Answer:
To solve for the amount of stretch in the nylon rope, we can use the equation for the elongation (stretch) of a rope under tension:
ΔL = FL / AE
where ΔL is the change in length of the rope, F is the force on the rope, L is the original length of the rope, A is the cross-sectional area of the rope, and E is the Young's modulus of the rope.
First, we need to find the force on the rope. This is equal to the weight of you and the rope:
F = mg = (75.0 kg + mass of rope)g
Next, we need to find the cross-sectional area of the rope. The diameter of the rope is given as 0.600 cm, so the radius is 0.300 cm = 0.00300 m. Therefore, the cross-sectional area is:
A = πr^2 = 2.83 × 10^-6 m^2
Now we can plug in the given values and solve for ΔL:
ΔL = FL / AE = [(75.0 kg + mass of rope)g](23.0 m) / [(2.83 × 10^-6 m^2)(5.00 × 10^9 N/m^2)]
We can simplify this expression by using the fact that the mass of the rope is much smaller than your mass, so we can assume that the force on the rope is equal to your weight:
ΔL = (mg)(23.0 m) / (A E) = (75.0 kg)(9.81 m/s^2)(23.0 m) / [(2.83 × 10^-6 m^2)(5.00 × 10^9 N/m^2)]
Solving for ΔL, we get:
ΔL = 0.068 cm
Therefore, the nylon rope stretches by approximately 0.068 cm when you hang 23.0 m below a rock outcropping.