Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_129

⚡ Mecademy AIENG정역학 · ch5  Problem Statement A mason stretches a string between two points 50 ft apart on the same level with a tension of 10 lb at each end. If the string weighs 0.1 lb, determine the sag h at the middle of the string. Problem 5/129 (a) Determine the sag at the middle of the string 1. Formula: Since the total weight of the string () is very small compared to the tension (), the sag will be small relative to the span . Under these conditions, the string configuration can be accurately approximated by a parabola. The relationship between the sag , horizontal tension , span , and total weight is given by: The relationship between the tension at the ends , the horizontal tension , and the total weight is: 2. Substitution: Given values: Span, End tension, Total weight, First, calculate the horizontal tension : Then, substitute into the sag formula: h W=0.1 lb T=10 lbhL hT 0 LW h= 8T 0 WL TT 0 W T= T + 0 2 ( 2 W ) 2 L=50 ft T=10 lb W=0.1 lb T 0 T = 0 = T − 2 ( 2 W ) 2 10− 2 ( 2 0.1 ) 2 h= 8×T 0 0.1×50 3. Calculation: Calculate : (Since is so small, is a valid approximation). Calculate in feet: Convert sag to inches: 4. Result: ● Final Conclusion: The sag at the middle of the string is (which is equivalent to ). ✨ Final Answer Summary (a) (or ) Mecademy AI Solution · ENGProblem 5/129 T 0 T = 0 = 100−0.0025 ≈ 99.99759.999875 lb W/2T ≈ 0 T=10 lb h h= ≈ 8×9.999875 5 = 80 5 0.0625 ft h h = in 0.0625 ft× 12 = ft in 0.75 in h=0.0625 ft or 0.75 in h0.0625 ft 0.75 in h=0.0