Help with pulley problem with sag

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SUMMARY

The discussion focuses on calculating the force in cord ABC when a 1.5 kg block is suspended from pulley B with a sag of 0.15 m. The angle formed by the sagging cord is determined to be 53.3 degrees using the tangent function. The user successfully resolves the forces in the x and y directions, leading to the conclusion that the total tension in the cord is 12.3 N. The confusion regarding the diagram's proportions and the pulley’s position is clarified, ensuring a better understanding of the problem.

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mike41
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Homework Statement



F3-3.jpg


If the 1.5 kg block is suspended from the pulley B and the sag of the cord is d = 0.15m, determine the force in cord ABC. Neglect the size of the pulley


The Attempt at a Solution



They don't even give an angle, I am not sure how I am supposed to do this.
 
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Draw a perpendicular from BD on AC. That bisects AC.
In triage BDC you can all the three sides.So fond the angles.
 
so for the angle i got 53.3 degrees. Tan^-1(0.2/0.15) = 53.3 degrees

Then i made components

Force x y
AB -ABsin53.3 ABcos53.3
CB CBsin53.3 CBcos53.3
14.715N 0 -14.715


X axis = 0 N

CBsin53.3 - ABsin53.3= 0

y axis = 17.715 N

CBcos53.3 + ABcos53.3 = 14.715 N


so then i substituted for CB = ([14.715 / cos53.3] - AB)
= 24.62

and got 12.3 as the tension total. thankyou
i was confused by how the lines in the diagram are different sizes and did not think that the pulley would be in the center with a diagram like that
 

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