Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_168

⚡ Mecademy AIENG정역학 · ch5  Problem Statement A channel-marker buoy consists of a 2.4-m hollow steel cylinder 300 mm in diameter with a mass of 90 kg and anchored to the bottom with a cable as shown. If at high tide, calculate the tension in the cable. Also find the value of when the cable goes slack as the tide drops. The density of sea water is . Assume the buoy is weighted at its base so that it remains vertical. Problem 5/168 (a) Tension in the cable at high tide () 1. Formula: From vertical equilibrium, the sum of forces must be zero. Where the buoyancy force and the weight . 2. Substitution: Cross-sectional area: Total weight: Buoyancy force: 3. Calculation: Submerged length: 4. Result: ● Final Conclusion: The tension in the cable at high tide is . (b) Value of when the cable goes slack h=0.6 m Th 1030 kg/m 3 h=0.6 m F = ∑ y 0⟹F − B W−T=0⟹T=F − B W F = B ρgA(L−h)W=mg A= = 4 πd 2 = 4 π(0.3 m) 2 0.070686 m 2 W=90 kg× 9.81 m/s= 2 882.9 N F = B 1030 kg/m× 3 9.81 m/s× 2 0.070686 m× 2 (2.4 m−0.6 m) L = sub 2.4−0.6=1.8 m F = B 1030×9.81×0.070686×1.8=1285.59 N T=1285.59 N−882.9 N=402.69 N T≈403 N 403 N h 1. Formula: When the cable goes slack, the tension , so the buoyancy force equals the weight. 2. Substitution: 3. Calculation: 4. Result: ● Final Conclusion: The value of when the cable goes slack is . ✨ Final Answer Summary (a) (b) Mecademy AI Solution · ENGProblem 5/168 T=0 F = B W⟹ρgA(L−h)=mg⟹ρA(L−h)=m (1030 kg/m)(0.070686 m)(2.4 m− 3 2 h)=90 kg 72.806×(2.4−h)=90 2.4−h= = 72