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posted by  obscur on 9/11/2008 6:19:31 PM  |  status: Live  

statics

Course Textbook Chapter Problem
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Question Details:
The 250 kg block is supported by linear spring BC and the flexible cable system,
Determine:
a) the tension in cable AC in Newtons
b) the tension in cable CE in Newtons
c) the tension in cable DE in Newtons
d) the spring constant in N/m of spring BC if its unstretched length is equal to distance AB in meters





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posted by jayborre on 9/12/2008 2:18:02 PM  |  status: Live
Asker's Rating: Lifesaver   
obscur's comment:
"thank you very much"
Response Details:
Isolate Joint E and C


ΣFV = 0]
-250 + EC sin 60 = 0
EC = 288.675 kg (tension in cable CE)

ΣFH = 0]
EC cos 60 - ED = 0
288.675 cos 60 - ED = 0
ED = 144.338 kg (tension in cable DE)




ΣFH = 0]
CB cos 30 - CA cos 40 - 288.675 sin 30 = 0
0.866 CB - 0.766 CA- 144.338 = 0       -------------> equation 1

ΣFV = 0]
CA sin 40 + CB sin 30 - 288.675 cos 30 = 0
0.5 CB + 0.643 CA - 245 = 0   ---------------------> equation 2

Solving 2 equations with 2 unknowns, it will yield:
CB = 298.434 kg (Tension in cable BC)
CA = 148.963 kg (Tension in cable AC)


To calculate the constant spring k:
According to Hooke's Law, the Force that elongates the spring is directly proportional to its deformation, we insert a constant k to eliminate the proportionality sign

F = kx

298.434 kg = k (1 m)

Therefore:       k = 298.434 kg/m

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 9/16/2008 3:05:13 AM  |  status: Live
Asker's Rating: Helpful   
obscur's comment:
"thanks again"
Response Details:
Good day obscur!
Regarding with ypur query about the change of the unstretched length of the spring from 1m to 0.46m, the same force will be applied on the spring and that is still be equal to 298.434kg, hence, only the spring constant k will be change. Since the unstretched length becomes smaller, then the constant spring k will yield a larger value.

Again, to calculate the constant spring k:
According to Hooke's Law, the Force that elongates the spring is directly proportional to its deformation, we insert a constant k to eliminate the proportionality sign

F = kx

298.434 kg = k (0.46 m)

Therefore:       k = 648.77 kg/m


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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