Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_195

⚡ Mecademy AIENG정역학 · ch5  Problem Statement The tapered body has a horizontal cross section which is circular. Determine the height of its mass center above the base of the homogeneous body. Problem 5/195 (a) Height of the mass center 1. Formula: For a homogeneous body, the height of the mass center coincides with the -coordinate of the centroid of its volume. It is calculated by: where is the volume element. For a body with circular cross-sections, . 2. Substitution: Let be the height from the base ( at the bottom, at the top). The radius varies linearly from to : Substituting this into the integrals: 3. Calculation: Total Volume (): h ˉ h ˉ h ˉ z =h ˉ dV∫ zdV∫ dVdV= π[r(z)]dz 2 zz=0z=H r(z)2RR r(z)=2R+ z = H R−2R R2− ( H z ) V= πR2− d z∫ 0 H 2 ( H z ) 2 z d V = ∫ z ⋅ ∫ 0 H πR2− d z 2 ( H z ) 2 V V=πR 4− + d z = 2 ∫ 0 H ( H 4z H 2 z 2 )πR4z− + 2 [ H 2z 2 3H 2 z 3 ] 0 H V=πR4H−2H+ = 2 ( 3 H ) πRH 3 7 2 First Moment of Volume (): Centroidal Height (): 4. Result: ● Final Conclusion: The height of the mass center for the homogeneous tapered body is (approximately ) above its base. ✨ Final Answer Summary (a) Mecademy AI Solution · ENGProblem 5/195 M xy M = xy πR 4z− + d z = 2 ∫ 0 H ( H 4z 2 H 2 z 3 )πR2z− + 2 [ 2 3H 4z 3 4H 2 z 4 ] M = xy πR2H− + = 2 ( 2 3 4H 2 4 H 2 )πRH = 22 ( 12 24−16+3 ) 12 11 h ˉ =h ˉ = V M xy = πRH 3 7 2 πRH 12 1122 ⋅ 12 11 H = 7 3 H 28 11 =h ˉ H 28 11 H 28 11 0.393H =h ˉ H 28 11