Question

Find the magnitude and direction of the resultant of the concurrent

forces of 8N, 12N, 15N and 20N making of 30, 70
, 120 ,101 and

155
respectively with a fixed line.

Answer 1

Your question is incomplete . Complete question is ------->Find the magnitude and direction of the resultant of the concurrent forces 8N,12N,15N and 20N making angles of 30°, 70° , 120°15' and 155° with a fix line?

Answer :- first of all all forces are in vector form
F₁ = 8N , direction 30° ⇒F₁ = 8cos30° i + 8sin30° j = 4√3i + 4j
F₂ = 12N , direction 70° ⇒F₂ = 12cos70° i + 12sin70° j = 12 ×0.34 i + 12 × 0.94 j = 4.08i + 11.28 j
F₃ = 15N, direction 120°15' ⇒F₃= 15cos120°15' i + 15sin120°15' j = 15 × (-0.5)i + 15 × 0.86j = -7.5 i + 12.90 j
F₄ = 20N, direction 155° ⇒F₄ = 20cos155° i + 20sin155° j = 20 ×(-0.9) i + 20 × 0.42 j = -18i + 8.4j

Now, resultant force , F = F₁ + F₂ + F₃ + F₄
= (4√3 + 4.08 -7.5 - 18)i + (4 + 11.28 + 12.90 + 8.4)j
= (6.92 + 4.08 - 25.5)i + (15.28 + 21.30)j
= -14.5i + 36.30j

And direction of resultant force ,
Tanθ = 36.30/-14.5 = tan111.774° e.g., θ = 111.74°