Question - Determining the Resultant Force and its Direction

Given:

\(\vec{F_1} = 800 \text{ N} \ (\text{upwards along the y-axis})\)

\(\vec{F_2} = 600 \text{ N} \ (\text{30 degrees below the -x-axis})\)

To solve for the resultant force \(\vec{R}\), break \(\vec{F_2}\) into its x and y components.

\(F_{2x} = 600 \cos(30^\circ)\)

\(F_{2y} = -600 \sin(30^\circ)\) (The y-component is negative because it is downward.)

Calculate \(F_{2x}\) and \(F_{2y}\):

\(F_{2x} = 600 \times \frac{\sqrt{3}}{2} = 300\sqrt{3}\) N

\(F_{2y} = -600 \times \frac{1}{2} = -300\) N

The resultant force \(\vec{R}\) in the x and y components is:

\(R_x = F_{2x}\) because there is no x-component for \(\vec{F_1}\).

\(R_y = F_{1} + F_{2y}\)

Calculate \(R_x\) and \(R_y\):

\(R_x = 300\sqrt{3}\) N

\(R_y = 800 - 300 = 500\) N

Now calculate the magnitude of the resultant vector \(|\vec{R}|\):

\(|\vec{R}| = \sqrt{R_x^2 + R_y^2}\)

\(|\vec{R}| = \sqrt{(300\sqrt{3})^2 + (500)^2}\)

\(|\vec{R}| = \sqrt{(900 \times 3) + (250000)}\)

\(|\vec{R}| = \sqrt{2700 + 250000}\)

\(|\vec{R}| = \sqrt{252700}\)

\(|\vec{R}| \approx 502.7\) N

To find the direction θ measured counterclockwise from the positive x-axis:

\(\tan(\theta) = \frac{R_y}{R_x}\)

\(\theta = \arctan\left(\frac{500}{300\sqrt{3}}\right)\)

\(\theta = \arctan\left(\frac{500}{300 \times \frac{\sqrt{3}}{2}}\right)\)

\(\theta = \arctan\left(\frac{500}{150\sqrt{3}}\right)\)

\(\theta \approx \arctan\left(\frac{500}{259.81}\right)\)

\(\theta \approx \arctan(1.925)\)

\(\theta \approx 62.5^\circ\) (above the negative x-axis)

Finally, from the positive x-axis, add 180 degrees because it is on the left half of the coordinate system:

\(\theta_{\text{from positive x-axis}} = 180^\circ + \theta\)

\(\theta_{\text{from positive x-axis}} = 180^\circ + 62.5^\circ\)

\(\theta_{\text{from positive x-axis}} = 242.5^\circ\)