The City Electric System(CES) recently constructed a major transmission line along parts of South Street in the city. The transmission lines are strung between power poles and the hanging lines form a catenary curve. For this Problem we will use the following equation y=(T/w)cosh((w/T)*(x))+k.
Where T is the tension on the, w is the unit weight of the line, x is zero at the lowest point of the hanging transmission line and k deals with the height of the low point above the ground.
The two power poles we are dealing with are 600ft apart and each pole is 150ft tall. The attachment point for the lowest line is 80ft above the ground (three other lines are placed above this lowest line). Place your coordinate system with its origin at the midpoint between the two poles and on the ground. Now answer the following questions.
1. a) If the tension on the transmission line is 500 pounds and the unit weight of the line is 1.5 pounds per foot. Find the equation for this catenary curve
b) What is the height of the lowest point of this transmission line?
c) What is the slope of the tangent line at the right attachment point?
Trees can be a problem for transmission lines and is for these ones. There are many varieties of trees growing alone the line that are currently as high as 50ft. Of these varieties reaching 50ft, the fastest growing is the silver maple, which normally grows 4 to 6 feet per year. Vegetation management practices dictate that 5 feet must be maintained between vegetation and the power line at all times-- so crews must trim back to a point where the vegetation will not grow to within 5 feet before the next trimming cycle. Assume the information above is from September of this year. Use this information and your work from question 1 to answer the following questions.
2.a) What is the earliest possible September the trimming crew will have to trim these trees?
b) The trimming crews are expensive and there are many areas in the City that need trimming not just the power lines. If the trimming crew would not want to return to trim these trees for two years, how much should they cut off the tops of the trees the first time to meet this requirement?
The values given for T,w,k indicate
y=(T/w)cosh((w/T)*(x))+k
= (500/1.5)cos(1.5/500 x) + 80
But this curve does not pass through (300,150) as explained in the beginning. To get that, we need
y = a*cosh(x/a)+80
where y(300)=150
But that has no solution.
1. a) To find the equation for the catenary curve, we can use the given parameters T, w, x, and k in the equation y=(T/w)cosh((w/T)*(x))+k.
Given:
T = 500 pounds (tension on the transmission line)
w = 1.5 pounds per foot (unit weight of the line)
x is zero at the lowest point of the hanging transmission line
k = 80ft (height of the low point above the ground)
Substituting these values into the equation, we get:
y=(500/1.5)cosh((1.5/500)*(x))+80
b) The height of the lowest point of the transmission line can be found by substituting x = 0 into the equation:
y=(500/1.5)cosh((1.5/500)*(0))+80
y=(500/1.5)cosh(0)+80
y=(500/1.5)+80
y=333.33+80
y=413.33ft
Therefore, the height of the lowest point of the transmission line is 413.33ft.
c) The slope of the tangent line at the right attachment point can be found by taking the derivative of the equation with respect to x.
Taking the derivative, we get:
dy/dx = (T/w)*(w/T)*sinh((w/T)*x)
At the right attachment point (x = 300ft), the slope can be found by substituting x = 300 into the derivative:
dy/dx = (500/1.5)*(1.5/500)*sinh((1.5/500)*300)
dy/dx = sinh(0.9)
dy/dx ≈ 0.745
Therefore, the slope of the tangent line at the right attachment point is approximately 0.745.
2. a) To find the earliest possible September the trimming crew will have to trim these trees, we need to calculate when the vegetation height will reach 5 feet from the power line.
Given:
Maximum tree height = 50ft
Silver maple growth rate = 4 to 6 feet per year
Minimum required clearance between vegetation and power line = 5 feet
Since trimming is done to maintain a 5-foot clearance, we need to find when the vegetation height will reach 55 feet (50ft + 5ft clearance).
Using the fastest growth rate of 6 feet per year, we can calculate the earliest possible September as:
Earliest possible September = [55ft - (current height - 50ft)] / growth rate + current year
If we assume the current year is Year 1, then the earliest possible September would be:
Earliest possible September = [55ft - (50ft - 0ft)] / 6ft/year + Year 1
Earliest possible September = 5ft / 6ft/year + Year 1
Earliest possible September ≈ 0.833 years + Year 1
Therefore, the earliest possible September the trimming crew will have to trim these trees would be approximately September of Year 2.
b) If the trimming crew does not want to return to trim these trees for two years, they need to cut off enough from the tops of the trees to ensure that the vegetation height remains below the 5-foot clearance for that period.
Considering a growth rate of 6 feet per year, the crew needs to cut off at least 2 years * 6 feet/year = 12 feet from the tops of the trees to meet this requirement.
Therefore, the crew should cut off at least 12 feet from the tops of the trees during the first trimming cycle.