There are two vertical poles, one of height 100 feet and the other 70 feet, positioned a horizontal distance of 80 feet apart on level ground. A rope of length 100 feet connects the tops of the two poles. A weight, placed on the rope so that it can slide freely along it, is allowed to come to rest.

Find the horizontal distance of the weight from the 100 feet pole and the vertical distance from the ground.

The value of the mass m is immaterial. For equilibrium, the horizontal components

of the rope's tension must cancel. These are T = mg cos(theta), where theta is the angle from the x-axis drawn through the weight to the respective pole top. So the problem boils down to finding the position where the two angles are equal, given the length of the rope. Yes, even with a modest mass, all lines are straight. It becomes a geometry problem.

If h is the horizontal offset from the 100' pole and v is the vertical ground distance, we have:

theta = arctan[(100-v)/h] = arctan[(70-v)/(80-h)]

and, adding the hypotenuses:

len = 100 = l1 + l1 = h/cos(theta) + (80-h)/cos(theta)

Rather than solve the system, I just plugged-in h and v's 'till I got close.

60 and 55 work for h and v and theta is 36.8699265

*Edited on ***June 26, 2019, 4:36 pm**