Calculating Length of a Cable Between Poles

• blumfeld0
In summary: The 15.1 pounds per linear foot is the "weight of the cable". That is, if you cut the cable at any point and weighed it, it would weigh 15.1 pounds for each foot of length. Since you are finding the "length of the curve", you need to multiply by the weight per foot. (39.25+ 24.9i)(15.1)= 593.775+ 375.39i. I don't see why you would get an imaginary answer.
blumfeld0
A cable hangs between two poles of equal height and 37 feet apart.
At a point on the ground directly under the cable and
x feet from the point on the ground halfway between the poles
the height of the cable in feet is
h(x)=10 +(0.4)( x^{1.5})
The cable weighs 15.1 pounds per linear foot.
Find the weight of the cable.

so i find h'(x) i square it and add one to find the length of the curve
i.e
(1+ h'(x)^2 ) ^(1/2)

my question is is that right? and what are the limits of integration??

blumfeld0

blumfeld0 said:
A cable hangs between two poles of equal height and 37 feet apart.
At a point on the ground directly under the cable and
x feet from the point on the ground halfway between the poles
the height of the cable in feet is
h(x)=10 +(0.4)( x^{1.5})
The cable weighs 15.1 pounds per linear foot.
Find the weight of the cable.

so i find h'(x) i square it and add one to find the length of the curve
i.e
(1+ h'(x)^2 ) ^(1/2)

my question is is that right? and what are the limits of integration??

blumfeld0
Well, more correctly "i find h'(x) i square it and add one" , take the square root and integrate "to find the length of the curve"

Since your variable, x, is the "from the point on the ground halfway between the poles", x= 0 there. At one pole x= -37/2 and at the other x= 37/2. Integrate with respect to x from -37/2 to 37/2.

ok so i integrate (1 + h'(x)^2 )^(1/2) with repect to x from -37/2 to 37/2
i use mathematica and i get 39.25+ 24.9 i

where "i" is imaginary
why do i get an imaginary answer??also what does the 15.1 pounds per linear foot have to do with it?

thanks

Ah! Misinterpretation on my part. Since x is a distance it is always positive. The height "h(x)=10 +(0.4)( x^{1.5})" isn't even defined for x negative. Integrate from 0 to 37/2 to find the length of 1/2 and then double.

1. How do I calculate the length of a cable between two poles?

To calculate the length of a cable between two poles, you will need to measure the distance between the two poles using a measuring tape or a surveying tool. Once you have the distance, you can use the Pythagorean theorem to calculate the length of the cable, which is the hypotenuse of the right triangle formed by the distance between the poles and the height of the poles.

2. What units should I use when calculating the length of a cable between poles?

The units you use will depend on the measurement of the distance between the poles. If you measured the distance in feet, then you should use feet as your unit of measurement for the cable length. If you measured the distance in meters, then you should use meters as your unit of measurement for the cable length.

3. How does the height of the poles affect the length of the cable?

The height of the poles is a crucial factor in calculating the length of the cable between them. The taller the poles are, the longer the cable will need to be to reach from one pole to the other. This is because the taller poles will create a larger right triangle, resulting in a longer hypotenuse (cable length).

4. Are there any other factors that can affect the length of the cable between poles?

Yes, there are other factors that can affect the length of the cable between poles. These can include the angle of the cable as it connects to the poles, the tension of the cable, and any sagging of the cable due to its weight. These factors may require more complex calculations to determine the precise length of the cable.

5. Can I use a formula to calculate the length of the cable between poles?

Yes, there are formulas that can be used to calculate the length of the cable between poles. One common formula is the Sag-Tension formula, which takes into account the factors mentioned in the previous question. However, this formula may require knowledge of advanced mathematics and engineering principles. It is always best to consult with a professional or use a specialized calculator for accurate results.

• Mechanics
Replies
14
Views
1K
• Other Physics Topics
Replies
1
Views
3K
• DIY Projects
Replies
2
Views
2K
• Calculus
Replies
20
Views
2K
• Mechanical Engineering
Replies
5
Views
2K
• Mechanics
Replies
11
Views
7K
• Introductory Physics Homework Help
Replies
1
Views
3K
• General Discussion
Replies
4
Views
2K
• Calculus
Replies
3
Views
574
• Introductory Physics Homework Help
Replies
8
Views
1K