To determine the direction of the net force $\mathbf{F}_{\text{net}}$ on charge $A$, we'll break down the problem into the following steps:

1. Identify the Forces:

   • The force exerted by charge $B$ on $A$ is repulsive and acts directly to the right (along the positive $x$-axis). Let's denote this force as $\mathbf{F}_B = 3.3 \, \text{N}$.
   • The force exerted by charge $C$ on $A$ is attractive and acts directly downward (along the negative $y$-axis). Let's denote this force as $\mathbf{F}_C = 6.6 \, \text{N}$.
   • The force exerted by charge $D$ on $A$ is attractive and acts diagonally toward $D$. Let's denote this force as $\mathbf{F}_D = 2.2 \, \text{N}$.

2. Determine the Components of the Forces:

   • $\mathbf{F}_B$ acts entirely along the $x$-axis: $\mathbf{F}_B = 3.3 \, \text{N} \, \hat{i}$.
   • $\mathbf{F}_C$ acts entirely along the $y$-axis: $\mathbf{F}_C = -6.6 \, \text{N} \, \hat{j}$.
   • $\mathbf{F}_D$ has both $x$ and $y$ components. The angle between the diagonal $AD$ and the horizontal axis can be determined using the dimensions of the rectangle.

3. Calculate the Angle for Force $D$:

   • The length $AB = 141 \, \text{cm}$ and $AC = 100 \, \text{cm}$.
   • The distance $AD$ (the diagonal) can be calculated using the Pythagorean theorem: $AD = \sqrt{AB^2 + AC^2} = \sqrt{141^2 + 100^2} = \sqrt{19881 + 10000} \approx 174.92 \, \text{cm}$
   • The angle $\theta$ between the diagonal $AD$ and the horizontal side $AB$ is: $\theta = \tan^{-1}\left(\frac{AC}{AB}\right) = \tan^{-1}\left(\frac{100}{141}\right) \approx 35.1^\circ$

   Therefore, the components of $\mathbf{F}_D$ are: \mathbf{F}_{Dx} = 2.2 \, \cos(35.1^\circ) \, \text{N} \quad \text{(towards the negative x-axis)} \mathbf{F}_{Dy} = -2.2 \, \sin(35.1^\circ) \, \text{N} \quad \text{(towards the negative y-axis)}

4. Sum the Components to Find the Net Force:

   • Net force in the $x$-direction: $F_{\text{net},x} = F_B + F_{Dx} = 3.3 + 2.2 \times (-\cos(35.1^\circ)) \approx 3.3 - 1.8 = 1.5 \, \text{N}$
   • Net force in the $y$-direction: $F_{\text{net},y} = F_C + F_{Dy} = -6.6 + 2.2 \times (-\sin(35.1^\circ)) \approx -6.6 - 1.3 = -7.9 \, \text{N}$

5. Determine the Angle of the Net Force:

   • The angle $\phi$ of the net force from the negative $y$-axis can be found using: $\tan(\phi) = \frac{F_{\text{net},x}}{-F_{\text{net},y}} = \frac{1.5}{7.9} \approx 0.19$ $\phi \approx \tan^{-1}(0.19) \approx 10.7^\circ$ This angle is measured clockwise from the negative $y$-axis.

Thus, the direction of the net force on charge $A$ is approximately 10.7° clockwise from the negative $y$-axis.

Would you like more details on this solution? Here are some related questions:

1. How would the direction change if the magnitude of force $C$ were reduced?
2. What if charge $D$ were moved closer to charge $A$?
3. How does the position of charge $B$ affect the net force on $A$?
4. What if charges $B$ and $C$ were switched?
5. How would the net force be affected if charge $A$ were also negative?

Tip: When dealing with forces in multiple directions, always break them into their components and handle each direction separately.