Question

Four forces of magnitudes 2N, 3N, 4N, and 5N are acting on a body at a point are inclined at 30°, 60°, 90°, and 120° respectively with the horizontal. Find their resultant.

Answer 1

Given, P₁ =2N, P₂ = 3N

P₃ = 4N and P₄ = 5N

These forces can be resolved into components along the X and Y directions

R_x = P₁ cosθ₁ + P₂ cosθ₂ + P₃ cosθ₃ + P₄ cosθ₄

= 2cos 30° + 3cos 60 + 4cos 90 + 5 cos 120

= 2\(\frac{\sqrt3}{2}\)+ 3 × \(\frac {1}{2}\)+ 4 × 0 + 5 × \(\frac{1}{2}\)= 0.732 N

R_y= P₁ sinθ₁ + P₂ sinθ₂ + P₃ sinθ₃ + P₄ sinθ₄

= 2 sin 30 + 3 sin 60 + 4 sin 90 + 5 cos 120

= 2 × \(\frac {1}{2}\) + 3 × \(\frac{\sqrt3}{2}\)+ 4 × 1 + 5 × \(\frac{\sqrt3}{2}\)

= 1 + 1.5\(\sqrt 3\) + 4 + 2.5\(\sqrt 3\)

= 1 + 2.59 + 4 + 4.33

= 11.92 N

∴ Resultant force R 

= \(\sqrt {R_x^2+R_Y^2}\)

= Vo.7322 +11.922

= 11.96 N.