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KEØM
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Homework Statement
A slender rod AB of weight W is attached to blocks A and B which move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when [tex]\theta[/tex] = 0. (a) Neglecting the weight of the blocks, derive an equation in W, k, l, and [tex]\theta[/tex] which must be satisfied when the rod is in equilibrium. (b) Determine the value of [tex]\theta[/tex] when W = 4.5 lb, l = 30 in., and k = 1.8 lb/ft.
Here is a picture of the problem. It is number 4.60
"[URL title="download file from Jumala Files"]http://jumalafiles.info/showfile2-15016951829111680748225395249314210/problem460.pdf [/URL]
Homework Equations
[tex]\Sigma F_{x} = 0[/tex]
[tex]\Sigma F_{y} = 0[/tex]
[tex]\Sigma M_{B} = 0[/tex]
The Attempt at a Solution
[tex]F_{s} = ks = lsin(\theta)[/tex]
[tex]\Sigma M_{B} = lsin(\theta)(klsin(\theta) - lcos(\theta)A_{y} + W(l/2cos)(\theta) = 0[/tex] where [tex]A_{y}[/tex] is the reaction force at A.
[tex]\Sigma F_{y} = A_{y} - W = 0[/tex]
[tex]\Sigma M_{B} = lsin(\theta)(klsin(\theta) - lcos(\theta)W + W(l/2cos)(\theta) = 0[/tex]
[tex]\Sigma M_{B} = l^2sin^2(\theta)k - (l/2)cos(\theta)W = 0[/tex]
[tex] \frac{2lk}{W} = \frac{cos(\theta)}{sin^2(\theta)} [/tex]
After inverting both sides and using the identity [tex]sin^2(\theta) + cos^2(\theta) = 1[/tex]
I get for an answer:
[tex]sec(\theta) - cos(\theta) = \frac{W}{2lk}[/tex]
The book's answer is [tex]tan(\theta) - sin(\theta) = \frac{W}{2lk}[/tex]
Can someone tell me where I went wrong?
Thanks in advance,
KEØM
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