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Christian is making a Tyrolean traverse. That is, he traverses a chasm by stringing a rope between a tree on one side of the ch...Asked by Precilla
Christian is making a Tyrolean traverse. 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 27kN before breaking, and use a "safety factor" of 10 (that is, the rope should only be required to undergo a tension force of 2.7kN ) at the center of the Tyrolean traverse.
Determine the distance x in meters that the rope must sag if it is to be within its recommended safety range and Christian's mass is 74.0kg.
If the Tyrolean traverse is incorrectly set up so that the rope sags by only one-fourth the distance found in part A, determine the tension force (in Newtons) in the rope.
Will the rope break?
repost that for me please
Determine the distance x in meters that the rope must sag if it is to be within its recommended safety range and Christian's mass is 74.0kg.
If the Tyrolean traverse is incorrectly set up so that the rope sags by only one-fourth the distance found in part A, determine the tension force (in Newtons) in the rope.
Will the rope break?
repost that for me please
Answers
Answered by
Precilla
How do I do it exactly....
Answered by
john
7.6
Answered by
Cris
For anyone who is looking to solve this in the future:
First figure out the angle
F= 2Tsin(theta)-mg
where, F is total force, and here it must be zero to be at equilibrium
T=tension of rope (2.7kN)
mg=mass*grav= 9.8*74
then use Tan(theta) = Opposite/ Adjacent to figure out the length (Opp)
the angle should be under 10-ish, adjacent is 25/2. So your final answer is Opposite side= should be under 2 meters. Mine was 1.7M with slightly different values
Then for the second part, you divide the length (<2meters) by four. then find the new angle where one leg is 12.5 and the other is (length/4) =. you plug that angle into
0=2Tsin(theta)-mg and solve for T :)
My answer was around 10238 N
First figure out the angle
F= 2Tsin(theta)-mg
where, F is total force, and here it must be zero to be at equilibrium
T=tension of rope (2.7kN)
mg=mass*grav= 9.8*74
then use Tan(theta) = Opposite/ Adjacent to figure out the length (Opp)
the angle should be under 10-ish, adjacent is 25/2. So your final answer is Opposite side= should be under 2 meters. Mine was 1.7M with slightly different values
Then for the second part, you divide the length (<2meters) by four. then find the new angle where one leg is 12.5 and the other is (length/4) =. you plug that angle into
0=2Tsin(theta)-mg and solve for T :)
My answer was around 10238 N
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