In a mountain-climbing technique called the "Tyrolean traverse," a rope is anchored on both ends (to rocks or strong trees) across a deep chasm, and then a climber traverses the rope while attached by a sling as in the figure . This technique generates tremendous forces in the rope and anchors, so a basic understanding of physics is crucial for safety. A typical climbing rope can undergo a tension force of perhaps 27 kN before breaking, and a "safety factor" of 12 is usually recommended. The length of rope used in the Tyrolean traverse must allow for some "sag" to remain in the recommended safety range.
Consider a 75kg climber at the center of a Tyrolean traverse, spanning a 25m chasm. To be within its recommended safety range, what minimum distance x in meters must the rope sag?
If the Tyrolean traverse is set up incorrectly so that the rope sags by only one-fourth the distance found in Part A, determine the tension in the rope in Newtons.
Will the rope break?
Please, someone help me understand this problem.
Take the angle of sag. if Then x/12.5 =tan Sagangle.
But look at force. The downward force is 1/2 75g (each side holds half the weight)
so 37.5g/Tension=Sin Sagangle
But the tension cannot be greater than 27kN/12
so you can then determine the sag angle. Put that back in to the tangent equation, and find x.
Physics 201
Why did the rope go to the circus? Because it wanted to see the high-wire act! Now, let's tackle this problem together.
In order to calculate the minimum distance that the rope must sag for the climber to be within the recommended safety range, we need to make use of some physics principles.
The tension in the rope can be found using the weight of the climber and the gravitational force acting on them. The weight (W) of the climber can be calculated using the equation W = mg, where m is the mass of the climber (75 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The tension in the rope (T) can be calculated using the equation T = 2W, since there are two segments of rope supporting the climber.
To find the minimum distance the rope must sag, we need to consider the forces acting on the rope. At the center of the Tyrolean traverse, the total tension force in the rope is equal to the weight of the climber, since the forces are balanced.
Now, let's calculate the minimum distance x in meters that the rope must sag:
T = 2W
T = 2mg
T = 2 * (75 kg) * (9.8 m/s^2)
T = 14700 N
To calculate the minimum distance x, we need to consider the sag in the rope. The sag forms a right triangle, with the rope acting as the hypotenuse. We can use Pythagoras' theorem to find the length of the sag:
x^2 = (25 m)^2 - (12.5 m)^2
x^2 = 625 m^2 - 156.25 m^2
x^2 = 468.75 m^2
x ≈ 21.63 m
So, the minimum distance x that the rope must sag to remain within the recommended safety range is approximately 21.63 meters.
Now, let's move on to the second part of the problem. If the Tyrolean traverse is set up incorrectly and the rope sags by only one-fourth the distance found in Part A (approximately x/4 = 5.41 meters), we can calculate the tension in the rope as before:
T = 2W
T = 2mg
T = 2 * (75 kg) * (9.8 m/s^2)
T = 14700 N
So, the tension in the rope is still 14700 N, regardless of the incorrect setup.
Will the rope break? Well, since the maximum tension the rope can withstand before breaking is given as 27 kN (27000 N), and the tension in the rope is only 14700 N, the rope will not break in this scenario. Phew!
I hope this clarifies the problem for you. Remember, always double-check your setup to ensure a safe climb.
To solve this problem, we'll need to use some basic physics principles. Let's break it down step-by-step:
Step 1: Calculate the weight of the climber
The weight of the climber can be calculated using the formula:
Weight = mass * acceleration due to gravity
Weight = 75 kg * 9.8 m/s^2 (acceleration due to gravity)
Weight = 735 N
Step 2: Calculate the tension force in the rope
In the Tyrolean traverse, the tension force in the rope is maximum at the anchor points and gradually decreases towards the center. However, to ensure safety, we'll consider the maximum tension force that the rope can handle.
The recommended tension force the rope can handle is 27 kN. However, we need to consider the safety factor of 12, so the actual tension force will be:
Tension force = 27 kN / 12
Tension force = 2250 N
Step 3: Calculate the sag distance
To calculate the minimum sag distance, we need to consider the weight of the climber and the tension force in the rope.
Since the rope forms a right triangle with the sag distance (x) as the vertical side and half the rope length (25 m / 2 = 12.5 m) as the horizontal side, we can use trigonometry to find the sag distance.
