Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_130

⚡ Mecademy AIENG정역학 · ch5  Problem Statement The left-to-right current in the stream causes a uniform drag of 4 lb per cross-stream foot on the floating cable which is attached to posts A and B. Determine the minimum and maximum tensions in the cable and the location of each. Problem 5/130 (a) Minimum Tension and its Location ● Calculation Process 1. Formula: For a cable under a uniform load distributed along a span perpendicular to the load direction, the shape is a parabola. If we place the origin at the vertex and let the current direction be and the cross-stream direction be , the equation is: where is the minimum tension (the horizontal component in the vertex frame, which is the cross-stream component here). 2. Substitution: From the diagram, the total cross-stream span between A and B is , so for a symmetric setup, . The sag (maximum distance along current) is . The load is . Substituting these into the formula for : 3. Calculation: 4. Result: at point . wL Cxy x= 2T 0 wy 2 T 0 24 fty = max = 2 24 12 ft h=16 ftw=4 lb/ft T 0 T = 0 2h wy max 2 T = 0 2⋅16 4⋅12 2 T = 0 32 4⋅144 T = 0 = 32 576 18 lb T = min 18 lbC ● Final Conclusion: The minimum tension in the cable is , and it occurs at point , which is at the vertex of the parabola (midway between the posts and downstream). (b) Maximum Tension and its Location ● Calculation Process 1. Formula: The tension at any point along the cable is given by: The maximum tension occurs at the supports where the vertical load componen