Please help When y is zero, the springs sustain a force of 60 lb. The applied vertical forces F and -F pull the point A away from B a distanceof y=2 ft. The cords CAD and CBD are attached to the rings at C and D. Draw the free-body diagrams for point A and ring C.k= 40 lb/ftiii.For C, they are caused by:i.2 ftC 2 ftB-F-2 ft2 ft OSolution1. Imagine A and C to be separated or detached from the system.2. Each of A and C is subjected to three external forces. For A, they are caused by:i.ii.k = 40 lb/ft3. Draw the free-body diagrams of A and C showing all these forces labeled with their magnitudes and directions. You shouldalso include any other available information e.g. lengths, angles etc. - which will help when formulating the equilibriumequations.. A

At y=0, The horizontal stretch is as follows-Now when y=2 ( the given condition), the horizontal…

Answered: of y= 2 ft. The cords CAD and CBD are…

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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Please help

When y is zero, the springs sustain a force of 60 lb. The applied vertical forces F and -F pull the point A away from B a distance
of y=2 ft. The cords CAD and CBD are attached to the rings at C and D. Draw the free-body diagrams for point A and ring C.
k= 40 lb/ft
iii.
For C, they are caused by:
i.
2 ft
C 2 ft
B
-F
-2 ft
2 ft O
Solution
1. Imagine A and C to be separated or detached from the system.
2. Each of A and C is subjected to three external forces. For A, they are caused by:
i.
ii.
k = 40 lb/ft
3. Draw the free-body diagrams of A and C showing all these forces labeled with their magnitudes and directions. You should
also include any other available information e.g. lengths, angles etc. - which will help when formulating the equilibrium
equations.
. A

Expert Solution

Step 1: Determine the distance as follows-

At y=0, The horizontal stretch is as follows-

$x equals F over e subscript e q end subscript equals fraction numerator 2 F over denominator k end fraction equals fraction numerator 2 cross times 60 over denominator 40 end fraction equals 3 space ft$

Now when y=2 ( the given condition), the horizontal distance becomes-

$table row cell sin 60 end cell equals cell fraction numerator O D over denominator 2 end fraction end cell row cell O D end cell equals cell 1.73 space ft end cell row blank blank cell Thus comma end cell row cell C D end cell equals cell 2 open parentheses O D close parentheses end cell row blank equals cell 3.46 space ft end cell row blank blank blank end table$

Here, O is at the center point.

Now net shift of length, when the external load is applied is as follows-

$table row cell increment x end cell equals cell 3.46 minus 3 end cell row blank equals cell 0.46 space ft end cell row cell open parentheses increment straight x close parentheses subscript each end cell equals cell fraction numerator 0.46 space ft over denominator 2 end fraction end cell row blank equals cell 0.23 space ft end cell row blank blank blank end table$

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