Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_61

⚡ Mecademy AIENG정역학 · ch6  Problem Statement For both conditions (a) and (b) as stated in Prob. 6/60, determine the magnitude and direction of the force required to begin lowering the 100-lb body. [From Prob. 6/60 context: The coefficient of static friction between the 100-lb body and the 15° wedge is 0.20. (a) Rollers of negligible friction are present under the wedge. (b) The rollers are removed and the coefficient of static friction applies at this surface as well. The 100-lb body is constrained to move vertically by frictionless rollers.] Problem 6/61 (a) Force with rollers of negligible friction under the wedge ● Calculation Process 1. Present Final Formula: First, determine the friction angle between the 100-lb body and the wedge: For impending downward motion of the 100-lb body (lowering), the wedge moves to the right. The resultant reaction from the wedge on the block is tilted from the normal by towards the direction of friction (which opposes the block's downward relative motion). The angle of from the vertical is . The horizontal force balance on the wedge (with no floor friction) gives: 2. Substitute Values: 3. Partial Operations: P ′ μ s μ = s 0.20 P ′ φ φ=tan(μ ) −1 s R 1 φ R 1 α−φ P= ′ Wtan(α−φ) φ=tan(0.20)≈ −1 11.31 ∘ α=15,W= ∘ 100 lb P= ′ 100tan(15− ∘ 11.31) ∘ α−φ=15− ∘ 11.3099= ∘ 3.6901 ∘ 4. Final Calculation: Since , the wedge is not self-locking and tends to slide out to the right. To maintain equilibrium at the point of impending motion, must be directed