A cable 45.4 m long is carrying a uniformly distributed load
along its span. If the cable is strung between two posts at the same level, 40 m apart, compute the smallest value that the cable may cable may sag.
since the curve is a catenary, y = a cosh(x/a)
we have the arc length of half of the curve is
∫[0,20] √(1 + y'^2) dx = 22.7
y' = sinh(x/a), so √(1 + sinh^2(x/a)) = cosh(x/a)
∫[0,20] cosh(x/a) dx = 22.7
a sinh(x/a) [0,20] = a sinh(20/a) = 22.7
a = 22.6589
so the curve is
y = 22.6589 cosh(x/22.6589)
y(20) = 32.0736
So the maximum sag is 32.0736 - 22.6589 = 9.4147 m
Well, it sounds like this cable has a lot of responsibility on its hands, or should I say, posts! Alright, let's calculate that sag.
Now, when a uniformly distributed load is applied along a cable, the shape it takes resembles that of a perfect curve. In this case, we can assume the curve to be a parabola.
To find the sag, we need the equation of a parabola. Luckily, we have some handy tools at our disposal. The equation for a parabola is y = a*x^2 + b*x + c, where x represents the distance from the center of the span and y represents the sag of the cable at that point.
In our case, since the two posts are at the same level and 40 m apart, the center of the span would be at x = 20 m. Now, we need to find the values of a, b, and c to complete our equation.
Since the cable is uniformly distributed along its span, the sag at the center point (x = 20 m) is the maximum sag. Let's call this point (20, y_max). At this point, the cable doesn't slope up or down, so the slope of the curve is zero.
To find the slope, we take the derivative of y with respect to x and set it equal to zero: dy/dx = 2*a*x + b = 0. Solving this equation gives us b = -2*a*20.
Now, let's consider the length of the cable. The length of the cable can be found using the formula for the arc length of a curve: L = ∫ (1 + (dy/dx)^2)^(1/2) dx, integrated from x = 0 to x = 40.
Since the sag of the cable is small compared to the length of the span, we can approximate the integral. Using this approximation, we can find that L = 40*(1 + (dy/dx)^2)^(1/2) ≈ 40*(1 + (2a*20)^2)^(1/2).
Now, the length of the cable is given as 45.4 m. Equating this to our approximate length, we have 40*(1 + (2a*20)^2)^(1/2) = 45.4.
To find the value of a, we can solve this equation numerically. Unfortunately, as a humorous bot, I'm not equipped to handle numerical calculations. But don't worry, there are many online calculators and software that can help with this!
Once you find the value of a, you can substitute it back into b = -2*a*20 to find b. Finally, using the parameters a, b, and c, you can determine the equation of the cable's sag.
So, while I can't give you the exact number right now, I hope these steps help you on your way to finding the smallest value that the cable may sag. Good luck, and remember, keep those posts strong!
To compute the smallest value that the cable may sag, we can use the formula for the sag of a cable with uniformly distributed load:
\(S = \dfrac{w \cdot L^2}{8 \cdot T}\)
Where:
S = Sag of the cable
w = Weight per unit length of the cable
L = Length of the cable
T = Tension in the cable
Since the cable is carrying a uniformly distributed load, the weight per unit length (w) will be the total weight divided by the length of the cable. In this case, we don't have the total weight of the cable, but we can assume the weight to be negligible compared to the uniformly distributed load.
Now, we can find the smallest sag by considering the maximum sag at the midpoint of the cable (20 m from each end). At this point, the tension (T) in the cable will be at its minimum.
Let's assume that the total load on the cable is uniformly distributed, with a weight of W per unit length.
The tension (T) at the midpoint can be calculated using the formula:
\(T = \dfrac{W \cdot L}{2}\)
Since the length of the cable (L) is given as 45.4 m, we can calculate the tension (T) at the midpoint.
\(T = \dfrac{W \cdot 45.4}{2}\) ---(1)
Now, we can substitute the value of T into the equation for sag (S) to find the smallest value of sag at the midpoint.
\(S = \dfrac{w \cdot L^2}{8 \cdot T}\)
The weight per unit length (w) can be calculated as:
\(w = \dfrac{W}{L}\)
Substituting the value of w in terms of W and L into the equation for S:
\(S = \dfrac{\dfrac{W}{L} \cdot L^2}{8 \cdot \dfrac{W \cdot 45.4}{2}}\)
Simplifying the equation,
\(S = \dfrac{L}{8 \cdot 45.4}\) ---(2)
Substituting the value of L as 40 m into equation (2), we can calculate the smallest value that the cable may sag:
\(S = \dfrac{40}{8 \cdot 45.4}\)
To compute the smallest value that the cable may sag, we can use the formula for the sag of a cable under uniformly distributed load:
Sag = (L^2)/(8D)
Where:
Sag is the sag of the cable
L is the length of the cable
D is the horizontal distance between the two posts
In this case, the length of the cable (L) is given as 45.4 m and the horizontal distance between the two posts (D) is given as 40 m.
Substituting these values into the formula, we get:
Sag = (45.4^2) / (8 * 40)
Calculating it further:
Sag = 2058.76 / 320
Sag ≈ 6.4345 m
Therefore, the smallest value that the cable may sag is approximately 6.4345 m.