Solved Problem on Vectors
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What are the components of the vectors with magnitudes 8 and 6 (see figure)?


Problem data:
  • Magnitude of vector 1:    v1 = 8;
  • Angle of vector 1:    θ1 = 60° with the positive x-axis;
  • Magnitude of vector 2:    v2 = 6;
  • Angle of vector 2:    θ2 = 30° with the negative x-axis.
Solution:
  • For vector v1:

The angle measured counterclockwise from the positive x-axis is positive (Figure 1). The component in the x direction is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {v_x=v\cos\theta} \tag{I} \end{gather} \]
\[ \begin{gather} v_{x1}=v_1\cos\theta_1\\[5pt] v_{x1}=8\cos60° \end{gather} \]
Figure 1

From Trigonometry:   \( \cos 60°=\dfrac{1}{2} \)

\[ \begin{gather} v_{x1}=8\times{\frac{1}{2}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {v_{x1}=4} \end{gather} \]

The component in the y direction is given by (Figure 1)
\[ \begin{gather} \bbox[#99CCFF,10px] {v_y=v\sin\theta} \tag{II} \end{gather} \]
\[ \begin{gather} v_{y1}=v_1\sin\theta_1\\[5pt] v_{y1}=6\sin60° \end{gather} \]

From Trigonometry:   \( \sin60°=\dfrac{\sqrt{3\;}}{2} \)

\[ \begin{gather} v_{y1}=6\times{\frac{\sqrt{3\;}}{2}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {v_{y1}\approx 5.2} \end{gather} \]

  • For vector v2:
The angle measured counterclockwise from the positive x-axis is positive, 180°+30°=210°. The angle measured clockwise from the positive x-axis is negative, −180°+30°=−150° (Figure 2). The component in the x direction is given by applying formula (I)
\[ \begin{gather} v_{x2}=v_2\cos\theta_2\\[5pt] v_{x2}=6\cos210° \end{gather} \]
Figure 2

From Trigonometry:   \( \cos 210°=\cos (-150°)=-\cos30°=-{\dfrac{\sqrt{3\;}}{2}} \)

\[ \begin{gather} v_{x2}=6\times\left(-{\frac{\sqrt{3\;}}{2}}\right) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {v_{x2}\approx -5.2} \end{gather} \]
The component in the y direction is given by applying formula (II) (Figure 2)
\[ \begin{gather} v_{y2}=v_2\sin\theta_2\\[5pt] v_{y2}=6\sin210° \end{gather} \]

From Trigonometry:   \( \sin210°=\sin(-150°)=-\sin30°=-{\dfrac{1}{2}} \)
\[ \begin{gather} v_{y2}=6\times\left(-{\frac{1}{2}}\right) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {v_{y2}=-3} \end{gather} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .