Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_162

⚡ Mecademy AIENG정역학 · ch5  Problem Statement The uniform 62-kg pole of 150-mm diameter is hinged at A, and its lower end is immersed in fresh water. Determine the tension T in the vertical cable required to maintain C at a depth of 1 m. Problem 5/162 (a) Tension T in the vertical cable 1. Formula: First, we determine the geometry of the system. Let be the angle the pole makes with the horizontal. The total vertical height from the hinge to the end is: where is the height of above the water surface and is the depth of . The angle is found using: The submerged length is: The buoyancy force is: The weight of the pole is: Moment equilibrium about the hinge : where is the distance to the center of gravity and is the distance to the center of buoyancy along the pole. 2. Substitution: θ AC H=h + A h C h = A 1.5 mAh = C 1 m C sinθ= L H L sub L = sub sinθ h C B B=ρ gV = wsub ρ g L w ( 4 πd 2 sub ) W W=mg A M = ∑ A 0⟹T⋅Lcosθ+B⋅s cosθ− B W⋅s cosθ= G 0 s = G L/2s = B L−L /2 sub Substituting into the moment equation (canceling ): 3. Calculation: 4. Result: ● Final Conclusion: The tension required in the vertical cable at to maintain the pole in the given orientation is . ✨ Final Answer Summary (a) Mecademy AI Solution · ENGProblem 5/162 H=1.5+1.0=2.5 m L=5 m sinθ= = 5 2.5 0.5⟹θ=30 ∘ L = sub = 0.5 1.0 2 m W=62×9.81=608.22 N B=1000×9.81× ×2( 4 π×0.15 2 ) s = G = 2 5 2.5 m s = B 5− = 2 2 4 m cos30 ∘ T(5)+B(4)−W(2.5)=0 B=9810×0.035343≈346.71 N 5T=2.5×608.22−4×346.71 5T=1520.55−1386.84=133.71