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 @ 50⁰ N of E F2= 650 N @ 22⁰ 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 4 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 F3=250 d 80° 3501 F2y F₂=150 d ANSWER NO. 2 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 @ 50⁰ N of E F2= 650 N @ 22⁰ 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 4 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 F3=250 d 80° 3501 F2y F₂=150 d ANSWER NO. 2 ONLY

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Description: Using the component method, determine the resultant force and theequilibrant force of each of the following sets of concurrent forces.1. F₁ = 380 N @ 50⁰ N of EF2 = 650 N @ 22⁰ E of SF3 = 720 N West3. FA = 20 lb SАFB = 24 lb NEFc = 16 lb WestFD = 25

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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 @ 50⁰ N of E
F2 = 650 N @ 22⁰ E of S
F3 = 720 N West
3. FA = 20 lb S
А
FB = 24 lb NE
Fc = 16 lb West
FD = 25 lb @ 32º E of S
A
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)
Fy
0
300(cos 35°)
=+246 (North)
45⁰
= -177 (South)
450(sin 80°)
=-443 (South)
-374 (South)
Sketch
F₁ = 200 d
F₂=300 d
FAY!
F3-250 d
45⁰
F
t
80⁰
35⁰1 F2y
F3v
F4=150 d
ANSWER NO. 2 ONLY

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