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 27 kN before breaking, and use a "safety factor" of 10 (that is, the rope should only be required to undergo a tension force of 2.7 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 74.0 kg.
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 the rope in newtons.
Will the rope break?
How can you do this problem? I can't even start it.
To determine the distance x that the rope must sag in order to be within its recommended safety range, we need to consider the weight of Christian and the tension force in the rope.
First, let's find the weight of Christian using the formula: weight = mass * acceleration due to gravity. Given that Christian's mass is 74.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight:
Weight = 74.0 kg * 9.8 m/s^2
Weight = 725.2 N
Next, let's consider the tension force in the rope. We know that the tension force at the center of the traverse is 2.7 kN (10 times less than the rope's maximum tension capacity). Since the Tyrolean traverse is symmetric, this tension force will be shared equally between the two sides of the rope. Therefore, the tension force in each side of the rope is:
Tension force = 2.7 kN / 2
Tension force = 1.35 kN
We can now consider the forces acting on the sagging rope. The weight of Christian acts vertically downwards, and the tension force in the rope acts horizontally towards the trees. These forces create a right triangle, and the sagging distance x is the height of this triangle.
Using Pythagoras' theorem, we can relate the tension force, the weight of Christian, and the sagging distance:
Tension force^2 = weight^2 + (x/2)^2
Rearranging the equation, we get:
(x/2)^2 = Tension force^2 - weight^2
Plugging in the values we have:
(x/2)^2 = (1350 N)^2 - (725.2 N)^2
(x/2)^2 = 1822500 N^2 - 526570.04 N^2
(x/2)^2 = 1295930 N^2
Taking the square root of both sides, we get:
x/2 = sqrt(1295930 N^2)
x/2 = 1138.358 N
Finally, multiplying both sides by 2, we find:
x = 2276.716 N
Therefore, the rope must sag approximately 2276.716 N to be within its recommended safety range.
To find the tension force in the rope if it sags only one-fourth of the distance found above, we can use similar calculations.
Tension force = weight^2 + (x/8)^2
Tension force^2 = (725.2 N)^2 + (2276.716 N/8)^2
Tension force^2 = 526570.04 N^2 + 8020.2135 N^2
Tension force^2 = 534590.2535 N^2
Taking the square root, we find:
Tension force = sqrt(534590.2535 N^2)
Tension force ≈ 729.9 N
Therefore, the tension force in the rope, if it sags only one-fourth of the recommended distance, is approximately 729.9 N.
As for whether the rope will break, we can compare the tension force in the rope with its maximum tension capacity. Given that the maximum tension capacity of the rope is 27 kN and the tension force required for safety is 2.7 kN, we can see that even at the maximum sagging distance, the tension force in the rope (1350 N) is well within the maximum tension capacity.
Therefore, the rope will not break if properly set up according to the given parameters.