Answer:
The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.
Explanation:
First, we must calculate the resultant force ([tex]\vec F[/tex]), in newtons, by vectorial sum:
[tex]\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j}[/tex] (1)
[tex]\vec F = 182.843\,\hat{i} + 356.048\,\hat{j}[/tex]
Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:
[tex]\|\vec F\| = \sqrt{(482.843\,N)^{2}+(356.048\,N)^{2}}[/tex]
[tex]\|\vec F\| \approx 599.923\,N[/tex]
Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:
[tex]\theta = \tan^{-1}\left(\frac{356.048\,N}{482.843\,N} \right)[/tex]
Where [tex]\theta[/tex] is the direction of the resultant force, in sexagesimal degrees.
[tex]\theta \approx 36.405^{\circ}[/tex]
The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.
