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posted by  obscur on 9/14/2008 4:52:26 PM  |  status: Live  

statics

Course Textbook Chapter Problem
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Question Details:


The 75 kg traffic light is supported by three cables, Determine:
a) the tension in cable A in newtons
b) the tension in cable B in newtons
c) the tension in cable C in newtons
d) the direction angle θx of cable OC in degrees
e) the direction angle θx of cable OA in degrees
f) the direction angle θx of cable OB in degrees
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posted by jayborre on 9/15/2008 8:04:34 AM  |  status: Live
Asker's Rating: Lifesaver   
obscur's comment:
"thanks, I was having a real problem with this one"
Response Details:


Tabulate the forces and their vectorial distances

FORCES

x

y

z

d

TOA

-8

5

4

10.25

TOB

-8

5

-6

11.18

TOC

8

5

0

9.43

ΣMAB = 0]
735.75 (8) - Cy (16) = 0
Cy = 367.875 N

ΣMAx = 0]

735.75 (4) - By (10) - Cy (4) = 0
735.75 (4) - By (10) - 367.875 (4) = 0
By = 147.15 N

ΣFy = 0
Ay + By + Cy - 735.75 = 0
Ay + 147.15 + 367.875 - 735.75 = 0
Ay = 220.725 N

By Proportion:



So:

(tension on cable OA)

(tension on cable OB)

(tension on cable OC)

Calculate unit vector:




Calculate R:



Finally, calculating the angles
(θx of force OA)
(θx of force OB)
  (θx of force OC)
If you still have some questions regarding this topic, Please don't hesitate to ask me. Also, please don't forget to rate me. Thank you and GOD bless



JAY ROBIN B. BORRE
BS Civil Engineering
Technological Institute of the Philippines - Manila
jayborre@yahoo.com
0917-9137085
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posted by jayborre on 9/17/2008 5:08:00 AM  |  status: Live
Asker's Rating: Helpful   
obscur's comment:
"thanks again"
Response Details:
Good day obscur,

It is ok that you keep on asking questions, it only shows that you are also solving the problem on your own rather than just copying on the answers that the cramster people provided.

This are the numbers highlighted in red in your query regarding the street light problem:

ex. 735.75 (8) -Cy (16) = 0
      735.75 (4) - By (10) -Cy (4) = 0

Anyway, the red marked numbers are the distances of the forces that we are rotating about a point where we take moment. Remember that moment of a force is equal to the force multiplied by its perpendicular distance with reference to the point you're taking into account.

Likewise, the Uoa, Uob and Uoc are the unit vectors of each of the forces. To get the R, you should multiply the calculated forces by its units vectors.

Forces in three dimension really is a confusing topic specially if this is the first time you encounter it, again, just like the saying goes, practice makes perfect, so keep on practicing, and in no time, you will find this topic very easy and enjoying.... Goodluck and GOD bless...
If you still have some questions regarding this topic, Please don't hesitate to ask me. Also, please don't forget to rate me. Thank you and GOD bless



JAY ROBIN B. BORRE
BS Civil Engineering
Technological Institute of the Philippines - Manila
jayborre@yahoo.com
0917-9137085
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Karma Points: 312
posted by Juan Colorado on 9/17/2008 3:47:49 PM  |  status: Live
Asker's Rating: Helpful   
obscur's comment:
"thank you"
Response Details:
You can indicate the tension in cables like this:
TOA = TOA( (-8, 5, 4)/ 10.25 ) =  (- 0.780i + 0.488j + 0.390 k ) TOA
TOB = TOB( (-8, 5, -6)/ 11.18 ) =  (- 0.715i + 0.447j - 0.537 k ) TOB
TOC = TOC( ( 8, 5, 0) / 11.18 )  =  ( 0.848i + 0.530j + 0.000 k ) TOC
And obtain the following ecuations:
 
ΣFx=0:
-0.780TOA - 0.715TOB + 0.848TOC      = 0
ΣFy=0:
+0.488TOA + 0.447TOB + 0.530TOC  - 736N  = 0
ΣFz=0:
+ 0.390TOA - 0.537TOB                          = 0
 
But the solution of this system is different... let me review...
 
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posted by Juan Colorado on 9/17/2008 4:01:25 PM  |  status: Live
Asker's Rating: Helpful