Answer :
To determine the magnitude of the vector sum \( \mathbf{V} = \mathbf{V}_1 + \mathbf{V}_2 \) and the angle \( \theta_x \) which \( \mathbf{V} \) makes with the positive \( x \)-axis, we'll break down the vectors into their components and then sum them.
Step-by-Step Solution:
1. Determine the components of \( \mathbf{V}_1 \) and \( \mathbf{V}_2 \):
- \( \mathbf{V}_1 \) has magnitude 21 units and is along the positive \( x \)-axis (\( \theta = 0^\circ \)).
[tex]\[ V_{1x} = V_1 \cos(0^\circ) = 21 \cos(0^\circ) = 21.0 \text{ units} \][/tex]
[tex]\[ V_{1y} = V_1 \sin(0^\circ) = 21 \sin(0^\circ) = 0.0 \text{ units} \][/tex]
- \( \mathbf{V}_2 \) has magnitude 23 units and makes an angle \( \theta = 59^\circ \) with the positive \( x \)-axis.
[tex]\[ V_{2x} = V_2 \cos(59^\circ) = 23 \cos(59^\circ) \approx 11.85 \text{ units} \][/tex]
[tex]\[ V_{2y} = V_2 \sin(59^\circ) = 23 \sin(59^\circ) \approx 19.71 \text{ units} \][/tex]
2. Sum the components to get the resultant vector \( \mathbf{V} \):
- Sum the \( x \)-components:
[tex]\[ V_x = V_{1x} + V_{2x} = 21.0 + 11.85 = 32.85 \text{ units} \][/tex]
- Sum the \( y \)-components:
[tex]\[ V_y = V_{1y} + V_{2y} = 0.0 + 19.71 = 19.71 \text{ units} \][/tex]
3. Calculate the magnitude of the resultant vector \( \mathbf{V} \):
[tex]\[ V = \sqrt{V_x^2 + V_y^2} = \sqrt{(32.85)^2 + (19.71)^2} \approx 38.31 \text{ units} \][/tex]
4. Determine the angle \( \theta_x \) which \( \mathbf{V} \) makes with the positive \( x \)-axis:
[tex]\[ \theta_x = \tan^{-1} \left( \frac{V_y}{V_x} \right) = \tan^{-1} \left( \frac{19.71}{32.85} \right) \approx 30.97^\circ \][/tex]
Conclusion:
The magnitude of the vector sum [tex]\( \mathbf{V} \)[/tex] is approximately 38.31 units, and the angle [tex]\( \theta_x \)[/tex] which [tex]\( \mathbf{V} \)[/tex] makes with the positive [tex]\( x \)[/tex]-axis is approximately 30.97 degrees.
Step-by-Step Solution:
1. Determine the components of \( \mathbf{V}_1 \) and \( \mathbf{V}_2 \):
- \( \mathbf{V}_1 \) has magnitude 21 units and is along the positive \( x \)-axis (\( \theta = 0^\circ \)).
[tex]\[ V_{1x} = V_1 \cos(0^\circ) = 21 \cos(0^\circ) = 21.0 \text{ units} \][/tex]
[tex]\[ V_{1y} = V_1 \sin(0^\circ) = 21 \sin(0^\circ) = 0.0 \text{ units} \][/tex]
- \( \mathbf{V}_2 \) has magnitude 23 units and makes an angle \( \theta = 59^\circ \) with the positive \( x \)-axis.
[tex]\[ V_{2x} = V_2 \cos(59^\circ) = 23 \cos(59^\circ) \approx 11.85 \text{ units} \][/tex]
[tex]\[ V_{2y} = V_2 \sin(59^\circ) = 23 \sin(59^\circ) \approx 19.71 \text{ units} \][/tex]
2. Sum the components to get the resultant vector \( \mathbf{V} \):
- Sum the \( x \)-components:
[tex]\[ V_x = V_{1x} + V_{2x} = 21.0 + 11.85 = 32.85 \text{ units} \][/tex]
- Sum the \( y \)-components:
[tex]\[ V_y = V_{1y} + V_{2y} = 0.0 + 19.71 = 19.71 \text{ units} \][/tex]
3. Calculate the magnitude of the resultant vector \( \mathbf{V} \):
[tex]\[ V = \sqrt{V_x^2 + V_y^2} = \sqrt{(32.85)^2 + (19.71)^2} \approx 38.31 \text{ units} \][/tex]
4. Determine the angle \( \theta_x \) which \( \mathbf{V} \) makes with the positive \( x \)-axis:
[tex]\[ \theta_x = \tan^{-1} \left( \frac{V_y}{V_x} \right) = \tan^{-1} \left( \frac{19.71}{32.85} \right) \approx 30.97^\circ \][/tex]
Conclusion:
The magnitude of the vector sum [tex]\( \mathbf{V} \)[/tex] is approximately 38.31 units, and the angle [tex]\( \theta_x \)[/tex] which [tex]\( \mathbf{V} \)[/tex] makes with the positive [tex]\( x \)[/tex]-axis is approximately 30.97 degrees.
