Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_19

⚡ Mecademy AIENG정역학 · ch6  Problem Statement Determine the range of mass for which the system is in equilibrium. The coefficient of static friction between the block and the incline is . Neglect friction associated with the pulley. Problem 6/19 (a) Range of mass for equilibrium 1. Formula: The system is in equilibrium when the net force on each mass is zero. For the hanging mass , the tension in the cord is: For the block of mass on the incline, the cord passes around a pulley attached to the block, resulting in a total force of pulling it up the incline. The equilibrium equation parallel to the incline () is: The normal force is: The maximum static friction force is: 2. Substitution: Substituting and into the equilibrium equation to find the boundaries of the range: For minimum (block on the verge of sliding down): For maximum (block on the verge of sliding up): 3. Calculation: Given , , and : m 2 μ = s 0.25 m 2 m 2 T T=m g 2 m 1 2T 20 ∘ F = ∑ ∥ 2T±f−m gsin20= 1 ∘ 0 N N=m gcos20 1 ∘ f = max μ N= s μ m gcos20 s1 ∘ T=m g 2 f=f max m 2 m 1 2m g+ 2 μm gcos20− s1 ∘ m gsin20= 1 ∘ 0 m 2 m 1 2m g− 2 μm gcos20− s1 ∘ m gsin20= 1 ∘ 0 μ = s 0.25sin20≈ ∘ 0.34202cos20≈ ∘ 0.93969 Lower Bound (): Upper Bound (): 4. Result: The range of mass is . ● Final Conclusion: The mass must be between and to maintain the system in static equilibrium. Below the minimum value, block will slide down the incline; above the maximum value, it will be pulled up the incline. ✨ Final Answer Summary (a) Mecadem