Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_156

⚡ Mecademy AIENG정역학 · ch5  Problem Statement A rectangular block of density floats in a liquid of density . Determine the ratio , where is the submerged depth of block. Evaluate for an oak block floating in fresh water and for steel floating in mercury. Problem 5/156 (a) General expression for the ratio 1. Formula: For a floating body, the weight of the body must be balanced by the buoyant force (Archimedes' Principle): 2. Substitution: Express weight and buoyant force in terms of densities and volumes: Where: (Total volume of the rectangular block) (Volume of the block below the liquid surface) 3. Calculation: Substitute the volume expressions into the equilibrium equation and simplify: Cancel common terms , , and : Rearrange to solve for the ratio : 4. Result: ρ 1 ρ 2 r=h/c hr r=h/c W F B W=F B ρ ⋅ 1 g⋅V = total ρ ⋅ 2 g⋅V submerged V = total a⋅b⋅c V = submerged a⋅b⋅h ρ ⋅ 1 g⋅(a⋅b⋅c)=ρ ⋅ 2 g⋅(a⋅b⋅h) gab ρ ⋅ 1 c=ρ ⋅ 2 h r=h/c = c h ρ 2 ρ 1 r= ρ 2 ρ 1 ● Final Conclusion: The ratio of the submerged depth to the total height is equal to the ratio of the block's density to the liquid's density, . (b) Evaluation for an oak block floating in fresh water 1. Formula: Use the derived ratio or in terms of specific gravity . 2. Substitution: From standard properties (Table D/1), for white oak and for fresh water . 3. Calculation: 4. Result: ● Final Conclusion: For an oak block floating in fresh water, approximately of its height is submerged, giving . (c) Evaluation for steel floating