### Problem Statement:Determine the magnitude of the resultant force acting on the corbel and its direction \( \theta \) measured counterclockwise from the x-axis.### Explanation of Diagram:The diagram above illustrates three forces acting at a point, represented as follows:- \( F_1 = 700 \) lb: This force is applied at an angle of 30 degrees above the positive x-axis.- \( F_2 = 400 \) lb: This force is applied vertically upward along the y-axis.- \( F_3 = 600 \) lb: This force is applied in the third quadrant. The direction forms a right triangle with sides 3, 4, and 5, indicating it makes an angle with the x-axis. The exact angle is not specified but can be calculated using the arctangent function or by recognizing the 3-4-5 triangle's properties.### Steps to Solve:1. **Resolve Each Force into Components:** - \( F_1 \) has both x and y components. - \( F_2 \) is along the y-axis. - \( F_3 \) should be broken into x and y components using the 3-4-5 rule.2. **Sum the X and Y Components Separately:** - \( \Sigma F_x = F_{1x} + F_{2x} + F_{3x} \) - \( \Sigma F_y = F_{1y} + F_{2y} + F_{3y} \)3. **Calculate the Magnitude of the Resultant Force:** - \( F_R = \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2} \)4. **Determine the Direction \( \theta \):** - \( \theta = \tan^{-1}\left(\frac{\Sigma F_y}{\Sigma F_x}\right) \)### Example Calculation:1. Resolve \( F_1 \) into components: - \( F_{1x} = 700 \, \text{lb} \cdot \cos(30^\circ) = 700 \, \text{lb} \cdot \approx 0.866 \approx 606.2 \, \text{lb} \) - \( F_{1y} = 700 \, \text{lb} \cdot \

break force into components to get the resultant.

Answered: F2=400L6 F3=60016 Fi=70olb 30°…

F2=400L6 F3=60016 Fi=70olb 30° Detremine the magnitude of the resultant force acting on the corbel and its direction 0 measured counterclockwise from the x axis.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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