Meriam, Kraige & Bolton — Introduction to Statics: Problem 1_4

⚡ Mecademy AIENG정역학 · ch1  Problem Statement 1/4 A force is specified by the vector N. Calculate the angles made by with the positive -, -, and -axes. Problem 1/4 (a) Angle with the positive x-axis () ● Calculation Process 1. Present Final Formula: The angle that a vector makes with the positive x-axis is given by the direction cosine: where . 2. Substitute Values: Identify components: , , . 3. Partial Operations: Calculate the magnitude of the force: Calculate the cosine of the angle: 4. Final Calculation: ● Final Conclusion: F=160i+80j−120k Fxyz θ x θ x F=F i+ x F j+ y F k z θ = x arccos ( ∣F∣ F x ) ∣F∣= F +F +F x 2 y 2 z 2 F = x 160 NF = y 80 NF = z −120 N ∣F∣= 160+80+(−120) 222 θ = x arccos ( 160+80+(−120) 222 160 ) ∣F∣= = 25600+6400+14400 ≈ 46400215.41 N cosθ = x ≈ 215.41 160 0.7428 θ = x arccos(0.7428)≈42.03 ∘ The angle made by the force with the positive -axis is . (b) Angle with the positive y-axis () ● Calculation Process 1. Present Final Formula: 2. Substitute Values: 3. Partial Operations: Calculate the cosine of the angle: 4. Final Calculation: ● Final Conclusion: The angle made by the force with the positive -axis is . (c) Angle with the positive z-axis () ● Calculation Process 1. Present Final Formula: 2. Substitute Values: 3. Partial Operations: Calculate the cosine of the angle: Fxθ = x 42.0 ∘ θ y θ = y arccos ( ∣F∣ F y ) θ = y arccos ( 215.41 80 ) cosθ ≈ y 0.3714 θ = y arccos(0.3714)≈68.20 ∘ Fyθ = y 68.2 ∘ θ z θ = z arccos ( ∣F∣ F z ) θ = z arccos ( 215.41 −