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Find the moment about 0 B с fft/ sft 1+1 3ft, A loft 100lbs 100165 0

Find the moment about 0 B с fft/ sft 1+1 3ft, A loft 100lbs 100165 0

Elements Of Electromagnetics
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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Title: Calculating Moments About a Point --- ## Problem Statement Find the moment about point O. ## Diagram Explanation The diagram depicts a horizontal beam with several key points (C, B, A) and forces applied at these points. Here is a detailed description: - The point \( O \) is the pivot point, located at the right end of the beam. - Point \( A \) is located 10 feet to the left of point \( O \). - Point \( B \) is located 3 feet to the left of point \( A \). - Point \( C \) is located 5 feet to the left of point \( B \). - There are two downward forces (100 lbs each): - One force is applied at point \( B \). - Another force is applied at point \( C \). ## Calculations To find the moment about point \( O \), we use the formula for the moment \( M \): \[ M = F \times d \] where \( F \) is the force applied and \( d \) is the perpendicular distance from the point \( O \) to the line of action of the force. ### Moment due to the force at C - Force \( F_C \) = 100 lbs - Distance from \( O \) to \( C \) = Distance from \( O \) to \( A \) + Distance from \( A \) to \( B \) + Distance from \( B \) to \( C \) \[ d_C = 10 \text{ ft} + 3 \text{ ft} + 5 \text{ ft} = 18 \text{ ft} \] \[ M_C = F_C \times d_C = 100 \text{ lbs} \times 18 \text{ ft} = 1800 \text{ lb-ft} \] ### Moment due to the force at B - Force \( F_B \) = 100 lbs - Distance from \( O \) to \( B \) = Distance from \( O \) to \( A \) + Distance from \( A \) to \( B \) \[ d_B = 10 \text{ ft} + 3 \text{ ft} = 13 \text{ ft} \] \[ M_B = F_B \times d_B = 100 \text{ lbs} \times 13 \text{ ft} = 130
Transcribed Image Text:Title: Calculating Moments About a Point --- ## Problem Statement Find the moment about point O. ## Diagram Explanation The diagram depicts a horizontal beam with several key points (C, B, A) and forces applied at these points. Here is a detailed description: - The point \( O \) is the pivot point, located at the right end of the beam. - Point \( A \) is located 10 feet to the left of point \( O \). - Point \( B \) is located 3 feet to the left of point \( A \). - Point \( C \) is located 5 feet to the left of point \( B \). - There are two downward forces (100 lbs each): - One force is applied at point \( B \). - Another force is applied at point \( C \). ## Calculations To find the moment about point \( O \), we use the formula for the moment \( M \): \[ M = F \times d \] where \( F \) is the force applied and \( d \) is the perpendicular distance from the point \( O \) to the line of action of the force. ### Moment due to the force at C - Force \( F_C \) = 100 lbs - Distance from \( O \) to \( C \) = Distance from \( O \) to \( A \) + Distance from \( A \) to \( B \) + Distance from \( B \) to \( C \) \[ d_C = 10 \text{ ft} + 3 \text{ ft} + 5 \text{ ft} = 18 \text{ ft} \] \[ M_C = F_C \times d_C = 100 \text{ lbs} \times 18 \text{ ft} = 1800 \text{ lb-ft} \] ### Moment due to the force at B - Force \( F_B \) = 100 lbs - Distance from \( O \) to \( B \) = Distance from \( O \) to \( A \) + Distance from \( A \) to \( B \) \[ d_B = 10 \text{ ft} + 3 \text{ ft} = 13 \text{ ft} \] \[ M_B = F_B \times d_B = 100 \text{ lbs} \times 13 \text{ ft} = 130
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