Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_158

⚡ Mecademy AIENG정역학 · ch5  Problem Statement The submersible diving chamber has a total mass of 6.7 Mg including personnel, equipment, and ballast. When the chamber is lowered to a depth of 1.2 km in the ocean, the cable tension is 8 kN. Compute the total volume displaced by the chamber. Problem 5/158 (a) Calculation of the total displaced volume 1. Formula: From the principles of static equilibrium for a submerged body, the sum of vertical forces must be zero. Where: is the cable tension (upward). is the buoyant force (upward), defined as . is the total weight of the chamber (downward), defined as . Rearranging to solve for : 2. Substitution: Substitute the given values into the formula. Note that , so . Also, . For seawater, we use the standard density provided in related problems of this chapter, , and acceleration due to gravity . 3. Calculation: Calculate total weight: Calculate buoyant force required: V V F = ∑ y 0⟹T+B−W=0 T BB=ρgV WW=mg V V= ρg mg−T 1 Mg= 1000 kgm=6.7×10 kg 3 T=8 kN=8000 N ρ= 1030 kg/m 3 g= 9.81 m/s 2 V= (1030 kg/m)(9.81 m/s) 32 (6.7×10 kg)(9.81 m/s)−8000 N 3 2 W=6700×9.81=65727 N B=W−T=65727−8000= 57727 N 3 Calculate the denominator (): Divide to find volume: 4. Result: ● Final Conclusion: The total volume displaced by the submersible diving chamber is . This volume represents the external size of the chamber required to support its weight such that the remaining 8 kN of downward force is carried by the cable tension. ✨ Final Answer Summary (a) Mecad