Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_7

⚡ Mecademy AIENG정역학 · ch6  Problem Statement The tongs are designed to handle hot steel tubes which are being heat-treated in an oil bath. For a 20° jaw opening, what is the minimum coefficient of static friction between the jaws and the tube which will enable the tongs to grip the tube without slipping? Problem 6/7 (a) Determination of the minimum coefficient of static friction ● Calculation Process 1. Present Final Formula: For the tube to be held securely without being "squeezed out" of the wedge-shaped jaws, the horizontal component of the friction forces must be equal to or greater than the horizontal component of the normal forces. Let be the total jaw opening angle. The half-angle is . Considering the equilibrium of the tube in the horizontal direction (at the verge of slipping): Where . Substituting this gives: 2. Substitute Values: The given jaw opening angle is . Therefore, the half- angle is: Substituting this into the formula for the minimum coefficient: 3. Partial Operations: Calculate the value of the tangent: 4. Final Calculation: Rounding to three significant figures: α θ= 2 α F = ∑ x 2(F cosθ− f Nsinθ)=0 F = f μ N s μ Ncosθ− s Nsinθ=0⟹μ = s tanθ α=20 ∘ θ= = 2 20 ∘ 10 ∘ μ = s tan(10) ∘ tan(10)≈ ∘ 0.176327 μ = s 0.1763 ● Final Conclusion: The minimum coefficient of static friction required for the tongs to grip the tube without slipping is . ✨ Final Answer Summary (a) Mecademy AI Solution · ENGProblem 6/7 μ= s 0.1763 μ = s 0.1763