Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_107

⚡ Mecademy AIENG정역학 · ch6  Problem Statement The 4600-lb lathe with mass center at is positioned with the aid of the steel wedge. Determine the horizontal force required to remove the wedge if the coefficient of friction for all contacting surfaces is 0.30. Also show that no horizontal movement of the lathe takes place. Problem 6/107 (a) Determination of the force to remove the wedge ● Calculation Process 1. Calculate vertical reactions on the lathe: The lathe is in equilibrium. We sum moments about support to find the vertical force at , . The reaction at is: 2. Analyze forces on the wedge: The wedge has an angle and the coefficient of friction for all surfaces is . The friction angle is . Removing the wedge (pulling it right) requires overcoming friction at the floor and the "wedging effect" at the lathe. Since the wedge thins to the right, the normal force from the lathe has a horizontal component to the left, which opposes removal. The total force from the lathe on the wedge, , is tilted from the vertical by an angle of to the left. The vertical component of this force must equal the vertical load . The horizontal component to be overcome at the top surface is: The horizontal friction at the floor (bottom surface) is: G5 ∘ P P ABN B M = ∑ A 0⟹N ⋅ B (92 in.)−W⋅(36 in.)=0 N = B = 92 in. 4600 lb×36 in. 1800 lb A N = A W−N = B 4600 lb−1800 lb=2800 lb α=5 ∘ μ=0.30φ=tan(0.30)≈ −1 16.70 ∘ R 1 (α+φ) N B H = B N tan(α+ B φ) F = 2 μN= 2 μN B The total force is the sum of these hor