1 The image presents a problem involving a beam with various loads applied to it. The diagram shows a horizontal beam supported at point A and connected by a cable to point B. The beam experiences five downward forces: \( F_1 = 15 \, \text{lb} \), \( F_2 = 12 \, \text{lb} \), \( F_3 = 13 \, \text{lb} \), \( F_4 = 16 \, \text{lb} \), and \( F_5 = 12 \, \text{lb} \).These forces are evenly distributed, with distances between them as follows:- \( F_1 \) is located 6 inches from the left end of the beam.- \( F_2 \) is 8 inches to the right of \( F_1 \).- \( F_3 \) is 8 inches to the right of \( F_2 \).- \( F_4 \) is 6 inches to the right of \( F_3 \).- \( F_5 \) is 8 inches to the right of \( F_4 \).Point \( B \) is connected vertically by cable \( BC \), which passes over a pulley to point \( C \).Beneath the diagram, there are three fill-in-the-blank fields with dropdown menus to determine:1. The force in cable \( BC \) in pounds.2. The vertical reaction force at \( A \), with options to indicate direction (up, down, or N/A).3. The horizontal reaction force at \( A \), with similar direction options.This exercise is intended to teach how to calculate reaction forces and tension in static equilibrium scenarios involving distributed forces on beams. The loading for a beam is as shown in the figure, where \( F_1 = 15 \, \text{lb} \), \( F_2 = 12 \, \text{lb} \), \( F_3 = 13 \, \text{lb} \), \( F_4 = 16 \, \text{lb} \), and \( F_5 = 12 \, \text{lb} \).Determine the reaction forces at \( A \) and the tension in cable \( BC \).- **Diagram Explanation:** - The beam is horizontal, with forces \( F_1 \) through \( F_5 \) acting vertically downward. - The distances between the forces are: 6 in. between \( F_1 \) and \( F_2 \), 8 in. between \( F_2 \) and \( F_3 \), 6 in. between \( F_3 \) and \( F_4 \), and 8 in. between \( F_4 \) and \( F_5 \). - Point \( A \) is a support, and Point \( B \) is connected to cable \( BC \), which runs to support \( C \).Input the following:- The force in cable \( BC \) is \(\_\_\_\_\_\_\_\_\) lbs tension.- The vertical reaction force at \( A \) is \(\_\_\_\_\_\_\_\_\) lbs \([ \text{click to select}]\).- The horizontal reaction force at \( A \) is \(\_\_\_\_\_\_\_\_\) lbs \([ \text{click to select}]\). - Options for direction: left, right, N/A.

F1 = 15 lbF2 = 12 lbF3 = 13 lbF4=16 lbF5 = 12 lbFrom free body diagramSummation of forces in x-direction is zero,ΣFx = 0Ax= 0 (Horizontal reaction force at A)Summation of forces in y-direction is zero,ΣFy = 0F1+ F2+ F3+ F4+ F5 - Ay + TBC = 015 + 12 + 13 + 16+ 12 - Ay + TBC = 0Ay - TBC = 68 lb. (1)Moment about B= 0ΣMB = 0F1 *(6+8+6+8) + F2 *(8+6+8) + F3 *(6+8) + F4 *8 - Ay *8 = 015*(6+8+6+8) + 12*(8+6+8) + 13*(6+8) + 16*8 - Ay*8 = 0Ay = 124.25 lb ( Vertical reaction force at A) Now from equation (1)Ay - TBC = 68TBC = 68 - 124.25 = - 56.25 lb Force in BC = 56.25 lb (Tension) Vertical reaction force at A = 124.25 lb (up) Horizontal reaction force at A = 0

Engineering

Mechanical Engineering

The loading for a beam is as shown in the figure, where F₁ = 15 lb, F2 =12 lb, F3 = 13 lb, F4 = 16 lb, and F5 = 12 lb. Determine the reaction forces at A and the tension in cable BC. F1 F2 F3 F4 Fs -6 in-8 in.6 in.--8 in- The force in cable BC is The vertical reaction force at A is The horizontal reaction force at A is B lbs tension. lbs (Click to select) (Click to select) up down N/A

The loading for a beam is as shown in the figure, where F₁ = 15 lb, F2 =12 lb, F3 = 13 lb, F4 = 16 lb, and F5 = 12 lb. Determine the reaction forces at A and the tension in cable BC. F1 F2 F3 F4 Fs -6 in-8 in.6 in.--8 in- The force in cable BC is The vertical reaction force at A is The horizontal reaction force at A is B lbs tension. lbs (Click to select) (Click to select) up down N/A

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

The image presents a problem involving a beam with various loads applied to it. The diagram shows a horizontal beam supported at point A and connected by a cable to point B. The beam experiences five downward forces: \( F_1 = 15 \, \text{lb} \), \( F_2 = 12 \, \text{lb} \), \( F_3 = 13 \, \text{lb} \), \( F_4 = 16 \, \text{lb} \), and \( F_5 = 12 \, \text{lb} \).

These forces are evenly distributed, with distances between them as follows:
- \( F_1 \) is located 6 inches from the left end of the beam.
- \( F_2 \) is 8 inches to the right of \( F_1 \).
- \( F_3 \) is 8 inches to the right of \( F_2 \).
- \( F_4 \) is 6 inches to the right of \( F_3 \).
- \( F_5 \) is 8 inches to the right of \( F_4 \).

Point \( B \) is connected vertically by cable \( BC \), which passes over a pulley to point \( C \).

Beneath the diagram, there are three fill-in-the-blank fields with dropdown menus to determine:
1. The force in cable \( BC \) in pounds.
2. The vertical reaction force at \( A \), with options to indicate direction (up, down, or N/A).
3. The horizontal reaction force at \( A \), with similar direction options.

This exercise is intended to teach how to calculate reaction forces and tension in static equilibrium scenarios involving distributed forces on beams.

The loading for a beam is as shown in the figure, where \( F_1 = 15 \, \text{lb} \), \( F_2 = 12 \, \text{lb} \), \( F_3 = 13 \, \text{lb} \), \( F_4 = 16 \, \text{lb} \), and \( F_5 = 12 \, \text{lb} \).

Determine the reaction forces at \( A \) and the tension in cable \( BC \).

- **Diagram Explanation:**
- The beam is horizontal, with forces \( F_1 \) through \( F_5 \) acting vertically downward.
- The distances between the forces are: 6 in. between \( F_1 \) and \( F_2 \), 8 in. between \( F_2 \) and \( F_3 \), 6 in. between \( F_3 \) and \( F_4 \), and 8 in. between \( F_4 \) and \( F_5 \).
- Point \( A \) is a support, and Point \( B \) is connected to cable \( BC \), which runs to support \( C \).

Input the following:

- The force in cable \( BC \) is \(\_\_\_\_\_\_\_\_\) lbs tension.
- The vertical reaction force at \( A \) is \(\_\_\_\_\_\_\_\_\) lbs \([ \text{click to select}]\).
- The horizontal reaction force at \( A \) is \(\_\_\_\_\_\_\_\_\) lbs \([ \text{click to select}]\).
- Options for direction: left, right, N/A.

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