Question

Mountain climbers use nylon safety rope whose elasticity plays an important role in cushioning the sharp jerk if a climber falls and is suddenly stopped by the rope.

(a) Suppose that a climber of 80 kg attached to a 10 m rope falls freely from a height of 10 m above to a height of 10 m below the point at which the rope is anchored to a vertical wall of rock. Treating the rope as a spring with k = 4.9 × 103 N/m (which is the appropriate value for a braided nylon rope of 9.2 mm diameter), calculate the maximum force that the rope exerts on the climber during stopping.

(b) Repeat the calculations for a rope of 5.0 m and an initial height of 5.0 m. Assume that this second rope is made of the same material as the first, and remember to take into account the change in the spring constant due to the change in length. Compare your results for (a) and (b) and comment on the advantages and disadvantages of long ropes versus short ropes.

Answer 1

Force by a spring =kx

for max force x should be maximum so

apply energy. Conservation

Loss in PE = stored energy in spring

mg(10+10+x) = 1/2*k*x^2

from here x = x_max = 2.69 m

Fmax = 4.9*10^3*2.69= 13204.9 N answer

part 2

same way

mg(5+5+x) = 1/2*k*x^2

80*9.8(5+5+x) = 1/2* 4.9*10^3*x^2

from here x = x_max ≈1.956 m ,

so Fmax = 9584.38N answer

compare = 13204.9 /9584.38 =1.4

so first case the force is 1.4 times bigger than to the second case