If someone could point me in the right direction I would greatly appreciate it. You can send me a link to a concept I'm missing here, anything that helps.
