Mathematics 72 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

solve for theta

OpenStudy (anonymous):

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I thought I just did....-)

OpenStudy (anonymous):

What don't u get?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Thank you by the way

OpenStudy (anonymous):

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

OpenStudy (anonymous):

OK?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

what is mu?

OpenStudy (anonymous):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

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