Question - Analysis of Resultant Forces in Different Configurations

The two forces can be added using vector addition. Let's break them down into their x and y components. \(F_{1x} = 6 \cos(45^\circ), F_{1y} = 6 \sin(45^\circ)\) \(F_{2x} = -2 \sin(45^\circ), F_{2y} = 2 \cos(45^\circ)\) \(R_x = F_{1x} + F_{2x}\) \(R_y = F_{1y} + F_{2y}\)

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: \(\theta = \arctan\left(\frac{R_y}{R_x}\right)\)

To find the resultant force acting on the hook, we add the two forces vectorially. \(F_{1x} = 200 \cos(30^\circ), F_{1y} = 200 \sin(30^\circ)\) \(F_{2x} = -500 \sin(60^\circ), F_{2y} = 500 \cos(60^\circ)\) \(R_x = F_{1x} + F_{2x}\) \(R_y = F_{1y} + F_{2y}\)

Thus, the magnitude of the resultant force \(R\) is: \(R = \sqrt{R_x^2 + R_y^2}\)

The resultant force is the vector sum of the three forces. \(F_{1x} = 800 \cos(30^\circ), F_{1y} = 800 \sin(30^\circ)\) \(F_{2x} = -600\) \(F_{3x} = -600 \cos(45^\circ), F_{3y} = -600 \sin(45^\circ)\) \(R_x = F_{1x} + F_{2x} + F_{3x}\) \(R_y = F_{1y} + F_{3y}\)

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: \(\theta = \arctan\left(\frac{R_y}{R_x}\right)\)