A cable is loaded and supported as shown in Fig.The vertical distance ve is 4 ft. DetermineThe length L of the cable.8ft-14 ft8 ft5 ftD4 ftУв600 IbCAB1000 lb2000 lb

Answered: A cable is loaded and supported as…

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

A cable is loaded and supported as shown in Fig.
The vertical distance ve is 4 ft. Determine
The length L of the cable.
8ft-
14 ft
8 ft
5 ft
D
4 ft
Ув
600 Ib
C
A
B
1000 lb
2000 lb

Expert Solution

Step 1

Given,

y_c = 4 ft

Step 2

$C D = \sqrt{4^{2} + 8^{2}} = 8.944 f t S i n θ_{1} = \frac{4}{8.944} = 0.447 C o s θ_{1} = \frac{8}{8.944 = 0.894} B C = \sqrt{{(y_{b} - 4)}^{2} + 14^{2}} f t S i n θ_{2} = \frac{y_{b} - 4}{\sqrt{{(y_{b} - 4)}^{2} + 14^{2}}} C o s θ_{2} = \frac{14}{\sqrt{{(y_{b} - 4)}^{2} + 14^{2}}} A B = \sqrt{{(y_{b} - 5)}^{2} + 8^{2}} f t S i n θ_{3} = \frac{y_{b} - 5}{\sqrt{{(y_{b} - 5)}^{2} + 8^{2}}} C o s θ_{3} = \frac{8}{\sqrt{{(y_{b} - 5)}^{2} + 8^{2}}} \tan θ_{3} = \frac{y_{b} - 5}{8}$

Step 3

Considering Horizontal and Vertical Equilibrium at C,

$\sum_{}^{} F_{x} = 0 C D C o s θ_{1} + 600 = B C C o s θ_{2} B C = \frac{1}{C o s θ_{2}} [C D C o s θ_{1} + 600] \sum_{}^{} F_{y} = 0 B C S i n θ_{2} + 1000 = C D S i n θ_{1} C D = \frac{1}{S i n θ_{1}} [\frac{S i n θ_{2}}{C o s θ_{2}} {C D C o s θ_{1} + 600} + 1000] C D = \frac{2237.13 + 1342.282 \tan θ_{2}}{1 - 2 \tan θ_{2}} . . . . (1)$