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Solution
Solution
1/ \alpha \frac{dt}{dz}[br]=1/ \alpha (\frac{h\alpha}{k}e^{\alpha y}+\alphae^{\alpha y}(T-\frac{\alpha hy}{k})-\frac{h\alpha}{k}e^{\alpha y})=e^{\alpha y}(T-\frac{\alpha hy}{k})=(e^{\alpha y}-1)\frac{h}{k}+e^{\alpha y}(T-\frac{\alpha hy}{k})-e^{\alpha y}(T-\frac{\alpha hy}{k})=t-(e^{\alpha y}-1)\frac{h}{k}
Mg=Tcos\phi[br]T(AD)=(L/2)sin\thetaMg
TLsin(\theta+\phi)=(L/2)sin\thetaMg \Rightarrow sin(\theta+\phi)=\frac{sin\thetacos\phi}{2}[br]\Rightarrow sin\theta cos\phi+cos\theta sin\phi=\frac{sin\theta cos\phi}{2}[br]\Rightarrow cos\theta sin\phi=-\frac{sin\theta cos\phi}{2}
y'=-2a\theta' sin2\theta \Rightarrow y''=-2a(\theta''sin2\theta+cos2\theta*2(\theta')^2)[br]=-2a[(g/4a)sec^2\theta Tan\theta sin2\theta + cos2\theta*\frac{g}{2a}Tan^2\theta][br][br]=-2g[(1/2)sec^2\thetaTan\theta sin\theta cos\theta +\frac{1}{2}cos2\theta Tan^2\theta][br][br]=-2g[(1/2)Tan^2\theta +\frac{1}{2}Tan^2\theta(2cos^2 \theta-1)][br][br]=-2g[(Tan^2\theta/2)(1+2cos^\theta-1)]=-2gsin^2\theta