Respuesta :
Answer:
A. [tex]F_x=-8.74N[/tex]
B. [tex]F_y=3.55N[/tex]
C. [tex]F=9.43N[/tex]
Explanation:
First, we break down the two vectors F₁ and F₂ into their components. In order to do that, we have to use trigonometry:
[tex]F_{1x}=F_1\cos\theta_1\implies F_{1x}=-9.60N\cos55.0\°=-5.50N\\\\F_{1y}=F_1\sin\theta_1\implies F_{1y}=9.60N\sin55.0\°=7.86N\\\\\\F_{2x}=F_2\cos\theta_2\implies F_{2x}=-5.40N\cos53.1\°=-3.24N\\\\F_{2y}=F_2\sin\theta_2\implies F_{2y}=-5.40N\sin53.1\°=-4.31N\\\\[/tex]
(Note: [tex]F_{1x}[/tex], [tex]F_{2x}[/tex] and [tex]F_{2y}[/tex] are negative because the vectors are in the second and third quadrant, respectively. So those components must be negative)
Next, we sum the x components and the y components to get the components of the resultant force:
[tex]F_x=F_{1x}+F_{2x}=-5.50N-3.24N=-8.74N\\\\F_y=F_{1y}+F_{2y}=7.86N-4.31N=3.55N[/tex]
So, the x component of the resultant force is -8.74N (A), and its y component is 3.55N (B).
Finally, to get the magnitude of the resultant force, we have to use the Pythagorean Theorem:
[tex]F=\sqrt{F_x^{2}+F_y^{2}}\\ \\F=\sqrt{(-8.74N)^{2}+(3.55N)^{2}}=9.43N[/tex]
In words, the magnitude of the resultant force is 9.43N (C).
