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Chapter 3

Q. 3.1

The owners of a zip line attraction want to know if they can increase the weight limit for riders from 100 kg to 120 kg. They recently purchased new equipment that is designed for the increase in rider weight. The question is whether the zip line and anchorages are strong enough.

Our team is analyzing the cable considering the live load at various locations along the zip line. The team has collectively decided that we should use a load factor of 2.0 for the live load rather than the code factor because of the lack of redundancy in the system. The resulting factored live load is 2.4 kN.

To perform a preliminary analysis, the team has decided to make a couple of simplifying assumptions:

• Although the problem is geometrically nonlinear, we are going to superimpose the results of analysis of the cable subjected to self-weight with the results of the cable with a live load.
• Although the rider is not in equilibrium, we are going to conservatively treat the problem as static. In reality, the rider’s weight vector remains vertical and the unresolved horizontal component at the pulley connection is what accelerates and decelerates the rider.

We have been assigned to analyze the cable with a rider halfway across the zip line. An experienced member of the team tells us that for this particular situation, the cable should drop by 3 m from the start point to the rider. Our task is to find the resulting force in the cable.

E3.22
E3.23

Step-by-Step

Verified Solution

Analysis by System of Nonlinear Equations:
Since there is only one applied load, we can solve for both the load angle and the cable force by hand without needing numerical methods.

Calculate the cable segment lengths:

l_{AB} = \sqrt{(20  m)²  +  (3  m)²} = 20.22  m

 

l_{BC} = \sqrt{(20  m)²  +  (3  m  –  2  m)²} = 20.02  m

FBD of B:

\underrightarrow{+} \sum{F_x} = 0 = -\frac{20}{20.22}F_{cable} + \frac{20}{20.02}F_{cable} + 2.4 kN sin\theta _B

-9.88\times 10^{-3}F_{cable} = 2.4  kN\sin \theta_B

-4.12 × 10^{-3}/kN  •  F_{cable} = sin  θ_B

+↑ΣF_y = 0 = \frac{3}{20.22}F_{cable} + \frac{1}{20.02}F_{cable}  –  2.4  kN  cos  θ_B

-0.1983F_{cable} = -2.4  kN  cos  θ_B

8.26×10^{-2}/kN  •  F_{cable} = cos  θ_B

Divide the two equations:

\frac{-4.12×10^{-3}/kN  •  F_{cable}}{8.26×10^{-2}/kN  •  F_{cable}} = \frac{sin θ_B}{cos  θ_B}

-4.99×10^{-2} = tan  θ_B

θ_B = -2.86°

Substitute to find the cable force:

8.26×10^{-2}/kN  •  F_{cable} = cos(-2.86°)

F_{cable} = 12.09  kN

E3.1;