Ex Score O of 1 pt HW Score 9585 2971 of 31 pts 31 of 31 com

Ex. Score: O of 1 pt HW Score: 95.85% (29.71 of 31 pts) 31 of 31 complete A climber anchors a rope at two points of equal height, separated by a distance yt Sag angle of 110 ft. The rope follows the catenary fx) 260 coshover the interval [- 55,55]. The climber clips onto the rope and pulls himself to the rope\'s midpoint. Because the rope is supporting the weight of the climber, it no longer takes the shape of a catenary. Instead, the rope (nearly) forms two sides of an isosceles triangle. Compute the sag angle illustrated in the figure, assuming that the rope does not stretch when weighted. The length of the rope is 111 ft. 260 - 55 55 X The sag angle is approximately. Round to two decimal places as needed.)

Solution

Distance b/w two rocks = 110 m

Equation of rope : y = 260 cosh(x/260)
Consider length l of the string
l = 260 sinh (110/260) = 113.31105 ft

When the load is applied:

Length of hypotenuse = l / 2 = 56.65 ft
Length of base of triangle = 55 ft
cos ( sag angle ) = 55 / 56.65
sag angle = 13.86 degrees