Q7. Coplanar forces act outwards from a common point. Three forces have the following magnitudes and directions. The angles being measured clockwise from a horizontal datum, 80N,0°; 500,60°; 100N, 150°. a. Draw the force diagram b. Determine the magnitude and direction of the resultant force

Angles are measured clockwise from the positive x-axis. For calculations, we use standard angles (counter-clockwise from positive x-axis), so clockwise angles $\theta_{cw}$ become $-\theta_{cw}$.
For $F_1 = 80 \, \text{N}$ at $0^\circ$ clockwise:

$$F_{1x} = 80 \cos(0^\circ) = 80 \, \text{N}$$ $$F_{1y} = 80 \sin(0^\circ) = 0 \, \text{N}$$

For $F_2 = 50 \, \text{N}$ at $60^\circ$ clockwise:

$$F_{2x} = 50 \cos(-60^\circ) = 50 \times 0.5 = 25 \, \text{N}$$ $$F_{2y} = 50 \sin(-60^\circ) = 50 \times \left(-\frac{\sqrt{3}}{2}\right) = -25\sqrt{3} \, \text{N}$$

For $F_3 = 100 \, \text{N}$ at $150^\circ$ clockwise:

$$F_{3x} = 100 \cos(-150^\circ) = 100 \times \left(-\frac{\sqrt{3}}{2}\right) = -50\sqrt{3} \, \text{N}$$ $$F_{3y} = 100 \sin(-150^\circ) = 100 \times (-0.5) = -50 \, \text{N}$$

$$R_x = F_{1x} + F_{2x} + F_{3x} = 80 + 25 - 50\sqrt{3} = 105 - 50\sqrt{3} \, \text{N}$$ $$R_y = F_{1y} + F_{2y} + F_{3y} = 0 - 25\sqrt{3} - 50 = -50 - 25\sqrt{3} \, \text{N}$$

Using approximate values:
$R_x \approx 105 - 50(1.732) = 105 - 86.6 = 18.4 \, \text{N}$
$R_y \approx -50 - 25(1.732) = -50 - 43.3 = -93.3 \, \text{N}$

$$R = \sqrt{R_x^2 + R_y^2}$$ $$R = \sqrt{(105 - 50\sqrt{3})^2 + (-50 - 25\sqrt{3})^2}$$ $$R \approx \sqrt{(18.4)^2 + (-93.3)^2}$$ $$R \approx \sqrt{338.56 + 8704.89} = \sqrt{9043.45}$$ $$R \approx 95.1 \, \text{N}$$

Let $\alpha$ be the angle measured counter-clockwise from the positive x-axis.

$$\tan \alpha = \frac{R_y}{R_x}$$ $$\tan \alpha = \frac{-50 - 25\sqrt{3}}{105 - 50\sqrt{3}} \approx \frac{-93.3}{18.4} \approx -5.0717$$ $$\alpha = \arctan(-5.0717) \approx -78.85^\circ$$

Since $R_x$ is positive and $R_y$ is negative, the resultant force is in the fourth quadrant. An angle of $-78.85^\circ$ counter-clockwise is equivalent to $78.85^\circ$ clockwise from the horizontal datum.

The magnitude of the resultant force is $\boxed{\text{95.1 N}}$ and its direction is $\boxed{\text{78.9}^\circ \text{ clockwise from the horizontal datum}}$.