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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