Walking by a pond, you find a rope attached to a tree limb 5.2 m off the ground.
ID: 1696678 • Letter: W
Question
Walking by a pond, you find a rope attached to a tree limb 5.2 m off the ground. You decide to use the rope to swing out over the pond. The rope is a bit frayed but supports your weight. You estimate that the rope might break if the tension is 74 N greater than your weight. You grab the rope at a point 4.6 m from the limb and move back to swing out over the pond. Assume your weight to be 650 N (about 145 lb).
(a) What is the maximum safe initial angle between the rope and the vertical so that it will not break during the swing?
°
(b) If you begin at this maximum angle, and the surface of the pond is 1.2 m below the level of the ground, with what speed will you enter the water if you let go of the rope when the rope is vertical?
m/s
Explanation / Answer
We'll have to assume the tree is reasonably close to thepond's surface, so that the rope is vertical when you grabit. You grab x =5.2 - 4.6 = 0.6 m off the ground. The distance from the limb is r = 4.6 m. The total rope length is L = 5.2 m W=145lb= .45*145 =65.25 kg Now, conservation of energy E = E' KE+ PE = KE' + PE' 0 + mgh = 1/2mv2 + mgx 2g(h - x) = v2 This gives the speed when traveling over the pond. Therope is vertical at that point, you are going horizontal. OK, but what about the initial height h? When therope pulls back, the bottom is L - Lcos above theground. You're holding on at x along the rope now, so tack onxcos to this height. h = L - Lcos + xcos Phew! But we're not close to being done. At the bottom of the circle, the tension force suppliescentripetal acceleration and supports weight. Fbottom = FT - mg =mv2 / r FT = mv2 / r + mg = [74+650] = 724 Now plug in the other expression and solve for . It'll be a monster. FT = [W / g * 2g (h - x) ] / r + 650 =724 2W (h-x) / r =74 2W(h-x) = 74 r 2W * ( L - Lcos + xcos - x) = 74r 2WL- 2WLcos + 2Wxcos - 2Wx = 74r 2WL - 74 r - 2Wx = (2WL - 2Wx) cos 259.9 / 600.3 = cos =25.65 degree Sounds reasonable, it might actually be right! (b) Now it becomes a projectile question. We'll plug in this angle and get the speed v, as alreadyderived. v2 = 2g (h - x) = 2g (L - Lcos + xcos- x) = 8.884 vx = 2.98 m/s (xcomponent) vy2 = 2 g (x +1.2) =39.2 vy = 6.25 m/s v = (vx2 + vy2)1/2 = 6.92 m/sRelated Questions
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