Solve F=ma Zip Line Problem: Angles & Sag

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In summary, the conversation discusses using a zip line to cross a chasm, and determining the sag in the rope to match its recommended safety range. The problem involves calculating the tension in the ropes and resolving the weight of the zipliner. The maximum allowable tension is 26kN, with a safety factor of 10, meaning the tension in each rope should not exceed 2.6kN. The angle theta is not needed to solve the problem, and the focus should be on finding the maximum sag 'x' to keep the tension below 2.6kN.
  • #1
Jae
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Question Info:
A fun way to cross a chasm is to use a so-called Zip Line as shown in the figure. Assume that the width of the chasm you are crossing is, 21m . The rope must sag sufficiently so it won't break. Assume the rope can provide a tension force of up to 26kNbefore breaking, and use a "safety factor" of 10 (that is, the rope should only be required to undergo a tension force of 2.6kN ) when you are in the position shown.(Figure 1)

Question:
Determine the sag in the rope, x , for the rope to match its recommended safety range. Assume that your mass is 75.0kg.

Diagram:
zipLine.PNG


As I am trying to solve this problem, I can't seem to find the angle theta that would allow me to calculate x.
 
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  • #2
You are given the mass of the Zipliner, and you also know the maximum tension force allowed in the rope.

Start by drawing a free body diagram of this problem and work out the tension produced in each rope while
they are holding up the Zipliner. For now, the angle theta will be treated as an unknown until further analysis
can be done.
 
  • #3
SteamKing said:
You are given the mass of the Zipliner, and you also know the maximum tension force allowed in the rope.

Start by drawing a free body diagram of this problem and work out the tension produced in each rope while
they are holding up the Zipliner. For now, the angle theta will be treated as an unknown until further analysis
can be done.
I tried drawing the free body diagram, but I don't see any way I can get theta without a spring constant. How can this be done without the spring constant? Thank you for your time.
 
  • #4
Yes. The stretch in the rope caused by the maximum allowable tension is assumed to be negligible, i.e. treat the lines as if they
are rigid bodies.
 
  • #5
SteamKing said:
Yes. The stretch in the rope caused by the maximum allowable tension is assumed to be negligible, i.e. treat the lines as if they
are rigid bodies.
If the stretch is negligible, how can I calculate x?
 
  • #6
Jae said:
If the stretch is negligible, how can I calculate x?

What you are trying to determine is the tension in the ropes, not the amount of stretch. You are over analyzing this problem.

You know the weight of the guy hanging off the line and the total distance between supports. It basically comes down to resolving the weight of the Zipliner into each rope, and figuring out the maximum sag 'x' which keeps the calculated tension in each rope below the allowable value.
 
  • #7
SteamKing said:
What you are trying to determine is the tension in the ropes, not the amount of stretch. You are over analyzing this problem.

You know the weight of the guy hanging off the line and the total distance between supports. It basically comes down to resolving the weight of the Zipliner into each rope, and figuring out the maximum sag 'x' which keeps the calculated tension in each rope below the allowable value.
So the 75 * 9.81 = 735.75N divided by 2 is 367.875N. The only way I can think to find an angle here is by doing arctan (736/368), but that's the angle of the forces. How would I find the negligible angle that would allow me to find x?
 
  • #8
Jae said:
So the 75 * 9.81 = 735.75N divided by 2 is 367.875N.

This is OK.

Now remember what the problem statement told you: you want to find the maximum sag 'x' so that the tension in each rope does not exceed 2.6 kN. How do you use this restriction in your calculations?
 

1. What is the formula for solving the F=ma zip line problem?

The formula for solving the F=ma zip line problem is F=ma+mgsinθ, where F is the total force, m is the mass of the object, a is the acceleration, g is the acceleration due to gravity, and θ is the angle at which the zip line is suspended.

2. How do angles affect the F=ma zip line problem?

Angles play a crucial role in determining the total force required to move an object on a zip line. The steeper the angle of the zip line, the greater the force needed to counteract the force of gravity and maintain a constant speed.

3. Why is sag important in the F=ma zip line problem?

Sag is the amount of downward dip in the zip line due to the weight of the object being transported. It is important to consider sag in the F=ma zip line problem because it affects the angle and, therefore, the total force needed to move the object.

4. How can the F=ma zip line problem be solved for multiple objects?

If there are multiple objects on the zip line, the total force required can be calculated by adding the individual forces for each object using the formula F=ma+mgsinθ. The angle and sag should be determined for each object separately.

5. What are some common challenges in solving the F=ma zip line problem?

One common challenge in solving the F=ma zip line problem is determining the correct angle and sag for the zip line, as these factors can vary depending on the weight and positioning of the object. Another challenge is accounting for external forces such as wind or friction, which can affect the total force needed to move the object.

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