Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_118

⚡ Mecademy AIENG정역학 · ch6  Problem Statement The 8-kg block is resting on the 20° inclined plane with a coefficient of static friction . Determine the minimum horizontal force which will cause the block to slip. Problem 6/118 (a) Minimum horizontal force to cause the block to slip ● Calculation Process 1. Present Final Formula: To determine the force required to initiate slipping, we analyze the block's equilibrium at the verge of motion. For the horizontal force directed as shown in the diagram (to the right), the block will tend to slip up the incline. Summing forces along the plane (-axis) and perpendicular to it (-axis), we derive: where: is the weight of the block. is the angle of the incline. is the coefficient of static friction. is the friction angle. 2. Substitute Values: Substitute the given values , , , and : Weight: Friction angle: Using the simplified tangent form: 3. Partial Operations: Calculate the trigonometric components: μ = s 0.50P P P P xy P=W = cosθ−μ sinθ s sinθ+μcosθ s Wtan(θ+φ ) s W=mg θ=20 ∘ μ= s 0.50 φ = s arctan(μ ) s m=8 kgg= 9.81 m/s 2 θ= 20 ∘ μ = s 0.50 W=8×9.81=78.48 N φ = s arctan(0.50)≈26.57 ∘ P=78.48×tan(20+ ∘ 26.57)= ∘ 78.48×tan(46.57) ∘ sin20≈ ∘ 0.3420 cos20≈ ∘ 0.9397 Numerator: Denominator: Alternatively, using the tangent of the combined angles: 4. Final Calculation: ● Final Conclusion: The minimum horizontal force required to cause the block to slip up the incline is . (Note: If the force were directed to the left, a smaller magnitude