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 = 16 lb West FD = 25 lb @ 32⁰ E of S 2. F₁ = 332 d @ 35⁰ W of N F2 = 420 d @ 75⁰ E of N F3 = 108 d SW
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 = 16 lb West FD = 25 lb @ 32⁰ E of S 2. F₁ = 332 d @ 35⁰ W of N F2 = 420 d @ 75⁰ E of N F3 = 108 d SW
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Transcribed Image Text: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. 3 ONLY
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