Question

two forces, \\(\vec{f_1}\\) and \\(\vec{f_2}\\), act at a point. the magnitude of \\(\vec{f_1}\\) is 9.50 n, and its direction is an angle 65.0° above the negative direction of x-axis in the second quadrant. the magnitude of \\(\vec{f_2}\\) is 5.50 n, and its direction is an angle 53.1° below the negative direction of x-axis in the third quadrant.

part b
what is the y-component of the resultant force?
express your answer in newtons.
\\(f_y = \square\\) n

Explanation:

Step1: Find \( F_{1y} \)

\( \vec{F}_1 \) is in the second quadrant, angle \( 65.0^\circ \) above negative x - axis. So the y - component \( F_{1y}=F_1\sin(65.0^\circ) \), \( F_1 = 9.50\ N \).
\( F_{1y}=9.50\times\sin(65.0^\circ)\approx9.50\times0.9063 = 8.61\ N \) (positive as it's in positive y - direction).

Step2: Find \( F_{2y} \)

\( \vec{F}_2 \) is in the third quadrant, angle \( 53.1^\circ \) below negative x - axis, so it's \( 180^\circ - 53.1^\circ=126.9^\circ \) from positive x - axis or we can calculate its y - component as \( F_{2y}=-F_2\sin(53.1^\circ) \) (negative as it's in negative y - direction), \( F_2 = 5.50\ N \), \( \sin(53.1^\circ)\approx0.8 \).
\( F_{2y}=-5.50\times0.8=-4.40\ N \)

Step3: Find resultant \( F_y \)

Resultant y - component \( F_y = F_{1y}+F_{2y} \)
\( F_y=8.61+( - 4.40)=4.21\ N \) (approximate value, more precise calculation: \( \sin(65^\circ)\approx0.906307787 \), \( 9.5\times0.906307787 = 8.609924\ N \); \( \sin(53.1^\circ)\approx0.800038 \), \( 5.5\times0.800038 = 4.400209\ N \); \( F_y = 8.609924-4.400209 = 4.2097\approx4.21\ N \))

Answer:

\( \boxed{4.21} \) (or more precise value depending on calculation, if we use \( \sin(65^\circ)=\sin(65)=\frac{\sqrt{2 + \sqrt{2 + \sqrt{3}}}}{2}\approx0.9063 \) and \( \sin(53.1^\circ)\approx\frac{4}{5} \), the answer is approximately \( 4.21\ N \))