Example Question - direction
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Determining the Resultant Force and its Direction
<p>Given:</p> <p>\(\vec{F_1} = 800 \text{ N} \ (\text{upwards along the y-axis})\)</p> <p>\(\vec{F_2} = 600 \text{ N} \ (\text{30 degrees below the -x-axis})\)</p> <p>To solve for the resultant force \(\vec{R}\), break \(\vec{F_2}\) into its x and y components.</p> <p>\(F_{2x} = 600 \cos(30^\circ)\)</p> <p>\(F_{2y} = -600 \sin(30^\circ)\) (The y-component is negative because it is downward.)</p> <p>Calculate \(F_{2x}\) and \(F_{2y}\):</p> <p>\(F_{2x} = 600 \times \frac{\sqrt{3}}{2} = 300\sqrt{3}\) N</p> <p>\(F_{2y} = -600 \times \frac{1}{2} = -300\) N</p> <p>The resultant force \(\vec{R}\) in the x and y components is:</p> <p>\(R_x = F_{2x}\) because there is no x-component for \(\vec{F_1}\).</p> <p>\(R_y = F_{1} + F_{2y}\)</p> <p>Calculate \(R_x\) and \(R_y\):</p> <p>\(R_x = 300\sqrt{3}\) N</p> <p>\(R_y = 800 - 300 = 500\) N</p> <p>Now calculate the magnitude of the resultant vector \(|\vec{R}|\):</p> <p>\(|\vec{R}| = \sqrt{R_x^2 + R_y^2}\)</p> <p>\(|\vec{R}| = \sqrt{(300\sqrt{3})^2 + (500)^2}\)</p> <p>\(|\vec{R}| = \sqrt{(900 \times 3) + (250000)}\)</p> <p>\(|\vec{R}| = \sqrt{2700 + 250000}\)</p> <p>\(|\vec{R}| = \sqrt{252700}\)</p> <p>\(|\vec{R}| \approx 502.7\) N</p> <p>To find the direction θ measured counterclockwise from the positive x-axis:</p> <p>\(\tan(\theta) = \frac{R_y}{R_x}\)</p> <p>\(\theta = \arctan\left(\frac{500}{300\sqrt{3}}\right)\)</p> <p>\(\theta = \arctan\left(\frac{500}{300 \times \frac{\sqrt{3}}{2}}\right)\)</p> <p>\(\theta = \arctan\left(\frac{500}{150\sqrt{3}}\right)\)</p> <p>\(\theta \approx \arctan\left(\frac{500}{259.81}\right)\)</p> <p>\(\theta \approx \arctan(1.925)\)</p> <p>\(\theta \approx 62.5^\circ\) (above the negative x-axis)</p> <p>Finally, from the positive x-axis, add 180 degrees because it is on the left half of the coordinate system:</p> <p>\(\theta_{\text{from positive x-axis}} = 180^\circ + \theta\)</p> <p>\(\theta_{\text{from positive x-axis}} = 180^\circ + 62.5^\circ\)</p> <p>\(\theta_{\text{from positive x-axis}} = 242.5^\circ\)</p>
Determining the Resultant Force and its Direction
<p>To find the magnitude of the resultant force, we can break the forces into their x and y components. For the 600 N force:</p> <p>\[ F_{x1} = 600 \cos(30^\circ) \]</p> <p>\[ F_{y1} = 600 \sin(30^\circ) \]</p> <p>For the 800 N force, which is purely in the y-direction:</p> <p>\[ F_{x2} = 0 \]</p> <p>\[ F_{y2} = 800 \]</p> <p>Now, add the x-components and y-components of the forces to find the resultant:</p> <p>\[ R_x = F_{x1} + F_{x2} \]</p> <p>\[ R_y = F_{y1} + F_{y2} \]</p> <p>Substitute the values to find \( R_x \) and \( R_y \):</p> <p>\[ R_x = 600 \cos(30^\circ) \]</p> <p>\[ R_y = 600 \sin(30^\circ) + 800 \]</p> <p>Use the Pythagorean theorem to find the magnitude of the resultant force (R):</p> <p>\[ R = \sqrt{R_x^2 + R_y^2} \]</p> <p>Calculate the angle (\( \theta \)) from the positive x-axis:</p> <p>\[ \theta = \arctan\left(\frac{R_y}{R_x}\right) \]</p> <p>Solve for \( R \) and \( \theta \) to get the magnitude and direction of the resultant force.</p>
Determination of Resultant Force Magnitude and Direction
For Prob. F2-1: <p>\(R_{x} = 6 \cos(45^\circ)N - 2N\)</p> <p>\(R_{x} = 4.2426N - 2N\)</p> <p>\(R_{x} = 2.2426N\)</p> <p>\(R_{y} = 6 \sin(45^\circ)N\)</p> <p>\(R_{y} = 4.2426N\)</p> <p>\(R = \sqrt{R_{x}^2 + R_{y}^2}\)</p> <p>\(R = \sqrt{2.2426^2 + 4.2426^2}N\)</p> <p>\(R = \sqrt{5.0291 + 18.0000}N\)</p> <p>\(R = \sqrt{23.0291}N\)</p> <p>\(R \approx 4.8001N\)</p> <p>\(\theta = \arctan\left(\frac{R_{y}}{R_{x}}\right)\)</p> <p>\(\theta = \arctan\left(\frac{4.2426}{2.2426}\right)\)</p> <p>\(\theta \approx 62.1021^\circ\) clockwise from the +x-axis</p> For Prob. F2-2: Isolated from the image and omitted based on the provided instructions. For Prob. F2-3: <p>\(R_{x} = 800 \cos(30^\circ)N - 600 \cos(45^\circ)N\)</p> <p>\(R_{x} = 692.82N - 424.26N\)</p> <p>\(R_{x} = 268.56N\)</p> <p>\(R_{y} = 800 \sin(30^\circ)N + 600 \sin(45^\circ)N\)</p> <p>\(R_{y} = 400N + 424.26N\)</p> <p>\(R_{y} = 824.26N\)</p> <p>\(R = \sqrt{R_{x}^2 + R_{y}^2}\)</p> <p>\(R = \sqrt{268.56^2 + 824.26^2}N\)</p> <p>\(R = \sqrt{72105.59 + 679186.67}N\)</p> <p>\(R = \sqrt{751292.27}N\)</p> <p>\(R \approx 866.77N\)</p> <p>\(\theta = \arctan\left(\frac{R_{y}}{R_{x}}\right)\)</p> <p>\(\theta = \arctan\left(\frac{824.26}{268.56}\right)\)</p> <p>\(\theta \approx 71.5651^\circ\) counterclockwise from the +x-axis</p>
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