Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_74

⚡ Mecademy AIENG정역학 · ch5  Problem Statement The lampshade shown is constructed of 0.6-mm-thick steel and is symmetric about the z-axis. Both the upper and lower ends are open. Determine the mass of the lampshade. Take the radii to be to midthickness. Problem 5/74 (a) Mass of the lampshade ● Calculation Process 1. Present Final Formula: The mass of a thin-walled shell of uniform thickness and density is given by: The lampshade consists of three parts symmetric about the z-axis: Part 1 (Top Cylinder): Part 2 (Conical Frustum): , where is the slant height. Part 3 (Bottom Cylinder): Total area: 2. Substitute Values: From the diagram: , , Standard density of steel Slant height : 3. Partial Operations: Calculating individual surface areas: mt ρ m=ρ⋅V=ρ⋅(A ⋅ total t) A = 1 2πr h 11 A = 2 π(r + 1 r )L 2 L= (r −r )+h 21 2 2 2 A = 3 2πr h 23 A = total A + 1 A + 2 A 3 r = 1 30mmh = 1 60mm r = 2 100mmh = 2 80mm h = 3 12mm t=0.6mm ρ= 7850kg/m= 3 7.85× 10kg/mm −6 3 L L= = (100−30)+80 22 = 70+80 22 = 4900+6400 m 11300 Total Area: Total Volume: 4. Final Calculation: ● Final Conclusion: The mass of the lampshade is approximately or . ✨ Final Answer Summary (a) Mecademy AI Solution · ENGProblem 5/74 A = 1 2π(30)(60)=3600π≈11309.73mm 2 A = 2 π(30+100)(106.301)=130π(106.301)≈43414.12mm 2 A = 3 2π(100)(12)=2400π≈7539.82mm 2 A = total 11309.73+43414.12+7539.82=62263.67mm 2 V=A ⋅ total t=62263.67⋅0.6=37358.20mm 3 m=ρ⋅V=(7.85× 10kg/mm)⋅ −6 3 (37358.20mm) 3 m≈0.29326kg 0.293kg293g m=0.293kg