Question

Resolve horezontally and vertically a force F= 8N which mokes an angle of 45^(@) with the horizontal.

Answer 1

Horizontal force=8cosx=8*1/√2

=4√2N

Vertical force=8sinx=8*1/√2

=4√3N

Answer 2

For the given Force, Horizontal Component Fx = 4√2 N and Vertical Component Fy = 4√2 N

- It is given that :

Force, F= 8N

Angle with horizontal, θ = 45°

- Horizontal component of force, Fx = F × cos θ

⇒Fx = 8 × cos 45

  Fx = 8 × 1/√2  = 4√2 N

- Vertical component of force, Fy = F × sin θ

⇒Fy = 8 × sin 45

  Fy = 8 × 1/√2  = 4√2 N