Using the formula for tension force:
Tension force = weight / sin(theta)
where theta is the angle made by the rope with the horizontal direction, we can rearrange the equation to solve for theta:
theta = arcsin(weight / tension force)
theta = arcsin(735 N / 2250 N)
theta ≈ 18.9 degrees
Using the formula for the sag distance:
sag distance = horizontal side * tan(theta)
sag distance = 12.5 m * tan(18.9 degrees)
sag distance ≈ 4.17 m
So, the minimum sag distance required is approximately 4.17 meters.
Step 4: Calculate the tension in the incorrectly set up traverse
If the rope only sags by one-fourth of the minimum sag distance (1/4 * 4.17m = 1.04m), we can calculate the tension in the rope using similar trigonometry principles.
Using the same right triangle formed by the sag distance (1.04m) as the vertical side and half the rope length (12.5m) as the horizontal side, we can find the angle theta:
theta = arcsin(1.04m / 12.5m)
theta ≈ 4.94 degrees
Using the formula for tension force, with weight = 735 N:
Tension force = weight / sin(theta)
Tension force = 735 N / sin(4.94 degrees)
Tension force ≈ 13,314 N
Step 5: Determine if the rope will break
The tension force in the incorrectly set up traverse is 13,314 N. Since the recommended tension force the rope can handle is 27 kN (or 27,000 N), and the tension force we calculated is significantly less, the rope will not break.
I hope this step-by-step explanation helps you understand the problem.
To determine the minimum distance the rope must sag in order to remain within the recommended safety range, we can analyze the forces acting on the rope when a climber is at the center of the Tyrolean traverse.
First, let's calculate the weight of the climber using the formula:
Weight = mass × acceleration due to gravity
Weight = 75 kg × 9.8 m/s^2
Weight = 735 N
The tension in the rope must be able to support the weight of the climber. To calculate the tension, we use the following formula:
Tension = Weight + Additional forces
In this case, the additional forces come from the rope's sag. The sagging rope creates an upward force due to its shape.
Now, let's consider the forces:
1. Weight of the climber = 735 N (downward force)
2. Upward force due to sagging (upward force)
To determine the second force, we'll use the property of a "catenary curve," which describes the shape of a hanging rope. The equation for the catenary curve is:
y(x) = a * cosh(x/a)
Where "y" is the vertical position, "x" is the horizontal position, and "a" is half the distance between the anchors.
In this case, the anchor points are 25 m apart, so a = 12.5 m.
To find the upward force, we need to differentiate the equation to find the slope of the catenary curve. Then, we can calculate the vertical component of the slope at x = 12.5 m.
Differentiating the equation, we get:
dy/dx = sinh(x/a)
Evaluating at x = 12.5 m, we find:
dy/dx = sinh(12.5/12.5) = sinh(1) ≈ 1.175
Now we know the vertical slope of the curve at x = 12.5 m, which represents the upward force due to sagging.
To calculate the upward force, we multiply the vertical slope by the weight of the climber:
Upward force = dy/dx × Weight
Upward force = 1.175 × 735 N
Upward force ≈ 863.125 N
The tension in the rope must be equal to the sum of the weight and the upward force due to sagging:
Tension = Weight + Upward force
Tension = 735 N + 863.125 N
Tension ≈ 1598.125 N
Now, let's move on to part B of the problem.
If the rope sags by only one-fourth the distance found in part A (x/4), we can repeat the calculations.
The new sag distance (x/4) = 25 m / 4 = 6.25 m
Using the catenary curve equation, we can find the vertical component of the slope at x = 6.25 m:
dy/dx = sinh(x/a)
dy/dx = sinh(6.25/12.5) = sinh(0.5) ≈ 0.521
Now we calculate the new upward force:
Upward force = dy/dx × Weight
Upward force = 0.521 × 735 N
Upward force ≈ 383.135 N
Finally, we calculate the new tension in the rope:
Tension = Weight + Upward force
Tension = 735 N + 383.135 N
Tension ≈ 1118.135 N
Comparing the new tension (1118.135 N) to the breaking strength of the rope (27 kN = 27,000 N), we see that the rope will not break since the tension is well below the breaking strength.
So, to summarize:
A) The minimum distance x the rope must sag is approximately 12.5 meters to remain within the recommended safety range.
B) If the rope sags by only one-fourth the distance found in part A, the tension in the rope is approximately 1118.135 N, and the rope will not break.