Answer every part please Class Management I Help HomeWork 11 Begin Date: 7/2/201

Question

Answer every part please

Class Management I Help HomeWork 11 Begin Date: 7/2/2018 12:01:00 AM -- Due Date: 7/31/2018 11:59:00 PM End Date: 8/1/2018 11:59:00 PM (10%) Problem 1: Consider a cylindrical cable with a hanging weight suspended from it. 33% Part (a) When the mass is removed, the length of the cable is found to be 10 4.51 m. After the mass is added, the length is remeasured and found to be 11 = 5.31 m. Determine Young's Modulus Yin N/m2 for the steel cable if the weight has a mass ,n = 75 kg and the cable has a radius r = 0.045 m 33% Part (b) If we were to double the radius of the wire and re-suspend the weight, which of the following would be correct Grade Summary Deductions Potential O The strain experienced by the wire would increase by a factor of 2 O The stress experienced by the wire would decrease by a factor of 4 o The strain experienced by the wire would increase by a factor of 4 o Young's modulus would increase by a factor of 4 O The strain experienced by the wire would increase by a factor of 4 0% 100% Submissions Attempts remaining: 5 Young's modulus would be reduced by a factor of 4 )% per attempt) detailed view Submit Hint Hints: 1 % deduction per hint. Hints remaining: 1 Feedback: 1% deduction per feedback. ? 33% Part (c) If this cable is pulled down a distance d in m from its equilibrium position it acts like a spring when released Write an expression determining the spring constant k of this material using the cable-specific variables Y, l1, lo, and r provided in part A All content © 2018 Expert TA, LLC

Explanation / Answer

1. consider a cylinderical cable hanging with weight suspended

a. lo = 4.51 m

l1 = 5.31 m

youngs modulus = Y

m = 75 kg

r = 0.045 m

hence

Y = straa/strain = mg*lo/pi*r^2*(l1 - lo)

Y = 651991.4872878735 N/m^2

b. if the radius is doubled, the stress becomes 0.25 times

hence strain becomes 0.25 times

so, option b

c. spring constant = k

now from force balance

F = kd ( restoring force)

but F = stress*area

stress = Y*(l1 - lo)/lo

hence

Y*(l1 - lo)*pi*r^2/lo = kd = k(l1 - lo)

hence

k = Y*pi*r^2/lo

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