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. what is the x-component of the resultant force? express your answer in newtons.

Explanation:

Step1: Find x - component of \( \vec{F_1} \)

The angle of \( \vec{F_1} \) with the negative x - axis is \( 65.0^{\circ} \), so the angle with the positive x - axis is \( 180^{\circ}- 65.0^{\circ}=115^{\circ} \). The x - component of a vector \( \vec{F} \) with magnitude \( F \) and angle \( \theta \) with the positive x - axis is given by \( F_x = F\cos\theta \). For \( \vec{F_1} \), \( F_1 = 9.50\space N \) and \( \theta_1=115^{\circ} \). So \( F_{1x}=9.50\cos(115^{\circ}) \). We know that \( \cos(115^{\circ})=\cos(180 - 65)^{\circ}=-\cos(65^{\circ})\approx - 0.4226 \). So \( F_{1x}=9.50\times(- 0.4226)\approx - 4.0147\space N \).

Step2: Find x - component of \( \vec{F_2} \)

The angle of \( \vec{F_2} \) is \( 53.1^{\circ} \) below the negative x - axis, so the angle with the positive x - axis is \( 180^{\circ}+ 53.1^{\circ}=233.1^{\circ} \). The x - component of \( \vec{F_2} \) is \( F_{2x}=F_2\cos\theta_2 \), where \( F_2 = 5.50\space N \) and \( \theta_2 = 233.1^{\circ} \). We know that \( \cos(233.1^{\circ})=\cos(180 + 53.1)^{\circ}=-\cos(53.1^{\circ})\approx - 0.6 \) (since \( \cos(53.1^{\circ})\approx0.6 \)). So \( F_{2x}=5.50\times(- 0.6)=- 3.3\space N \) (more accurately, \( \cos(53.1^{\circ})\approx\frac{3}{5} = 0.6 \), so \( \cos(233.1^{\circ})=- 0.6 \)).

Step3: Find resultant x - component

The resultant x - component \( F_x=F_{1x}+F_{2x} \). Substituting the values of \( F_{1x} \) and \( F_{2x} \): \( F_x=-4.0147-3.3=-7.3147\space N\approx - 7.31\space N \) (or more accurately, let's recalculate \( \cos(115^{\circ}) \): \( \cos(115^{\circ})=\cos(90 + 25)^{\circ}=-\sin(25^{\circ})\approx - 0.4226 \), \( 9.5\times(- 0.4226)=- 4.0147 \); \( \cos(233.1^{\circ})=\cos(180 + 53.1)^{\circ}=-\cos(53.1^{\circ}) \), \( \cos(53.1^{\circ})=\frac{3}{5}=0.6 \) (since \( 3 - 4 - 5 \) triangle, \( \cos\theta=\frac{3}{5} \) for \( \theta = 53.1^{\circ} \)), so \( 5.5\times(- 0.6)=- 3.3 \). Then \( F_x=-4.0147-3.3=-7.3147\approx - 7.31\space N \). If we use more precise value of \( \cos(53.1^{\circ}) \): \( \cos(53.1^{\circ})\approx0.600 \) (exact value from calculator: \( \cos(53.13^{\circ})\approx0.6 \)), and \( \cos(115^{\circ})\approx\cos(115)= - 0.4226182617 \), so \( 9.5\times(- 0.4226182617)=- 4.014873486 \), \( 5.5\times(- 0.6)=- 3.3 \), sum is \( - 4.014873486-3.3=-7.314873486\approx - 7.31\space N \).

Answer:

\( \boxed{-7.31} \) (or more accurately, if we calculate with more precision, the value can be around - 7.31 N. If we use \( \cos(65^{\circ})\approx0.4226 \) and \( \cos(53.1^{\circ})\approx0.6 \), the calculation is as above)