mg sin theta = mu mg cos theta ( M+m/m)

\[mg \sin \theta = \mu mg \cos \theta (M+m/m)\]

solve for theta

Please help

Divide both sides by mg cos theta -> Tan theta = mu (M+m)/m theta = arctan (mu (M+m)/m)

can you explain to me how you did that?!?!

I thought I just did....-)

What don't u get?

lol , my physics teacher just threw this at us. I have no idea where you got tan from. what is arctan?

Thank you by the way

sin/cos = tan If tan x = something then x = arctan something (both these are on your calculator)

OK?

how did you get rid of the greek u? (sorry i'm not familiar with any of these)

I didn't, it's still there theta = arctan (mu (M+m)/m) mu is just friction constant, right?

what is mu?

Okay can you go like step by step on how you did it please, I'm not as guru as you , actually not even close lol

\[\lambda \sin(\theta)=\mu \lambda \cos(\theta) (\frac{M+m}{m})\] divide both sides by \[\lambda \cos(\theta)\] \[\frac{\lambda \sin(\theta)}{\lambda \cos(\theta)}=\mu \frac{M+m}{m}\] \[\frac{\not{\lambda} \sin(\theta)}{\not{\lambda}\cos(\theta)}=\mu \frac{M+m}{m}\] \[\frac{\sin(\theta)}{\cos(\theta)}=\mu \frac{M+m}{m}\] \[\tan(\theta)=\mu \frac{M+m}{m}\] \[\tan^{-1}(\tan(\theta))=\tan^{-1}(\mu \frac{M+m}{m})\] \[\theta=\tan^{-1}(\mu \frac{M+m}{m})\]

Wow, thank you so very very much. It looks so clear now!!!!

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