Question - Analysis of Resultant Forces in Various Configurations

For F2-1:

Using vector addition, the resultant force \( R \) is the vector sum of the two forces.

\( R_x = 6\cos(45^\circ) + 2\cos(45^\circ) \)

\( R_x = (6 + 2) \cdot \frac{\sqrt{2}}{2} \)

\( R_x = 8 \cdot \frac{\sqrt{2}}{2} \)

\( R_x = 4\sqrt{2} \text{ N} \)

\( R_y = 6\sin(45^\circ) - 2\sin(45^\circ) \)

\( R_y = (6 - 2) \cdot \frac{\sqrt{2}}{2} \)

\( R_y = 4 \cdot \frac{\sqrt{2}}{2} \)

\( R_y = 2\sqrt{2} \text{ N} \)

The magnitude of the resultant force is:

\( R = \sqrt{R_x^2 + R_y^2} \)

\( R = \sqrt{(4\sqrt{2})^2 + (2\sqrt{2})^2} \)

\( R = \sqrt{32 + 8} \)

\( R = \sqrt{40} \)

\( R = 2\sqrt{10} \text{ N} \)

The direction is given by:

\( \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) \)

\( \theta = \tan^{-1}\left(\frac{2\sqrt{2}}{4\sqrt{2}}\right) \)

\( \theta = \tan^{-1}\left(\frac{1}{2}\right) \)

The direction is \( \theta \) degrees above the +x-axis.

For F2-2:

This part of the question corresponds to a different problem about forces acting on a hook.

For F2-3:

This part of the question corresponds to yet another different problem about determining the magnitude of the resultant force and its direction.

Note: Since the question is about F2-1, only the solution for F2-1 is provided, and not for F2-2 or F2-3.