Exercise 11.9 - Enhanced with Solution 1 of 3 Constants Part A A 301-N, uniform,
ID: 2039579 • Letter: E
Question
Exercise 11.9 - Enhanced with Solution 1 of 3 Constants Part A A 301-N, uniform, 1.50-m bar is suspended horizontally by two vertical cables at each end Cable A can support a maximum tension of 483.0 N without breaking, and cable B can support up to 433.0 N. You want to place a small weight on this bar What is the heaviest weight you can put on without breaking either cable? You may want to review (Pages 343-347) For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Locating your center of gravity while you work out. Submit Request Answer Part B Where should you put this weight? Im from the cable A Submit uest AnswerExplanation / Answer
A] Heaviest weight w = T1+T2- w1 = 483+433-301 = 615 N answer
b] Balancing torque about A, 433*1.5 = 615*d + 301*1.5/2
d = [433*1.5 - 301*1.5/2]/615 = 0.689 m answer
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.