Question

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

Answer 1

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

Answer 2

solve for theta

Answer 3

Please help

Answer 4

Divide both sides by mg cos theta ->

Tan theta = mu (M+m)/m

theta = arctan (mu (M+m)/m)

Answer 5

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

Answer 6

I thought I just did....-)

Answer 7

What don't u get?

Answer 8

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

Answer 9

Thank you by the way

Answer 10

sin/cos = tan

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

Answer 11

OK?

Answer 12

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

Answer 13

I didn't, it's still there theta = arctan (mu (M+m)/m)

mu is just friction constant, right?

Answer 14

what is mu?

Answer 15

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

Answer 16

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

Answer 17

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