We have a vector [tex]\vec F[/tex] with a magnitude [tex]F[/tex] of 86.4 N.
a. Let [tex]\theta[/tex] be the angle [tex]\vec F[/tex] makes with the positive [tex]x[/tex]-axis. The [tex]x[/tex]-component of [tex]\vec F[/tex] is
[tex]F_x=(86.4\cos\theta)\,\mathrm N[/tex]
and has a magnitude of 72.3 N, so
[tex]72.3=86.4\cos\theta\implies\cos\theta=0.837\implies\theta=\boxed{33.2^\circ}[/tex]
b. The [tex]y[/tex]-component of [tex]\vec F[/tex] is
[tex]F_y=(86.4\cos33.2^\circ)\,\mathrm N=\boxed{47.3\,\mathrm N}[/tex]
a) The angle between the vector and the +x axis is approximately 33.196°.
b) The component of the force along the +y axis is approximately 47.304 newtons.
In this question we should apply the concepts of magnitude and direction of a vector to solve each part. The magnitude ([tex]\|\vec F\|[/tex]), in newtons, is a application of Pythagorean theorem and direction ([tex]\theta[/tex]), in degrees, is an application of trigonometric functions.
a) The angle between the vector and the component along the x axis ([tex]F_{x}[/tex]), in newtons, is found by means of the following expression:
[tex]\theta = \cos^{-1} \frac{F_{x}}{\|\vec F\|}[/tex] (1)
([tex]\|\vec F\| = 86.4\,N[/tex], [tex]F_{x} = 72.3\,N[/tex])
[tex]\theta = \cos^{-1} \left(\frac{72.3\,N}{86.4\,N} \right)[/tex]
[tex]\theta \approx 33.196^{\circ}[/tex]
The angle between the vector and the +x axis is approximately 33.196°. [tex]\blacksquare[/tex]
b) The magnitude of the +y component of the vector force ([tex]F_{y}[/tex]), in newtons, is determined by the following Pythagorean expression:
[tex]F_{y} = \sqrt{(\|\vec F\|)^{2}-F_{x}^{2}}[/tex] (2)
([tex]\|\vec F\| = 86.4\,N[/tex], [tex]F_{x} = 72.3\,N[/tex])
[tex]F_{y} = \sqrt{(86.4\,N)^{2}-(72.3\,N)^{2}}[/tex]
[tex]F_{y} \approx 47.304\,N[/tex]
The component of the force along the +y axis is approximately 47.304 newtons. [tex]\blacksquare[/tex]
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