Problem 1: Find the reactions at O for the beam below.B100 lbf10 in10 in 10 in10 in

Answered: B 100 lbf 10 in AC - 10 in- - 10 in- 10…

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

Problem 1: Find the reactions at O for the beam below.
B
100 lbf
10 in
10 in 10 in
10 in

Expert Solution

Step 1:-Given data

$(P) = (100 l b f)$

Step 2:-Finding the reaction force at point O

The beam is in equilibrium so we can apply the equilibrium condition.
Applying the equilibrium condition.

$\sum_{n} F_{Y} = 0;$

$(R_{O y}) + (T \times \sin 45) - (100 l b_{f}) = 0; (R_{O y}) + (T \times \sin 45) = (100 l b_{f}); . . . . . . . . . . . . . . . . . . . . . . . (1)$

$\sum_{n} F_{X} = 0;$

$(R_{O X}) - (T \times \cos (45)) = 0; (R_{O X}) = (T \times \cos (45)); . . . . . . . . . . . . . . . . . . . . . . (2)$

$\sum_{n} M = 0$

$((T \times \sin (45)) \times (10 i n)) - (100 l b f \times (30 i n)) = 0; ((T \times \sin (45)) \times (10 i n)) = (100 l b f \times (30 i n)); (T) = (\frac{(30 i n)}{(10 i n)}) \times (\frac{(100 l b f)}{(\sin (45))}) (T) = (3) \times (\frac{(100 l b f)}{(0.7071068)}) (T) = (3) \times (141.4213) (T) = (424.2641) l b f . . . . . . . . . . . . . . . . . . . . . . . . (3)$

Putting equation (3) in equation (1) ;

$(R_{O y}) + (T \times \sin 45) = (100 l b f) (R_{O y}) = [(100 l b f) - (T \times \sin 45)] (R_{O y}) = [(100 l b f) - (300)] t h a t (R_{O y}) = [(- 200 l b f)] \to N e g a t i v e s i g n r e p r e s e n t t h a t i t' s m a g n i t u d e i s o p p o s i t e t o i t s a s s u m p t i o n .$

Putting equation (3) in equation (2) ;

$(R_{O X}) = (T \times \cos (45)) (R_{O X}) = (300 l b f)$

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