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Old Oct 13th 2010, 09:55 PM   #1
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Difficult equilibrium problem

1. The problem statement, all variables and given/known data
A vertical force P=10 N is applied to the ends of the 0.2 m cord AB and spring AC. If the spring has an unstretched length of 0.2 m. determine the angle theta for equilibrium. Take k = 150 N/m.

[IMG]http://i49.photobucket.com/albums/f272/dranseth/****.png?t=1287031089[/IMG]

2. Relevant equations

Fspring=ks
Fy=0
Fx=0

3. The attempt at a solution

Any of my attempts at this solution have resulted in pages and pages of work. I really am not seeing a short cut as to find the angle theta. Thanks for your help!
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Old Oct 14th 2010, 08:36 AM   #2
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Your image is not appearing and the link leads to nowhere near the image... Could you try to have another link (look for "Link for message boards")?
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Old Oct 15th 2010, 02:44 PM   #3
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Originally Posted by Unknown008 View Post
Your image is not appearing and the link leads to nowhere near the image... Could you try to have another link (look for "Link for message boards")?
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Old Oct 16th 2010, 12:50 AM   #4
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Hm... let me clarify, that's it?



I can't find any shortcut.

The length of the spring is given by l + e, where e = F/k, e = F/150

So, length of spring = (0.2 + F/150) m

Let Tc be the tension in the cord and Ts be the tension in the spring. (I'll use B as the angle instead)

Tc sinB + Ts sinC = 10

where C is the angle BCA.

Tc cosB = Ts cosC

From the cosine rule, we get:

AC^2 = BC^2 + AB^2 - 2(AB)(BC)cos B

(0.2 + F/150)^2 = 0.4^2 + 0.2^2 - 2(0.4)(0.2)cos B

(0.2 + F/150)^2 = 0.2 - 0.16cos B

Now, F = Ts

Using the first two equations, we get:

((Ts cosC)/cosB)sinB + Ts sinC = 10

Ts cosC tanB + Ts sinC = 10

Ts = 10/(cosC tanB + sinC)

Hence,

[0.2 + 10/{150(cosC tanB + sinC)}]^2 = 0.2 - 0.16cos B

Hm... not very 'pretty'. Let's try the sine rule.

[0.2 + 10/{150(cosC tanB + sinC)}/sinB = 0.2/sinC = 0.4/sinA

[0.2 + 10/{150(cosC tanB + sinC)} = (0.2 sinB)/sinC

So,

0.2 - 0.16cos B = ((0.2 sinB)/sinC)^2

0.2 - 0.16cos B = (0.04 sin^2B)/sin^2C

5 - 4cos B = sin^2B/sin^2C

We have:
5 - 4cos B = sin^2B/sin^2C
0.2/sinC = 0.4/sinA
A + B + C = 180

Hmm....

Maybe trial and error?
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