Questions: 3. (I) If Vx=9.80 units and Vy=-6.40 units, determine the magnitude and direction of V. 4. (II) Graphically determine the resultant of the following three vector displacements: (1) 24 m, 36° north of east; (2) 18 m, 37° east of north; and (3) 26 m, 33° west of south.

Solution Steps

The magnitude of the vector \(\overrightarrow{\mathbf{V}}\) can be found using the Pythagorean theorem:

\[ |\overrightarrow{\mathbf{V}}| = \sqrt{V_x^2 + V_y^2} \]

Substituting the given values \(V_x = 9.80\) units and \(V_y = -6.40\) units:

\[ |\overrightarrow{\mathbf{V}}| = \sqrt{(9.80)^2 + (-6.40)^2} = \sqrt{96.04 + 40.96} = \sqrt{137.00} \approx 11.70 \text{ units} \]

The direction (angle \(\theta\)) of the vector \(\overrightarrow{\mathbf{V}}\) with respect to the positive \(x\)-axis can be found using the arctangent function:

\[ \theta = \tan^{-1}\left(\frac{V_y}{V_x}\right) \]

Substituting the given values:

\[ \theta = \tan^{-1}\left(\frac{-6.40}{9.80}\right) \approx \tan^{-1}(-0.6531) \approx -33.0^\circ \]

Since the angle is negative, it indicates that the vector is below the positive \(x\)-axis. Therefore, the direction is \(33.0^\circ\) below the positive \(x\)-axis.

Final Answer