Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_93

⚡ Mecademy AIENG정역학 · ch6  Problem Statement The 180-lb tree surgeon lowers himself with the rope over a horizontal limb of the tree. If the coefficient of friction between the rope and the limb is 0.60, compute the force which the man must exert on the rope to let himself down slowly. Problem 6/93 (a) Force required to lower himself slowly 1. Formula: Consider the man and the rope as a system in equilibrium (moving at a constant slow speed). Let be the weight of the man, be the tension in the rope segment held by his hands (force ), and be the tension in the rope segment attached to his harness. The total upward force on the man must balance his weight: When lowering himself, the harness side () moves downward while the hand side ( ) moves upward. Friction between the rope and the tree limb opposes this motion, making the harness side the "larger tension" side: Combining these equations, we solve for the force : 2. Substitution: Substitute the given values into the formula: Weight Coefficient of friction Contact angle (since the rope goes over the limb) 3. Calculation: Show intermediate calculation steps: Calculate the exponent: WT 1 PT 2 T + 1 T = 2 W T 2 T 1 T = 2 T e 1 μβ P=T 1 T + 1 T e = 1 μβ W⟹P= 1+e μβ W W=180 lb μ=0.60 β=180= ∘ π rad P= 1+e 0.60×π 180 0.60×π≈1.8850 Calculate the exponential term: Calculate the denominator: Calculate final force: 4. Result: ● Final Conclusion: To let himself down slowly, the tree surgeon must exert a force of on the rope. ✨ Final Ans