12:46 PM Edit P FR 150 N 115° $15 100 N (c) Fig. 2-11 88 4G B/s FR = √(100 N)2 + (150 N)² - 2(100 N)(150 N) cos 115° V10 000 +22 500 - 30 000(-0.4226) = 212.6 N = 213 N Applying the law of sines to determine 0, 150 N 212.6 N sin 0 sin 115° 150 N sin (sin 115°) 212.6 N 0 = 39.8° 3 50 Х Ans. Thus, the direction (phi) of FR, measured from the horizontal, is 39.8° 15.0° = 54.8° Ans. EXERCISE: EXAMPLE 2.2 Resolve the horizontal 600-lb force in Fig. 2-12a into components acting along the u and v axes and determine the magnitudes of these components. (a) 30 130° 600 lb Fig. 2-12 D DO w Tools Mobile View Share PDF to DOC Edit on PC

12:46 PM Edit P FR 150 N 115° $15 100 N (c) Fig. 2-11 88 4G B/s FR = √(100 N)2 + (150 N)² - 2(100 N)(150 N) cos 115° V10 000 +22 500 - 30 000(-0.4226) = 212.6 N = 213 N Applying the law of sines to determine 0, 150 N 212.6 N sin 0 sin 115° 150 N sin (sin 115°) 212.6 N 0 = 39.8° 3 50 Х Ans. Thus, the direction (phi) of FR, measured from the horizontal, is 39.8° 15.0° = 54.8° Ans. EXERCISE: EXAMPLE 2.2 Resolve the horizontal 600-lb force in Fig. 2-12a into components acting along the u and v axes and determine the magnitudes of these components. (a) 30 130° 600 lb Fig. 2-12 D DO w Tools Mobile View Share PDF to DOC Edit on PC

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter7: Analytic Trigonometry

Section7.6: The Inverse Trigonometric Functions

Problem 87E

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