Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_143

⚡ Mecademy AIENG정역학 · ch5  Problem Statement In setting its anchor in 100 ft of water, a small power boat reverses its propeller, which gives a reverse thrust . A total of 400 ft of anchor chain from anchor to bow has been released. The chain weighs , and the upward force due to water buoyancy is . Calculate the length of chain in contact with the bottom. Problem 5/143 (a) Calculation of the length of chain in contact with the bottom 1. Formula: First, determine the net weight per unit length of the chain when submerged. This is the difference between its weight in air and the buoyant force. In equilibrium, the reverse thrust equals the constant horizontal component of tension in the chain, . The catenary parameter is then: The relationship between the vertical depth , the parameter , and the arc length of the suspended portion (from the point of contact to the boat) is: The total released length consists of the length in contact with the bottom and the suspended length : 2. Substitution: Substitute the given values: , , , , and . P=800 lb 1.63 lb/ft 0.21 lb/ftl w w=w − air w buoyancy P T 0 c c= = w T 0 w P hc s (c+h)= 2 s+ 2 c⟹ 2 s= h+2ch 2 Ll s l=L−s h=100 ftP=800 lbL= 400 ftw = air 1.63 lb/ftw = buoyancy 0.21 lb/ft w=1.63−0.21=1.42 lb/ft c= ≈ 1.42 800 563.38 ft s= ft 100+2(563.38)(100) 2 3. Calculation: Calculate the suspended length : Calculate the length of the chain in contact with the bottom : 4. Result: ● Final Conclusion: The length of the anchor chain in contact wi