(20%) Problem 2: Suppose a 63-kg mountain climber has a 0.78 cm diameter nylon r

Question

(20%) Problem 2: Suppose a 63-kg mountain climber has a 0.78 cm diameter nylon rope Randomized Variables ? 63 kg d 0.78 cm l 31 m By how much does the mountain climber stretch her rope, in centimeters, when she hangs 31 m below a rock outcropping? Assume the Young's modulus of the rope is 5 x 109 N/m2 Grade Summary Deductions Potential 0% 100% sinO Submissions Attempts remaining: 10 (5% per attempt) detailed view tanO Jt( cos cotanO asinO acosO atanO acotan sinh(0 cosh0 t cotnhO tanh0 0 Degrees O Radians 0 Submit Hint I give up! Hints: 3% deduction per hint. Hints remaining: 3 Feedback: 3% deduction per feedback

Explanation / Answer

Problem 2

The question is pretty straightforward. We will use equation for change in length

l = FL/AY

where F force = mg = 63 x 9.8
L original length = 31m (because only 31m rope is under tension)
A area = 3.14 r^2 = 3.14 d^2 / 4 = 3.14 (0.0078)^2 / 4
Y = 5 x 10^ 9

l = 0.08 m

Problem 3

Shear force 480N
Area = 3.14 x r^2 = 3.14 x (0.0315)^2 / 4 = 7.8 x 10^-4 m^2
height = 2.5 cm = 0.025m
shear modulus = G = 80 x 10^9 Pa

G = shear stress / shear strain

Shear strain = shear stress/ G

x/h = (F/A) / G

x = Fh/AG = 480 x 0.025 / (7.8 x 10^-4 x 80 x 10^9)

x = 1.92 x 10^-7 m

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