Example Question - resultant force
Here are examples of questions we've helped users solve.
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>
Analysis of Resultant Forces in Various Configurations
<p>For F2-1:</p> <p>Using vector addition, the resultant force \( R \) is the vector sum of the two forces.</p> <p>\( R_x = 6\cos(45^\circ) + 2\cos(45^\circ) \)</p> <p>\( R_x = (6 + 2) \cdot \frac{\sqrt{2}}{2} \)</p> <p>\( R_x = 8 \cdot \frac{\sqrt{2}}{2} \)</p> <p>\( R_x = 4\sqrt{2} \text{ N} \)</p> <p>\( R_y = 6\sin(45^\circ) - 2\sin(45^\circ) \)</p> <p>\( R_y = (6 - 2) \cdot \frac{\sqrt{2}}{2} \)</p> <p>\( R_y = 4 \cdot \frac{\sqrt{2}}{2} \)</p> <p>\( R_y = 2\sqrt{2} \text{ N} \)</p> <p>The magnitude of the resultant force is:</p> <p>\( R = \sqrt{R_x^2 + R_y^2} \)</p> <p>\( R = \sqrt{(4\sqrt{2})^2 + (2\sqrt{2})^2} \)</p> <p>\( R = \sqrt{32 + 8} \)</p> <p>\( R = \sqrt{40} \)</p> <p>\( R = 2\sqrt{10} \text{ N} \)</p> <p>The direction is given by:</p> <p>\( \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) \)</p> <p>\( \theta = \tan^{-1}\left(\frac{2\sqrt{2}}{4\sqrt{2}}\right) \)</p> <p>\( \theta = \tan^{-1}\left(\frac{1}{2}\right) \)</p> <p>The direction is \( \theta \) degrees above the +x-axis.</p> <p>For F2-2:</p> <p>This part of the question corresponds to a different problem about forces acting on a hook.</p> <p>For F2-3:</p> <p>This part of the question corresponds to yet another different problem about determining the magnitude of the resultant force and its direction.</p> <p>Note: Since the question is about F2-1, only the solution for F2-1 is provided, and not for F2-2 or F2-3.</p>
Analysis of Resultant Forces in Different Configurations
For problem F2-1: <p>The two forces can be added using vector addition. Let's break them down into their x and y components. \(F_{1x} = 6 \cos(45^\circ), F_{1y} = 6 \sin(45^\circ)\) \(F_{2x} = -2 \sin(45^\circ), F_{2y} = 2 \cos(45^\circ)\) \(R_x = F_{1x} + F_{2x}\) \(R_y = F_{1y} + F_{2y}\)</p> <p>The magnitude of the resultant force \(R\) can be found using Pythagoras' theorem: \(R = \sqrt{R_x^2 + R_y^2}\) The direction \(\theta\) relative to the x-axis is: \(\theta = \arctan\left(\frac{R_y}{R_x}\right)\)</p> For problem F2-2: <p>To find the resultant force acting on the hook, we add the two forces vectorially. \(F_{1x} = 200 \cos(30^\circ), F_{1y} = 200 \sin(30^\circ)\) \(F_{2x} = -500 \sin(60^\circ), F_{2y} = 500 \cos(60^\circ)\) \(R_x = F_{1x} + F_{2x}\) \(R_y = F_{1y} + F_{2y}\)</p> <p>Thus, the magnitude of the resultant force \(R\) is: \(R = \sqrt{R_x^2 + R_y^2}\)</p> For problem F2-3: <p>The resultant force is the vector sum of the three forces. \(F_{1x} = 800 \cos(30^\circ), F_{1y} = 800 \sin(30^\circ)\) \(F_{2x} = -600\) \(F_{3x} = -600 \cos(45^\circ), F_{3y} = -600 \sin(45^\circ)\) \(R_x = F_{1x} + F_{2x} + F_{3x}\) \(R_y = F_{1y} + F_{3y}\)</p> <p>The magnitude of the resultant force \(R\) is: \(R = \sqrt{R_x^2 + R_y^2}\) And the direction \(\theta\) measured counterclockwise from the positive x-axis is: \(\theta = \arctan\left(\frac{R_y}{R_x}\right)\)</p>
Determining the Resultant Force in Various Configurations
For Prob. F2-1: <p>\( R_x = 6\cos(45^\circ) + 2\sin(45^\circ) \)</p> <p>\( R_y = 6\sin(45^\circ) - 2\cos(45^\circ) \)</p> <p>\( R = \sqrt{R_x^2 + R_y^2} \)</p> <p>\( \theta = \arctan\left(\frac{R_y}{R_x}\right) \)</p> For Prob. F2-2: <p>\( R_x = 500\cos(30^\circ) - 200\sin(60^\circ) \)</p> <p>\( R_y = 500\sin(30^\circ) + 200\cos(60^\circ) \)</p> <p>\( R = \sqrt{R_x^2 + R_y^2} \)</p> For Prob. F2-3: <p>\( R_x = 800\cos(30^\circ) - 600\sin(45^\circ) \)</p> <p>\( R_y = 800\sin(30^\circ) + 600\cos(45^\circ) \)</p> <p>\( R = \sqrt{R_x^2 + R_y^2} \)</p> <p>\( \theta = \arctan\left(\frac{R_y}{R_x}\right) \)</p>
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com