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enter image description here I'm not sure how to approach this problem. I don't know the magnitude of F1 or F2 and I'm not sure how to get it from F. Any help would be appreciated.

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You have two forces, which you can reduce to a single point like this for simplicity given that you do not forget the couple moment. For now, let's assume we only have the two forces.

enter image description here

Now, we know that the sum of these two vectors gives us the resultant:

$$\vec{\mathbf{F}}_1+\vec{\mathbf{F}}_2 = \vec{\mathbf{F}}$$

No we resolve the force into components and we get

$$F_1\cos(45^\circ)-F_2\sin(30^\circ)=10\\F_1\sin(45^\circ)+F_2\cos(30^\circ)=0$$

Solving these two equations gives us the following results

$$F_1\approx8.965\,\text{N} \\ F_2\approx-7.321\,\text{N}$$

Now you form the vectors from the resultants. Make sure you maintain consistency with the signs (+,-) you use on the component resolutions.

$$ \vec{\mathbf{F}}_1 = 8.965\cos(45^\circ)\hat{\mathbf{i}} + 8.965\cos(45^\circ)\hat{\mathbf{j}} \approx 6.339\hat{\mathbf{i}}+6.339\hat{\mathbf{j}} \\ \vec{\mathbf{F}}_2 = 7.321\sin(30^\circ)\hat{\mathbf{i}} - 7.321\cos(30^\circ)\hat{\mathbf{j}} \approx 3.660\hat{\mathbf{i}}-6.339\hat{\mathbf{j}}$$

To compute the moment you need the position vectors as well. Consider the following.

enter image description here

Now all you need is to come up with the position vectors

$$\vec{\mathbf{r}}_1 = 2\hat{\mathbf{j}} \\ \vec{\mathbf{r}}_2 = 4\hat{\mathbf{i}} $$

The moment is simply the sum of moments

$$\vec{\mathbf{M}} = \vec{\mathbf{r}}_1\times\vec{\mathbf{F}}_1 + \vec{\mathbf{r}}_2\times\vec{\mathbf{F}}_2 = 18.039\hat{\mathbf{k}}$$

Therefore the couple moment is $M=18.039\,\text{N m}$.

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