Answer:
F₁ = <-10, -10√3> N and F₂ = <6√3, -6> N
Explanation:
To express the given vectors into their horizontal and vertical components, we'll use trigonometric relationships.
The horizontal component 'F_x' and vertical component 'F_y' can be found using:
- F_x = ||F||cos(θ)
- F_y = ||F||sin(θ)
Our given vectors are:
- F₁ = 20 N at 240°
- F₂ = 12 N at 330⁰
The horizontal and vertical components for the given vectors are:
[tex]\vec F_1 = \big < F_1\cos(\theta_1), \ F_1\sin(\theta_1) \big > \\\\\\\\\Longrightarrow \vec F_1 = \big < (20 \text{ N})\cos(240^\circ), \ (20 \text{ N})\sin(240^\circ) \big > \\\\\\\\\therefore \vec F_1 = \boxed{\big < -10, \ -10\sqrt{3} \big > \text{ N}}[/tex]
[tex]\vec F_2 = \big < F_2\cos(\theta_2), \ F_2\sin(\theta_2) \big > \\\\\\\\\Longrightarrow \vec F_2 = \big < (12 \text{ N})\cos(330^\circ), \ (12 \text{ N})\sin(330^\circ) \big > \\\\\\\\\therefore \vec F_1 = \boxed{\big < 6\sqrt{3} , \ -6 \big > \text{ N}}[/tex]
Thus, the problem has been solved.
