Question
I have gone through the steps by the expert (it is problem 84GF forthe Physics: Principles with Applications (6th) book by Giancoli).I just cant seem to get this problem, please help! Here is theproblem:
Two rock climbers, Bill and Karen, use safety ropes of similarlength. Karen's rope is more elastic, called a dynamic rope byclimbers. Bill has a static rope, not recommended for safetypurposes in pro climbing. Karen falls freely about 2.2 m and then the rope stops her over a distanceof 1.2 m.
(a) Estimate, assuming that the force is constant, how large aforce she will feel from the rope. (Express the result in multiplesof her weight.)
(b) In a similar fall, Bill's rope stretches by 30 cm only. How many times his weight will therope pull on him?
Any help or direction would be greatly appreciated.
-BioHazard789
Explanation / Answer
The final velocity (v), initial velocity (u)and height of fall (h) are related as v2 = u2 + 2* g *h, here u = 0 (freefall), h = 2.2 m v2 = 02 + 2* 9.8 *2.2 => v = 6.56 m/s a. Let Karen bedeacceleratedat a1 m/s2. vf2 = v2 - 2* a1 *s, s = distanceover which she stops completely. 02 = 6.562 - 2* a1 *1.2 => a1 = 43.03/2.4 = 17.93 m/s2 Pull ofrope F1 = m* a1 = m* 17.93 N F1/ W1 = m * 17.93 / m *g = 17.93 /9.8 = 1.83 force onthe rope will be 1.83 times her weight. b. Let Bill bedeacceleratedat a2 m/s2. vf2 = v2 - 2* a2 *s, s = distanceover which she stops completely. 02 = 6.562 - 2* a2* 0.30 => a2 = 43.03/ 0.6 = 71.72 m/s2 Pull ofrope F2 = m* a2 = m* 71.72 N F2/ W2 = m * 71.72/ m *g = 71.72/9.8 = 7.32 force onthe rope will be 7.32 times her weight. = m* 17.93 N F1/ W1 = m * 17.93 / m *g = 17.93 /9.8 = 1.83 force onthe rope will be 1.83 times her weight. b. Let Bill bedeacceleratedat a2 m/s2. vf2 = v2 - 2* a2 *s, s = distanceover which she stops completely. 02 = 6.562 - 2* a2* 0.30 => a2 = 43.03/ 0.6 = 71.72 m/s2 Pull ofrope F2 = m* a2 = m* 71.72 N F2/ W2 = m * 71.72/ m *g = 71.72/9.8 = 7.32 force onthe rope will be 7.32 times her weight. vf2 = v2 - 2* a2 *s, s = distanceover which she stops completely. 02 = 6.562 - 2* a2* 0.30 => a2 = 43.03/ 0.6 = 71.72 m/s2 Pull ofrope F2 = m* a2 = m* 71.72 N F2/ W2 = m * 71.72/ m *g = 71.72/9.8 = 7.32 force onthe rope will be 7.32 times her weight. = m* 71.72 N F2/ W2 = m * 71.72/ m *g = 71.72/9.8 = 7.32 force onthe rope will be 7.32 times her weight.