Physic Question #3

Question

By how much does a 66-kg mountain climber stretch her 0.596-cm diameter nylon rope when she hangs 30 m below a rock outcropping? (For nylon, Y=1.35 × 10⁹Pa.) 

Round your answer to 4 decimal places

Answer 1

The cross sectional area of the rope is found:

A = π*r²

=> r = 0.298x10^-3m

<=> π*(0.298x10^-3)²

A = 2.79x10^-5m²

The force experienced by the rope is the tension, which is equal to the weight of the climber:

∑F = T - F_g = ma = 0

T - F_g = 0 => T = F_g = mg

<=> (66kg)(9.8m*s^-2)

T = 646.8N

We can calculate the elongation using the following equation:

F/A = Y(ΔL/L₀) => ΔL = (1/Y)(F/A)L₀

<=> (1/1.35x10⁹ Pa)(646.8N/2.79x10^-5m²)(30m)

ΔL = 0.5150m