**Equation 1:**\[\vec{F}_R = (277 \, \hat{\imath} + \, \text{{[missing value]}} \, \hat{\jmath} + \, \text{{[missing value]}} \, \hat{k}) \, \text{{N}}\]**Explanation:**- This equation represents a vector force \(\vec{F}_R\) in a 3D coordinate system.- The components of the vector are specified along the \( \hat{\imath} \), \( \hat{\jmath} \), and \( \hat{k} \) directions.- The known component is \(277\) in the \( \hat{\imath} \) direction.- Two components are missing for the \( \hat{\jmath} \) and \( \hat{k} \) directions.---**Equation 2:**\[(\vec{M}_R)_O = (\text{{[missing value]}} \, \hat{\imath} + \, \text{{[missing value]}} \, \hat{\jmath} + \, \text{{[missing value]}} \, \hat{k}) \, \text{{N-m}}\]**Explanation:**- This equation describes a moment or torque vector \((\vec{M}_R)_O\) based around a point \(O\).- The vector components are represented in the \( \hat{\imath} \), \( \hat{\jmath} \), and \( \hat{k} \) directions as well.- All components are missing and need to be provided or calculated.These representations are commonly used in physics and engineering to describe forces and moments in vector form. **Title: Analysis of Force Systems on a Beam****Introduction:**This section covers the application of forces on a beam, specifically focusing on determining the equivalent force and moment for a given system of forces.**Force Details:**Three forces are applied to the beam:- \( \vec{F_2} = (-250 \hat{i} + 650 \hat{j} - 300 \hat{k}) \, \text{N} \)**Diagram Explanation:**The 3D diagram depicts a beam oriented along the x, y, and z axes, where the origin is labeled as point \( O \). The forces \( F_1 \), \( F_2 \), and \( F_3 \) are applied at various points along the beam. The distances \( d_1 \), \( d_2 \), and \( d_3 \) represent the positioning of the forces along and between the x, y, and z axes.**Values for the Diagram:**The given values in the table correspond to force magnitudes and distance measurements:- \( F_1 = 300 \, \text{N} \)- \( F_3 = 350 \, \text{N} \)- \( d_1 = 2 \, \text{m} \)- \( d_2 = 2.5 \, \text{m} \)- \( d_3 = 0.5 \, \text{m} \)**Tasks:**a. **Equivalent Force:** Replace the force system with an equivalent force, \( \vec{F_R} \), and express it as a Cartesian vector.b. **Equivalent Couple Moment:** Replace the force system with an equivalent couple moment acting at point \( O \), \( (\vec{M_R})_O \), and express it as a Cartesian vector.**Instructions:**- Round your final answers to three significant digits/figures.

Given:Required:Resultant force in Cartesian vector formEquivalent resultant moment at O in Cartesian…

Three forces are applied to the beam. F₂ = (-250 + 650 + -300 ) N. Variable F₁ F3 d₁ d₂ d3 Value Z 300 N 350 N 2 m 2.5 m 0.5 m F2 d3. d2 d1 F₁ Values for the figure are given in the following table. Note the figure may not be to scale. F3 У → a. Replace the force system with an equivalent force, FR, express as a cartesian vector. b. Replace the force system with an equivalent couple → moment acting at point 0, (MR)o, express as a cartesian vector.

Three forces are applied to the beam. F₂ = (-250 + 650 + -300 ) N. Variable F₁ F3 d₁ d₂ d3 Value Z 300 N 350 N 2 m 2.5 m 0.5 m F2 d3. d2 d1 F₁ Values for the figure are given in the following table. Note the figure may not be to scale. F3 У → a. Replace the force system with an equivalent force, FR, express as a cartesian vector. b. Replace the force system with an equivalent couple → moment acting at point 0, (MR)o, express as a cartesian vector.

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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