Meriam, Kraige & Bolton — Statics of Particles: Problem 2_127

⚡ Mecademy AIENG정역학 · ch2  Problem Statement 2/127 SS A 1.8-lb vertical force is applied to the knob of the window-opener mechanism when the crank BC is horizontal. Determine the moment of the force about point A and about line AB. Problem 2/127 1. Coordinate System and Vector Definitions First, we define a Cartesian coordinate system with the origin at point . The and axes are in the horizontal plane as shown in the diagram, and the -axis is vertical (pointing upwards). From the geometric details in the figure: Point is at the origin: . Segment has length and makes an angle of with the -axis. Thus, its position vector is: Segment has length and is horizontal, parallel to the -axis. Its position vector is: The knob offset has length and is horizontal, parallel to the -axis. Its position vector is: The total position vector from to the point of application of the force is: The applied force is vertical with a magnitude of , directed downwards: A(0,0,0)x yz AA=(0,0,0) AB3 ′′ 30 ∘ x r = B/A 3(cos30i+ ∘ sin30j)= ∘ (2.598i+1.5j) in. BC2 ′′ y r = C/B 2j in. CD1 ′′ x r = D/C 1i in. AD r = D/A r + B/A r + C/B r = D/C (2.598+1)i+(1.5+2)j=(3.598i+3.5j) F1.8 lb F=−1.8k lb (a) Moment of the Force about Point A ● Calculation Process 1. Present Final Formula: The moment about point is calculated using the cross product: 2. Substitute Values: 3. Partial Operations: Using the distributive property and cross product identities ( and ): 4. Final Calculation: The magnitude of the moment is: ●