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 @ 22º E of S F3 = 720 N West 3. FA = 20 lb S FB = 24 lb NE Fc = 16 lb West FD=25 lb @ 32⁰ E of S 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) 450(cos 80°) =-78.1 (West) 2. F₁ = 332 d @ 35⁰ W of N F2=420 d @ 75⁰ E of N F3 = 108 d SW -273 (West) 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 F4y! F3-250 d 80⁰ 35⁰1 F2y F₁=150 d ANSWER NO. 3 ONLY

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 @ 22º E of S F3 = 720 N West 3. FA = 20 lb S FB = 24 lb NE Fc = 16 lb West FD=25 lb @ 32⁰ E of S 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) 450(cos 80°) =-78.1 (West) 2. F₁ = 332 d @ 35⁰ W of N F2=420 d @ 75⁰ E of N F3 = 108 d SW -273 (West) 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 F4y! F3-250 d 80⁰ 35⁰1 F2y F₁=150 d ANSWER NO. 3 ONLY