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A cable with linear density $ p = kg/m $ is strung from the tops of two poles that are $ 200 m $ apart.

(a) Use Exercise 52 to find the tension $ T $ so that cable is $ 60 m $ above the ground at its lowest point. How tall are the poles?

(b) If the tension is doubled, what is the new low point of the cable? How tall are poles now?

A. $f(100)=\frac{T}{\rho g} \cosh \left(\frac{\rho g \cdot 100}{T}\right)=60 \cosh \left(\frac{100}{60}\right) \approx 164.50 \mathrm{m}$

B. $f(100)=\frac{2 T}{\rho g} \cosh \left(\frac{\rho g \cdot 100}{2 T}\right)=120 \cosh \left(\frac{100}{120}\right) \approx 164.13 \mathrm{m},$ just a slight decrease

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Missouri State University

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

okay, Recalled. The minimum height is 60. Therefore, we have 60 equals. T over. PG plug in what we know. 60 equals cheese over two times 9.8. Therefore, t is 1176 men. Therefore, why have 100 is two times 9.8 times 100 over 1.176 co sign a tch Then we have 1176 over two times 9.8, which gives us the height of the polls is 164.5 meters. Part B. We know that the new lows tide of the cable is two times 60 Because it's doubled. This is the same thing as 120 meters. And then we know the new tension is again doubled two times 117 Sex, which is 235 to Newton's, Therefore plugging the send 2352 over two times 9.8 curse on age two times 9.8 times 100 over 2352 we get 164.13 meters is the new head of the pools