Question 1
The statement explains the mathematical relationship between the sag of the cable, horizontal tension inside the cable, weight per unit length, and the distance from the support to the lowest point (sag point). The relation arises from the pressures functioning in balance on the cable, is observable as a catenary curve, which is the shape that a flexible cable adopts when it is supported on both ends.
The sag of the cable y can be stated mathematically as:
Where:
Where y is the vertical sag.
The weight of the cable, expressed in newtons per meter, is denoted by w.
The horizontal distance (in meters) between the support and the cable’s lowest point is denoted by x.
T is the tension in the cable’s horizontal component (measured in newtons).
Explanation
The formula demonstrates that the sag is inversely proportional to the horizontal tension. Besides, the formula shows that the sag directly relates to the weight per unit length and the square of the distance to the low point. It accurately depicts the physical fact that sag increases as the cable weight or span rises and decreases with increasing strain.
Question 2
The weight of the entire cable would increase if a little mass was added in the middle. To maintain equilibrium, the additional weight would raise the vertical component of the tension in the cable at the middle point, which would cause a slight increase in the whole tension in the cable. However, this effect would be negligible if the mass is tiny in relation to the total mass of the cable.
Explanation
If the increased mass is much smaller than the mass of the cable, the effect of the added mass pulling the cable’s center downward will be extremely tiny and the sag may increase little as a result.
Question 3
The speed v can be determined using the formula:
Where:
- v is the speed of the wave
- T is the tension in the cable (in newtons),
- μ is the mass per unit length of the cable
The tension is given in the question as 200,000N and mass per unit length as 1.63kg/m
= 350.28 m/s
Speed of the wave is therefore 350.28 m/s.
Question 4
The lowest natural frequency is obtained using the formula (“How to Calculate Fundamental Frequency,” 2020):
Where:
- f is the lowest natural frequency
- v is the speed of the wave along the cable, v = 350.28
- L is the length of the cable (in meters). L = 1000m
= 0.1752 Hz
Therefore, the frequency at which the cable naturally vibrates is 0.1752 Hz.
Question 5
The mass per unit length (μ) of the cable would rise if ice built up on the wires. The rise would result in:
- Tension: A higher vertical tension component would result from an increase in the cable’s overall weight.
- Sag: The extra weight dragging the cable lower would cause the sag to grow with the increased weight but before changing the tension.
- Resonant Frequency: Since the lowest natural frequency (f=v/2L) is directly proportional to the wave speed, an increase in mass per unit length would result in a decrease in the speed at which waves propagate along the cable (v=T/μ) (Saric, 2016). Therefore, the resonant frequency of vibration is decreased.
References
Saric, S. (2016). Sag and tension of conductor. Unsa-ba. https://www.academia.edu/25942985/Sag_and_Tension_of_Conductor
How to calculate fundamental frequency. (2020, December 13). Sciencing. Retrieved March 29, 2024, from https://sciencing.com/calculate-fundamental-frequency-6005910.html
