Using the component method, determine the resultant force and the equilibrant force of each of the following sets of concurrent forces. 1. F₁ = 380 N @ 50° N of E F2 = 650 N @ 22⁰ E of S F3 = 720 N West 3. FA = 20 lb S FB = 24 lb NE Fc: FD = 25 lb @ 32⁰ E of S = 16 lb West FOLLOW THE FORMAT BELOW Example 2: F₁ = 200 d West F₂ = 300 d @ 35⁰ W of N F3 = 250 d SE F₁ = 450 d @ 80⁰ S of W Σ Fx -200 (West) 300(sin 350) =-172 (West) 250(cos 45°) = +177 (East) 2. F₁ = 332 d @ 35⁰ W of N F2 = 420 d @ 75⁰ E of N F3 = 108 d SW 450(cos 80°) = -78.1 (West) -273 (West) F₁ 0 300 (cos 35°) =+246 (North) 250(sin 45°) = -177 (South) 450(sin 80°) =-443 (South) -374 (South) Sketch F₁ = 200 d F₂=300 d FAY! 35⁰1 F2y F3x 45⁰ |F3=250 ! 80⁰ F3y F₁=150 d

Using the component method, determine the resultant force and the equilibrant force of each of the following sets of concurrent forces. 1. F₁ = 380 N @ 50° N of E F2 = 650 N @ 22⁰ E of S F3 = 720 N West 3. FA = 20 lb S FB = 24 lb NE Fc: FD = 25 lb @ 32⁰ E of S = 16 lb West FOLLOW THE FORMAT BELOW Example 2: F₁ = 200 d West F₂ = 300 d @ 35⁰ W of N F3 = 250 d SE F₁ = 450 d @ 80⁰ S of W Σ Fx -200 (West) 300(sin 350) =-172 (West) 250(cos 45°) = +177 (East) 2. F₁ = 332 d @ 35⁰ W of N F2 = 420 d @ 75⁰ E of N F3 = 108 d SW 450(cos 80°) = -78.1 (West) -273 (West) F₁ 0 300 (cos 35°) =+246 (North) 250(sin 45°) = -177 (South) 450(sin 80°) =-443 (South) -374 (South) Sketch F₁ = 200 d F₂=300 d FAY! 35⁰1 F2y F3x 45⁰ |F3=250 ! 80⁰ F3y F₁=150 d

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter1: Units And Measurement

Section: Chapter Questions

Problem 79P: Perform the following calculations and express your answer using the correct number of significant...

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