Example Question - force magnitude
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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>
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