Question

Find the magnitude and direction of resultant of 8N.
12 N. 16N and 200 making an angle 30°. 120°, and
330 with +ve x axis measured in anticlockwise
respectively
​

Answer 1

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°

Answer 2

Answer:

200-30=170

320+170=490