Given: A force of 40 N is acting on the end of a 'L' shaped beam at angle [tex]\theta[/tex]--beam is attached to a wall at point 'A' and projects straight out and turns up at the very end--that is 8m long and 2m high, find the angle [tex]\theta[/tex] corresponding to the min and max moment at point 'A'. Given: [tex]0^\circ\leq\theta\leq180^\circ[/tex](adsbygoogle = window.adsbygoogle || []).push({});

This is the moment eq I came up with:

[tex]f(\theta)=320\sin{\theta}+80\cos {\theta}[/tex]

I took the first derivative:

[tex]f'(\theta)=320\cos\theta-80\sin\theta[/tex]

[tex]0=320\cos\theta-80\sin\theta[/tex]

[tex]\sin\theta=4\cos\theta[/tex]

[tex]\sin\theta=4\sqrt{1-\sin^2\theta}[/tex]

[tex]\sin\theta=\sqrt{\frac{16}{17}[/tex]

[tex]\theta=\arcsin{\sqrt{\frac{16}{17}}[/tex]

[tex]\theta\approx76^\circ[/tex]

thus

[tex]f(0)= 320\sin{\theta}+40\cos{\theta}=40[/tex]

[tex]f(180)= 320\sin{\theta}+40\cos{\theta}=-40[/tex]

[tex]f(76)= 320\sin{\theta}+40\cos{\theta}\approx330[/tex]

Its been a while since I've taken calc so my question is "how do I find the angle corresponding to the minimum force?" I know its not 0 or 180 because the force is not collinear with point 'A'; moreover, I know the angle I'm looking for should be about 166 degrees (I solve the problem using trig already but my professori wants us to use calc when and where possible). How can I justify 166 degrees as an answer using calculus?

Thanks a-bunch

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# Should be easy but its been a while

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