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 content not fully visible part c: what is the magnitude of the resultant force? express your answer in newtons.

Explanation:

Step1: Resolve \( \vec{F_1} \) into components

\( F_{1x} = -F_1\cos(65.0^\circ) \), \( F_{1y} = F_1\sin(65.0^\circ) \)
\( F_{1x} = -9.50\cos(65.0^\circ) \approx -9.50\times0.4226 \approx -4.015 \, \text{N} \)
\( F_{1y} = 9.50\sin(65.0^\circ) \approx 9.50\times0.9063 \approx 8.610 \, \text{N} \)

Step2: Resolve \( \vec{F_2} \) into components

\( F_{2x} = -F_2\cos(53.1^\circ) \), \( F_{2y} = -F_2\sin(53.1^\circ) \) (since it's below negative x - axis, y - component is negative)
\( F_{2x} = -5.50\cos(53.1^\circ) \approx -5.50\times0.600 \approx -3.30 \, \text{N} \) (using \( \cos(53.1^\circ)\approx0.6 \))
\( F_{2y} = -5.50\sin(53.1^\circ) \approx -5.50\times0.800 \approx -4.40 \, \text{N} \) (using \( \sin(53.1^\circ)\approx0.8 \))

Step3: Find the net x - component (\( F_x \))

\( F_x = F_{1x}+F_{2x}=-4.015 - 3.30=-7.315 \, \text{N} \)

Step4: Find the net y - component (\( F_y \))

\( F_y = F_{1y}+F_{2y}=8.610-4.40 = 4.210 \, \text{N} \)

Step5: Calculate the magnitude of the resultant force (\( F \))

Using the formula \( F=\sqrt{F_x^{2}+F_y^{2}} \)
\( F=\sqrt{(-7.315)^{2}+(4.210)^{2}}=\sqrt{53.51 + 17.72}=\sqrt{71.23}\approx8.44 \, \text{N} \)

Answer:

\( \approx 8.44 \, \text{N} \)