Determine the magnitudes of the reactions R₁ and R₂. (d) Calculate the magnitude and direction of the system of concurrent forces below

We will measure angles counter-clockwise from the positive x-axis.

• Force 1 ($F_1 = 200 \, \text{N}$): Angle $\theta_1 = 30^\circ$.

  $$F_{1x} = 200 \cos(30^\circ) = 200 \times \frac{\sqrt{3}}{2} \approx 173.21 \, \text{N}$$ $$F_{1y} = 200 \sin(30^\circ) = 200 \times \frac{1}{2} = 100.00 \, \text{N}$$

• Force 2 ($F_2 = 300 \, \text{N}$): The angle is $70^\circ$ from the negative x-axis towards the positive y-axis. Angle $\theta_2 = 180^\circ - 70^\circ = 110^\circ$.

  $$F_{2x} = 300 \cos(110^\circ) \approx 300 \times (-0.3420) \approx -102.61 \, \text{N}$$ $$F_{2y} = 300 \sin(110^\circ) \approx 300 \times (0.9397) \approx 281.91 \, \text{N}$$

• Force 3 ($F_3 = 450 \, \text{N}$): The angle is $65^\circ$ from the negative y-axis towards the negative x-axis. This means it's in the third quadrant. The angle from the negative x-axis is $90^\circ - 65^\circ = 25^\circ$. Angle $\theta_3 = 180^\circ + 25^\circ = 205^\circ$.

  $$F_{3x} = 450 \cos(205^\circ) \approx 450 \times (-0.9063) \approx -407.84 \, \text{N}$$ $$F_{3y} = 450 \sin(205^\circ) \approx 450 \times (-0.4226) \approx -190.18 \, \text{N}$$

$$\sum F_x = F_{1x} + F_{2x} + F_{3x}$$ $$\sum F_x = 173.21 \, \text{N} - 102.61 \, \text{N} - 407.84 \, \text{N} = -337.24 \, \text{N}$$ $$\sum F_y = F_{1y} + F_{2y} + F_{3y}$$ $$\sum F_y = 100.00 \, \text{N} + 281.91 \, \text{N} - 190.18 \, \text{N} = 191.73 \, \text{N}$$

$$R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$$ $$R = \sqrt{(-337.24)^2 + (191.73)^2}$$ $$R = \sqrt{113731.8 + 36760.4}$$ $$R = \sqrt{150492.2}$$ $$R \approx 387.93 \, \text{N}$$

First, find the reference angle $\alpha$:

$$\tan \alpha = \left| \frac{\sum F_y}{\sum F_x} \right| = \left| \frac{191.73}{-337.24} \right| \approx 0.5685$$ $$\alpha = \arctan(0.5685) \approx 29.62^\circ$$

Since $\sum F_x$ is negative and $\sum F_y$ is positive, the resultant force is in the second quadrant.
The angle $\theta$ measured counter-clockwise from the positive x-axis is:

$$\theta = 180^\circ - \alpha$$ $$\theta = 180^\circ - 29.62^\circ = 150.38^\circ$$

The magnitude of the resultant force is $\boxed{387.93 \, \text{N}}$ and its direction is $\boxed{150.38^\circ}$ counter-clockwise from the positive x-axis.